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\ as“

This is to certify that the
thesis entitled
Optimal Allocation of Investments in a

Leontief Input—Output Mode 1

presented by

Amor Farouk Benghezal

has been accepted towards fulfillment
of the requirements for

Ph. D. degree inManagemenL Science

lewd-=1? 4 fies/.43. 5

Major professor

Date May 9, 1979

04639

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© 1979

AMOR FAROUK BENGHEZAL

ALL RIGHTS RESERVED

OPTIMAL ALLOCATION OF INVESTMENTS IN A
LEONTIEF INPUT-OUTPUT MODEL

By

Amor Farouk BenghezaT

A DISSERTATION

Submitted to
Michigan State University
in partia] fu1fi11ment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Management

1979

ABSTRACT

OPTIMAL ALLOCATION OF INVESTMENTS IN A
LEONTIEF INPUT-OUTPUT MODEL

By

Amor Farouk Benghezal

The purpose of this studylwas to develop a model based on Leon-
tief input-output relationships that would help economists formulate
plans of economic development. A predetermined amount of funds is
available for investment in the planning period. Economic planners
are faced with the question: l) in which sectors funds should be al-
located; 2) howrnuchshould be invested, and 3) when investments should
be carried out. A mathematical progranlwas formulated to answer these
questions, where the objective function wasto maximize value added
by the whole economy in the planning period. Three types of constraints
wereused, namely, demand, capacity, and investment.

The contribution to economic development depends upon the se-
quence of investments made during the planning period. A theory of
economic development recommends to underdeveloped countries to under-
take projects in upstream industries (steel, chemicals, etc.).

The implementation of the latter ones will provide inputs for down-
stream industries and thus will favor their development. To take into
account the preponderance of foreign trade in underdeveloped economies
two types of input-output coefficients are used, namely, domestic and
foreign coefficients. In Chapter 2, a model with both types of coef-
ficients is formulated, and a numerical example is solved by dynamic

programming. Two elements prevented the author from applying the

 

Amor Farouk Benghezal

dynamic programming solution to the Algerian plan; first, data used

to compute both types of input coefficients arenot available; second
the tediousness of computations required to write a computer program
that dealt with all details of the solution procedure. In Chapter 3,
a mixed integer prograniwas formulated for the Algerian plan and was
solved using the APEX package available on the CDC 6500. In the

mixed integer program the input-output coefficients used were those
obtained from l969 input-output table. The program was run first for
a four year plan (1970-l973), and second for an eight year plan (l970-
T977). The results obtained in the l970-l973 period disagree with the
claim that underdeveloped economies should invest in upstream indus—
tries (high capital ratio industries). 0n the contrary, they showed
that investments must be made in downstream industries (low capital
ratio andlaborintensivesectors). The author thought this theory
claim could be valid in the long run. So it was tested in the eight
year plan (l970-l977). Here again (section 3.3.3),the results dis-
approved the implementation of projects in upstream industries in the
early years of the plan. However after l974 high capital ratio pro-
jects showed up in the optimal solution. Two elements explained this
situation. First, funds available for investment in the 1974-1977
period were two and a half times greater than in the preceeding period.
Second, low capital ratio sectors did not need that many funds for
investment to satisfy the demand requirements imposed on them. There-
fore nore funds were available for spending in upstream industries.

In Chapter 4,limits to the mixed integer program results were reviewed

and some suggestions were made.

To my father Mohammed, who taught me
that education is the best asset one

can have -- 190l—l970

ii

ACKNOWLEDGMENTS

I would like to express my gratitude to some of the people who
contributed to the completion of this study.

I am grateful to my academic adviser Dr. Richard C. Henshaw,
Department of Management, for his encouragement,support, guidance
since my enrollment in the Ph.D. program at Michigan State University.
I would like to thank my other committee members, Dr. Phillip L. Car-
ter and Dr. Ram Narasimhan for their professional criticisms and en—
couragement.

I would also like to thank the Algerian "Ministere de l'Enseign—

ment Superieur et de la Recherche Scientifique for granting me a

scholarship, and my sponsors from AMIDEAST for their continuous help.
Special thanks go to my brother-in-law, Abdelhamid, who pro-
vided me with statistical data, I could not get otherwise, and to my
family for their moral support.
Particular recoginition is due my wife, Nadia, for her moral
and material support throughout this phase of our life, and my son
Mohanmed Amin, at the age of five, who understood somehow that this

task simply had to be completed.

iii

 

TABLE OF CONTENTS

Page
LIST OF TABLES ........................ vi
INTRODUCTION ........................ . 1
CHAPTER
I INPUT-OUTPUT MODELS . . .T. .............. 4
1.1 Leontief Model .................. 4
1.2 Maximization of the Growth Rate of Net Output . . 7
1.3 Introduction of Capacity Building in a Dynamic
Leontief Model .................. 16
1.4 A Model for Resource Allocation ......... 22
1.5 Objectives of this Research ........... 33
II THE DYNAMIC PROGRAMMING SOLUTION TO THE DYNAMIC
LEONTIEF MODEL FOR AN INVESTMENT ALLOCATION ...... 38
2.1 Definition of the Model ............. 38
2.2 Assumptions and Dynamics of the Model ...... 43
2.2.1 Assumptions ................ 43
2:2.2 Dynamics of the Model ........... 44
2.3 The Dynamic Programming Solution ...... . . . 51
2.3.1 Stages and States of the Problem ..... 51
2.3.2 Recursion Formula ............. 52
2.3.3 Principle of Optimality .......... 53
2.3.4 The Dynamic Programming Approach ..... 54
2.4 Insights in the Model .............. 62
2.4.1 Reduction of Computations by Marginal
Analysis ................. 52
2.4.2 Further Insights in the Model ....... 64
2.5 Dynamic Programming Solution in a New Light . . . 67
2.6 An Example .................... 73
III THE MIXED INTEGER PROGRAMMING SOLUTION TO THE DYNAMIC
LEONTIEF MODEL FOR AN INVESTMENT ALLOCATION ...... 98
3.1 The MIP Formulation ............... 98

iv

3.2

3.3

IV CONCL
APPENDIX
BIBLIOGRAPHY

Estimation of the Parameters and Data ......

3.2.1 Estimation of the vi ‘5 and a1 '5 .....
3. 2. 2 The Estimation of the Right Hand Side of
Demand Constraints ............

3.2.3 Estimation of the Right Hand Side of
Capacity Constraints ..........
3.2.4 Estimation of the Projects Costs gi's

3.2.4.1 General Assumptions .......
3.2.4.2 Method of Estimation ......
3.2.4.3 Estimation of a Project Cost in
Agriculture ...........
3.2.4.4 Estimation of the Cost of a Pro—
ject in the Mining Sector . . . .
3.2.4.5 Estimation of the Cost of a Pro-
ject in the Construction Indus—
tries Sector ..........
3.2.4.6 Estimation of Projects Costs for
All Other Sectors ........

3.2.5 Estimation of the Capacity Increase Due
to Investment . . ..........
3. 2. 6 Estimation of the Investment Expenditures.

Solution by MIP .................

3.3.1 Introductory Conments ..........
3.3.2 Analysis of the MIP Results .......
3.3.3 The MIP Solution to the Eight Year Plan .

USIONS AND SUGGESTIONS .............

oooooooooooooooooooooooo

oooooooooooooooooooooooo

Page

101
103

105

108
108

110
111

113
114

115
115

Table

NW

'N N N N N N

.1

0143-0)

.10

.11
.12

 

LIST OF TABLES

The Leontief Input-Output Table ...........
Stage (1,1) .....................
Stage (i,l) ....... ' ..............

Stage (i,k) .....................

Results of Stage (

Results of Stage (

Results of Stage (1,2) .............
(

N
U

N
V

Results of Stage

Investment Per Year and Sector ...........
The MIP Formulation .................

Estimates of ail's and vj's .............

Rate of Increase of the Different Types of Consump—

tion ........................

Value of Exports in 1969 and 1973 (as Projected)

The Right Hand Side of Demand Constraints . . . .

The Right Hand Side of Capacity Constraints .....
Sum of Years' Digits ................

Cost of Projects: gi (Based on U.S. Data) .....

The Value of Capacity Increases: diw. (Based on

American Data) ...................

Available Investment in Algeria ...........

Structure of Investments in the Four Year Plan

The Right Hand Side of Demand Constraints for 1974-

1977 ........................

vi

...............

Page

68
69
72‘
77
79
86
95
97
102
104

105
106
107
109
111
113

118
119
124

Table
3.13
3.14
3.15
A.l
A.2
A.3
A.4
A.5
A.6

Available Investment in the Eight Year Plan ......
Structure of Investments in the Eight Year Plan . . . .
Structure of Investments in the Eight Year Plan . .

Input-Output Table of 1969 ..............
Data Used to Compute Projects Costs ........

Listing of Input Cards of the Four Year Plan .....
Computer Output for the Four Year Plan ........
Listing of Input Cards of the Eight Year Plan .....

Computer Output for the Eight Year Plan ........

vii

169
186

INTRODUCTION

Economic theory has described underdeveloped countries as non-
integrated, dominated, lacking savings for investment purposes, having
food shortages, and so on.1 It has also suggested different economic
policies to answer the key question underdeveloped countries face;
how to allocate resources to achieve economic development? Each of
these policies has its own strength and weakness. It is not this re-
search purpose to examine these policies.

Many investment opportunities exist in underdeveloped countries,
because many economic activities are lacking, or, are at a primitive
stage. Since funds (available for investment) are in short supply,
all of the investment possibilities cannot be funded. At a micro-
economic level, they could be ranked according to their rates of re-
turn. But at a macro econOmic level, this would not work, because
relationships among activities are ignored. If it were used, it would
be likely that most of the funds would be spent in few economic activ—

ities belonging to the same sector (the clothing sector is an example)2.

1Samir Amin, Accumulation on a world scale; a criti ue of the
theory of underdevelopment (New York: Monthly Rev1ew Press, [974).
Charles Bettelheim, Planification et croissance acceleree (Paris:
F. Maspero, 1965).

2This does not mean that the rate of return method could not be
used once the funds allocated to a given sector are known.

However, we would like to suggest a model that could help in

answering the question of allocating resources--especia11y investment
to bring about economic growth. The Leontief‘s input-output analysis
provides the framework of the model.1 Input-output analysis shows the
relationships among the different sectors of the economy. Some econ-
omists have suggested the use of Leontief's system for the formulation
of economic development plans.2

The purpose of the model that will be presented is to find the
optimal sequence of investments that will achieve the goals economic
planners set. The modelhas been formulated in a mathematical program.
Its solution can provide planners with an answer to the following
three questions:

1) how much must be invested?

2) when must investments be made?

3) in which sectors must investments be made?

This research contains three chapters:

- the first one presents the Leontief's model, some of the re-
search on resource allocation, and the research goals.

- the second chapter deals with the dynamic programming approach
to solving the model. A nunerical example is carried out to show

how the method of solution works.

- the third one presents another approach to the same model. It

1Wassily w. Leontief, The structure of the American econom :
1919—1939 (2d ed., New York: Oxford Univer51ty Press, I951).

2Oskar R. Lange, Essa s on economic lannin (Bombay, New York:
Asia Publishing House, 2d ed., Calcuta: Stat1stical Pub. Society, 1967).

Mohamed Dowidar, Schemas de la re roduction et la methodologie de la
planification socialiste (Algiers: Editions du Tiers Monde, 1964).

is formUlated as a mixed integer program. It is applied to the Al-

gerian planning problem.

CHAPTER I

INPUT-OUTPUT MODELS

Input—output models have received a great deal of attention
since the publication of Leontief's work on the American economy.1
Leontief was interested in presenting a sort of balance sheet of the
economy: the output and its distribution among the different sectors
of the economy. Leontief prepared these balance sheets in table form
for the United States for 1919, 1929, and 1939. The input-output
tables are very useful in economic planning. In the following, firstly,
the Leontief model is presented; secondly, some research works made in
the field of input-output analysis are reviewed; thirdly, the purpose

of the present research is discussed.

1.1 The Leontief Model

Leontief model views the economy as being composed of several
sectors such as, agriculture, mining, steel, chemicals, etc. Let us
suppose there are n sectors in the economy. The main idea of input-
output models is to show the general dependence among sectors. This
dependence among sectors is quantified and is expressed by the flow of
goods from one sector to the others.

Let xi be the total output of sector i. Let xii denote that

1Nassily w. Leontief, The Structure of the American Econom 1919-
1939 (2d ed. enl.; New York: Oxford Un1versity Press, 1951).

part of the total output retained by sector 1 for its own use.

In general, let x.. be the value of goods of sector i consumed by

13
sector j for production in sector j and yi be the value of net output.

One has the following equality
n
x-=Zx..+y.i i=1,...n (1.1)

Equality (l.l) says the output of sector 1 is the sum of the
flows of sector 1 goods to all sectors and the net output.

Viewed differently, to get output Xj’ sector j must obtain goods
n
from all sectors i in an amount equal to z Xij and sector j must
i=1
add salaries and return on capital. Let w. be the wages paid in sector

J
j and pj the return on capital.
The following equality is obtained.
n
xj = E] Xij + wj + pj j = l, . . . n (1.2)

The sum of wj and pj is the value added of sector j. With Equalities

(1.1) and (1.2) one is able to construct the Leontief Input—Output

Table. Let x be the total output of the economy, in other words, the

Gross National Product.

I1 11 n n n n n
z z x.. + z y. = z z x.. + z. w. + z p.
i=1 j= ‘3 i=1 1 3:1 i=1 ‘3 j=l J j=l J
n n n n
Since 2 2 x1. = Z 2 x35, one obtains
i=1 j=l J j=l i=1

Table 1.1 The Leontief Input-Output Table

net total

sectors 1 2 "' n output output
1 X11 X12 x1n yl X1
2 x21 x22 X2n y2 x2
“ an an Xnn y" X"
Wages w1 ”2 W"
Return on p p P
Capital 1 2 "
Total x x x x
Output 1 2 n

n n n

zy-=zw-+zp- (1'4)

i=1 ‘ j=l J j=1 3

Equality (1.4) states the net output of the economy (Zyi) is
equal to the sum of the value added of all sectors.

Now, let us define aij as the ratio of the value of inputs 1

consumed in sector j to the output of sector j.

x..
=——u A:
aij x. 1 l n (1.5)
.1
From the above ratio, it results
= a.. x. (1.6)

xii 11 .1
Making use of Equation (1.6) in Equality (1.1) one gets:

(1.7)

II
—J
3

n
. = .. . + . '
x E a1J xJ y], 1

or, in matrix form

X = AX + Y, (1.8)

where X is an n-component column vector, A is an n x n matrix
with aij as elements, and Y is an n-component column vector.

If a matrix A of input coefficients and a set of values for Y
are given, the level of output X will be obtained right away, by solv-
ing

x = (I - A)“ Y (1.9)

This model has n equations and n unknowns (Xi)’ and has a unique
solution if (I — A)'] exists. If the yi are the n unknowns, one can
use Equation (1.8) to find them.

The Leontief model is very useful in planning of economic activ-
ities. If the planning authorities decide upon a given level of out-
put for each activity (Xi)’ they can obtain the levels of net output

for all activities (yi). In the following sections, some of the re-

search works made on the input-output models are presented.

1.2 Maximization of the growth rate of net output
Lange formulated a model where the goal is to maximize the growth
rate of net output by the end of the planning period using a Leontief
type of model.1
His objective was to maximize Z = 9%, where AY is the increase
of net output between the year zero and the ending year of the plan

and Y is the net output in year zero. The growth rate of net output

can be expressed as

1

Oskar R. Lange, Lecons d'Econometrie (Paris: Gauthier-Villars,
1970).

Av AM.
I Y

where I is the total investment that should be made during the plan-
ning period.

If one is given values of I and Y, the only way to achieve the
highest growth rate of net output is to maximize the ratio %¥~.

The constraints of the maximization problem are: the final
consumptions of goods, produced by each sector, should be at least
equal to some predetermined level for every period of the plan.

Suppose there are n sectors. Let Cit be the final consumption

of goods of sector 1 in period t. Let Kit be the minimum level of

final consumption of sector i goods in period t. The following must

hold for each period:

Cit :Kit 1

II
—I
3

(1.11)

Net output of sector i in period t (Yit)’ must equal the sum of

the final consumption of sector 1 goods, Cit’ and the amount invested

in sector i goods, Iit'

it + Iit i = 1 . . . n (1.12)
Inequality (1.11) and Equality (1.12) imply that

IitiYit' Kit ‘

II
—I
3

(1.13)

Hence,Lange's problem reduces to the following linear program.

max Z = %¥

subject to Ii < Y. - K.

In this LP (linear program), it is not clear what the decision
variables are. Also it does not show any resemblance to Leontief
model. In order to arrive at an LP containing the input-output coef-
ficients, Lange developed a new formulation by modifying the objective
function and the left hand sides of the constraints inequalities along
the following lines.

Prior to discussing these modifications, it is useful to define
the concept of investment cycle. It is the time elapsed between the
moment the investment is made and the moment an increase of the output
is obtained. Let Iijst be the investment needed by sector j, in terms
of sector i net output, in period t, and of cycle 5 years.

AX. is the increase in output of sector j obtained after 5

jst
years due to an investment made in period t. By definition

ijst = Xj,t+s — th (1.14)
and let
b.. = 27—111“ (1 15)
1jst stt
bijst is the ratio of investment of cycle 5 made by sector j in

sector 1 net output in period t, to the increase of sector j output
after 5 years.

Let It be the total investment made by the whole economy in

period t.

Let I(JSt) be the total investment of cycle 5 made in sector j
in period t.

Let

10

j = 1 . . . n (1.16)

 

Ajst is the ratio of sector j investment (of cycle 5) in period

t,to the total investment of year t. From Ratio (1.15) one gets

n n
21.. = 2 b.. AX.
i=1 1jst i=1 1jst jst
n .
. = (jst)
S1nce .2 Iijst I
1-1
. (jst)- "
one has. I - iEl bijst Astt
= (jst) . = l
or AXjst I Bjst’ where B jst _TF.—_—_—_' (1.17)
X b.. t
i=1 ‘35
Let ajit be the ratio of the amount of sector j output needed in sector

1, to the output of sector 1 in period t

X..

 

_ 1jt
a.. — (1.18)
j1t Xit
Let Ojst be that part of sector j increase of output in 5 years

from year t, used to achieve the increase in output in the other sec—

tors during the same period.

n
= Z a.. AX.
ojst 1:1 1j,t+s 1st (1.19)
Astt
Let AYjst be the increase in net output of sector j, obtained

after 5 years, due to investments made in year t.

11

n
AY331: = ijst ‘ 1:1 aji,t+s AXist “'20)
Using Ratio (1.19), Equality (1.20) becomes
AYjst = ijst (1 - Ojst) (1.21)
Making use of Equality (1.17), the equality above becomes
= _ (jst)
AYjst (l Ojst) Bjst I (1.22)
or
A = - (JSt) . = -
Yjst B jst I , where B jst (l Ojst) Bjst (1.23)
After using Ratio (1.16) and Equality (1.23)
A ‘2‘ A " A
= = Z I
Yst i=1 Yjst It i=1 8 jst jst (1'24)

To obtain the increase in net output during the planning period,
a k year plan of investments is considered. These investments have
different cycles 5 and are made in the various sectors j (j = 1 . . . n)
in year t, t = 1, 2, . . . k - 1. In fact, it is assumed that no in-
vestment is made in year k, since the increase in net output yielded

by investments in year k will only show up in year k + 1. Hence in

the first year, only investments having cycles 5 - l, 2 . . . k-l,
need be considered in the second year, only investments of cycles

5 = l, 2 . . . k—2 are taken into account. Likewise for the (k - 1)th
year, only investments of cycle 5 = 1 need be considered.

With the above considerations in mind, the increase of net output,

12

for the whole economy in the second year of the plan, is AY1 = Y2 - Y].
This is the result of investments of cycle 5 = 1. Now, using the no-

tation of Equality (1.24), one has

II
H
II
._a

where 5
Following the same reasoning, the increase of net output, in the
third year of the plan, is due to investments of cycle 5 = 1 made in

the second year and of cycle 5 = 2 made in the first year.

AY2 = AY + AY

12 21

In general, the 1ncrease of net output: AYt = Yt+l - Yt’

obtained in year (t+l) is

AY (1.25)

AY = 1 Sgt-5+1 I

t s

"Mt‘f'

The increase of net output, after k years, is the sum of the increases

during the different years. Let this increase be

Ak v = :2: AYt (1.26)
After using Equality (1.25), Equality (1.26) becomes

k-l t
Ak Y = til 5:1 AYS,t_S+] (1.27)

The former relation shows the increase in net output after k

years depends on the increases of net output, due to investments of

13

different cycles made in the various years of the plan.

Now, introducing Equality (1.24) into Equality (1.27), one ob—

 

tains
k—l t n
A Y= z z 21 A. _ s'. (1.28)
k t=l 5:1 j=1 t-s+1 js,t s+l js,t-s+1
k—l
I = E It is the total investment, and also the sum of the
t=l
investments made up to year k-l.
I(.ist) .
Let ajst - I j - 1 . . . n, s,t — l . . . k-l (1.29)
k—l t n

with Z Z Z a.
t=l s=l j=l M

II
_.1

since A. = 1(j5t)
jst I

 

, from Ratio (1.16)
t

Ratio (1.29) is rewritten as

Ajst It
ajst = ___T__— (1.30)

From Ratio (1.30), it follows

I o. I (1.31)

A =
t jst jst

Making use of Equality (1.31) in Equality (1.28), one gets

k-l t n (
A v=12 z 2‘. on. 8'. 1.32)
k t=1 s=1j=1 J53t-S+] JS,t-S+]
Since the objective is to maximize ég-, the objective function
becomes
AkY k-l t n
T: Z Z Z 01 (1.33)

t=l 3:] j=1 js,t-s+1 B js,t—s+l

14

Next, the constraints expressed by Inequality (1.13) are dis-

cussed; these are Ii Y1 — Ki

t 5- t t‘
k-] n n l u -
S1nce Iit = SE] jEl Iijst, the 1nvestment, 1n per1od t 1n sector

1 net output, is equal to the sum of all investments of cycles

5 = 1 . . . k - 1 made by all sectors j in period t, and in sector i

net output.

Using ratio (1.15), this equality becomes

k—1

n
I. = z z b.. AX. (1.34)
1t s=1 i=1 1jst jst

Substituting for ijst from Equality (1.17), this relationship

becomes

k-l n
I. = Z E b

(jst)
1t _ ._ B I
s—l j-l

ijst jst (1'35)

Combining Ratios (1.16) and (1.30), it results

1(jSt) = Iajst (1.36)

Using Equality (1.36) into Equality (1.35), one obtains

k-l n
I. = I Z E b

B
‘t s=1 j=l

ijst jst “jst (1'37)

Now, the modified linear program that must be solved can be stated as:

k-l k-1 n

Max Z = Z Z 2 B'. a.
t=l s=1 j=1 js,t—s+l js,t-s+l

15

subject to
k-l n

I SE, jg, bijst Bjst O‘jst i Yit ' Kit
1 = l n
t = l . k - 1

k-l k—l n

E Z Z d- = 1

t=l s=1 j=1 3“

“jst z_0 j = l . n; s,t = 1 . . . k—l

djst are the decision variables of this LP. They correspond

to the part of total investment, in the k year plan, made in sector j

in period t, and having a cycle of 5 years. The investment coeffic-

ients bijst appear in the constraints. The input-output coefficients
ajit do not appear directly in the LP, but must be used to compute
B|jst in the objective function (see Ratio (1.19)).

Once a set of ajst

one can determine how much must be invested, in each sector every year,

is found that maximizes the objective function,

by computing 1(j5t) = I dist from Ratio (1.29).

Since

8 jst = ETTEF)’ from Equa11ty (1.23)

B'jst can be interpreted as the increase of net output, in sec—
tor j, due to an investment of cycle 5 made in period t. B'jst
defined as the investment efficiency of cycle 5 in sector j. In fact,
the objective is to maximize 8' €%—I, i.e.,the investment efficiency

in the whole economy.

16

In summary, Lange developed a model, using Leontief type of
relationships between sectors, that maximizes the investment efficiency
in the whole economy subject to the constraints that investments in
each sector goods, in each year, should not exceed some predetermined

Y - K. ).

it: it 1t
In the next section, a were general type of Leontief model is

level (i.e. I

presented that includes sector capacity.

1.3 Introduction of Capacity Building in a Dynamic Leontief Model
Wagner formulated a Leontief model1 that included both in—
ventories and capacity of sectors.

Assume, for any period t, there is an n x n matrix of input-

th column of A represents the inputs

from each of the n sectors required to produce one unit of the jth sec-

output coefficients A, where the j

tor output. Also, consider an n dimensional square matrix B of capi-

th column of B represents the inputs

th

tal coefficients, where the j
from each sector needed to build one unit of capacity for the j
sector.

In the following vector notation the subscript t denotes the

relevant time period. All vectors are n demensional column vectors:

f = final demand

c1 = initial capacity
xt = level of production
lt = level of capacity being built (so Blt are the direct cap-

ital requirements for the creation of 1t)

1Harvey M. Wagner, “A linear programming solution to dynamic
Leontief type model,“ Management Science, Vol. 3, No. 3, April 1957.

17

st = level of inventories at the end of period t

u =leve1 of unused capacity.

t
The output of a sector and its beginning inventories are equal

to the sum of intermediate demand, final demand, investment require-

ments and ending inventories. In vector notation, this can be ex-

pressed by the following set of linear equations.

xt + St-l = Axt + Blt + ft + st t = 1 . . . T (1.38)

By definition, unused capacity in a sector is the difference

between available capacity and capacity used up for production.
U = C - X t = 1 . . . T (1.39)

In any time period, the capacity level is the sum of the capa—
city existing in the previous period and any increase in capacity

undertaken in the previous period:

Ct = Ct-1 + 1t-1
t-l
or ct = c1 + 2 1r t = 2 T (1.40)
r-l
Combining Equalities (1.39) and (1.40), one obtains
t-l
c1 = xt + ut - 2 1r t = 2 . . . T (1.41)
r—l
Assuming values for final demand ft, beginning inventories
so and initial capacity c], the model given by Equalities (1.38)

and (1.41) can be represented as follows:

(—+
II
——‘
_|

(1.42)

(1.41)

O
-—I
ll
)<
+
C
l
M
_1
('9'
II
N
.._|

Because of the very nature of the economic activities, it is

further required that

x l > 0 (1-43)

t’ 12’ St’ ”t —
Equations (1.42), (1.41) and restrictions (1.43) in the model
satisfy the standard form required for the application of linear pro—

gramming. The following figure gives a starting simplex tableau.

Decision Variables

 

Note, each line of symbols in the tableau actually represents n
separate linear constraints, one for each sector in period t. Hence
there are 2nT constraints. Similarly, each column heading is a vec-

tor of n elements, one for each sector in period t. There are 4nT

 

19

unknowns. The previous tableau nay be regarded as a matrix equation
Y = HZ where Y is the entire column on the left of the tableau, H is
the body of coefficients and Z is the entire top of column headings.

An inspection of the model reveals that if c1 3_(I—A)'1 ft, for

all t, then feasibility is obviously possible by setting xt = (I-A)']

f because production is undertaken under capacity (c1 3_xt). How—

t!

ever, if c1 < (I-A)‘1 (f1-so) then feasibility is not possible. The

reason for this becomes clear by examining Equation (1.42)

fl'so = (I—A) x1 - Bl1 - s1 §_(I-A) x], since 11 and 51 > 0;

hence, x1 > (I-A)']

(f1-so).

Since (I—A)'1 (f1-s ) > c], it follows that x1 > c]: the production

0

level in period 1 is above capacity. One is faced with an inconsis-
tency since xt §.ct (see Equality (1.41)).
A real problem is the case in which the first period demand is

1 ff'exceeds initial capacity for some t > 1.

feasible, but (I-A)'
It is here that the economy must be able either to stockpile in earlier
periods or to build additional capacity to meet future demand. If a
feasible solution of this case exists, it may not be unique. There-
fore, by defining an appropriate objective function, the optimal so—
lution, with respect to the objective function, can be found.

Let us define an m dimensional vector of constants K, where m is
equal to 4nT, and V as a vector of these 4nT decision variables
(Xt’ 1t, St’ ut).

The objective function is: Z = KV (1.44)

K can be specified as a vector of manpower input figures

20

associated with x and 1t.

t

K = (K11, K12, 0, K21, K22, 0, . . . KTl’ KT2’ 0),

where each K is a row vector in n manpower coefficients for period

ti
t with either xt (i = l) or 1t (1 = 2) and each 0 is a 2n dimensional

vector of zeros. An optimal solution nay be found by

T
min Z = E (K x + K 1

)
t=1t1t t2 t
subject to constraints (1.42) and (1.41).
KTl can be thought of as the cost of producing an output level
of xt and KT2 as the cost of creating a capacity equal to 1t.
th

If maximizing the total anount of new capacity in the i sector

by the end of period T is sought, K in (1.44) will become K = (O, 1, O,

1, . . . 0), where the scalar one appears in all positions correspond-
ing in V to lti’ the ith component of 1t. 50 the objective function
becomes
T
Z = til 1ti'

The model, represented by equations (1.42) and (1.41) contains
2nT equations and 4nT unknowns. Wagner explored the possibility of
reducing the model, to one containing only :nT equations in 3nT
unknowns and hence, one that is computationally easier to handle than
the larger model. For this, Equation (1.41) is multiplied by -(I-a)
and added to Equation (1.42);

t-l

+ r21 (I-A) 1r — (I-A) ut (1.45)

f - s

t t'l - (I-A) c1 = -B1t - St

21

Thus, xt is eliminated in the model. Since the model is less
constrained than the former one, a larger maximum (or smaller minimum)
value for Z should be expected in the abbreviated system than would
be in the larger and more restrictive system. If the optimal solu-
tion to the abbreviated system, when substituted in Equation (1.41),
yields all x 3_o, then by definition this solution is feasible in the
general model. It is also optimal, since the optimal Z in the general
model is less than or equal to the optimal Z in the abbreviated system.

In the case, where some components of xt are negative, say th’
Wagner showed that the input-output production relationship in sector
t1" t1 ”1
column of A) becomes the output. The author explained

j is reversed in period t, so, x
th

becomes the input and ij
being the j

the negativity of x .as follows.

tJ
th

”An input of the j good implied by the negativity of x must

th

tJ'
come from an existing stockpile of the j good and not from any pro-
duction in period t. The stockpile has only one possible origin: it
must have been created by production during previous periods in the
model. The correctness of the value of Z despite the fact that xtj
is negative can be explained as follows. The abbreviated model has
used a ”short-hand" way of storing raw materials. In period t, the
model indicated used the vector of inputs of industry j's product in
the amount equal to ijtj as raw materials for other industries' pro-
duction. Instead of storing these raw material inputs in the earlier
periods when they became available, the model embodied them in the

form of the jth industry's product and in period t reversed the produc-

tive process. Since the stockpile used in t was actually produced

earlier, one may easily translate the ”short-hand" program of the

22

abbreviated system to not produce the stockpile of the jth industry's
good in previous periods but rather to save the inputs for use in
t."1

Moreover, Wagner discussed a general algorithm, which involved
the dual problem for proceeding from an optinal solution in the ab-
breviated system to an optimal feasible solution in the more restric-
tive system. He argued the computational savings, due to utilizing
the abbreviated system, should more than offset the inconvenience of
having to switch at the end to the dual algorithm.

Wagner's model is very general and can accommodate any type
of objective function, as long as the vector K in the objective func-
tion is properly defined. Next, a model of economic development based
on Leontief's type of relationships is presented. It introduces other

constraints in the model such as foreign exchange, a limited resource

of underdeveloped countries.

1.4 A Model for Resource Allocation
The primary purpose of Chenery and Kretschmer's study2 is to

develop a model, based on mathematical programming and input-output
analysis, which will assist in the formulation of economic development
programs. The authors introduce imports and exports in their analysis;
these are important activities when underdeveloped economies are con-
sidered. Their model tries to minimize the investment needed to
achieve certain goals in the target year.

1Harvey M. Wagner, ”A linear programming solution to dynamic
Leontief type model.” Management Science, 3, No. 3, 1957, pp. 245-246.

2Hollis Chenery and Kenneth Kretschmer, "Resource allocation for
economic development,” Econometrica, Vol. 24, No. 4, October 1956.

23

Let n be the number of sectors, and let us use their notation.

Yi' is the total final demand for sector 1 in the target year;

Yi is the net final demand for sector 1 to be satisfied from
increased capacity or imports;

R} is the existing productive capacity in sector 1;
Xi is the increase in production in sector 1 above existing
capacity;

M1 is the level of imports in sector 1 measured in domestic
prices;
E1 is the level of exports in sector 1 measured in domestic
prices;

L* is the labor available for increasing production;

0* is the maximum foreign trade deficit;
di is the ratio of increased production to increased production
plus imports d1 = Xi/(Xi + Mi);

mi is the ratio of imports to increased production for sector 1

aij is the amount of input i currently used in producing one

unit of output from sector j (current input coefficient);

aij is the increase in use of input i required to increase the
production of sector j by one unit (marginal input coefficient);
mi is the marginal labor input coefficient for sector 1;

Ci is the amount of capital needed per unit increase in capacity
for domestic use in sector i (marginal capital coefficient);

k1 is the amount of capital needed per unit increase in capacity
for export in sector i;

hi is the price of exports from sector i in foreign currency;

24

gi is the ratio of import prices to domestic prices in sector i
at the given exchange rate;

It is assumed that the goods and services required in the economy
are divided into n commodity groups corresponding to the n sectors of
production. The final demands Y1 are taken as given. The net final
demand that must be satisfied in each sector from new capacity or im-

ports (after allowing for full operation of existing capacity) is

n
Y. = Y.' - (TI - z 51. —- . _
1 1 1 j=l 1j Xj)’ 1 - 1 . . . n (1.46)
— n — —
where (Xi - Z aij Xj) is the amount that could be supplied for final
i=1
use from existing capacity after deducting intermediate uses
(2 a_.. 1.).
j 1J J

The authors made the following assumptions. First, an economic
sector can be thought of as an aggregate of productive subsectors cor-
responding to a single commodity. Second, in each subsector there is
a choice between imports and only one domestic activity. Third, the
input coefficients, for all domestic activities in a given sector, are
identical for each input except capital. So, the only difference be-

~ tween subsectors is due to capital coefficients. The authors argue,
for any given set of shadow prices, it is efficient to import those
conmodities for which the corresponding domestic activity is unprofit—
able and to produce the rest. However, under the second and third
assumption, it is possible to rank subsectors according to increasing
capital coefficients alone, since the other coefficients are the same.

Therefore, the capital coefficient of a sector can be derived from the

25

coefficients of the subsectors, taken in a definite order.

In general, the authors conclude, the average capital coeffic-
ient is an increasing function of the ratio dj’ of total needs sup-

plied domestically in sector j. For simplicity, this function is

assumed linear.

.=.+. . , . . .
cJ onJ SJ dJ a > 0 B > 0 (1 47)

The total supply of a given output (Xi + Mi) less the inter—
mediate uses in domestic and export activities is equal to the net

final demand:
n n
X.+M. - z a..X.— 2: a..E.=Y. (1.48)

The last equation is the familiar balance equation of the
Leontief system, except that, production for export is considered as
a separate sector from domestic production of the same commodity.

The price received for exports is assumed to vary with the

quantity exported. This function is also supposed to be linear.

.= .+ . . . . .
hJ yJ oJ EJ YJ > 0. pJ < 0 (149)

. . . . 1
Furthermore, the authors cons1der two resource 11m1tat1ons:

labor and foreign exchange. The labor constraint is

The authors stated: ”The introduction of more resource limita—
tions such as natural resources can be handled quite easily because they
will usually affect only a small number of sectors, often only one. In
this latter case the resource limitation can be incorporated in the
suboptimization procedure for the sector in question without the neces-
sity of adding another equation.” Chenery, Kretschmer, p. 376.

26

n n
*
151 mi Xi + iEl mi E1 5 L (1.50)
The maximum trade deficit constraint is
n n
Z 91 Mi - ‘2 h. E. < D* (1.51)

The purpose of the economic development program is to choose increases
in capacity (Xi)’ import levels (Mi) and export levels (E1) in such a
way the given final demand will be supplied in the target year with a
minimum total investment, without exceeding the specified resource

limitations. So, the problem consists of

 

subject to Constraints (1.48), (1.50) and (1.51). Xi > O, M
E1 3_0. This is a non-linear programming problem, since the objective
function is non-linear.

If a solution exists, as Kuhn and Tucker have shown, there is a
set of nonnegative numbers (Lagrange multipliers) which may be inter—
preted as unit prices of the scarce resources.. It is assumed that a

th product, Pw is imputed to labor and a

value Pj is imputed to the j
value of Pf is imputed to foreign exchange. These values are the
shadow prices and must satisfy the following properties.

- “They must be nonnegative.

- If the supply of labor or foreign exchange is not fully

27

utilized for an optimum production plan, then its price should
be zero.

- Each connodity should be assigend a value equal to the largest
decrease in investment cost which could be obtained if one
less unit of that comnodity were demanded, labor (or foreign
exchange) should be assigned a value equal to the largest de-
crease in investment cost which could result if one more unit
of labor (or foreign exchange) were made available.“1

The authors assume that shadow prices with the above properties

exist and are known. Chenery and Kretschmer determine the values of
ii’ M1 and E1 minimizing the objective function as follows.
They claim, the optimum export levels Ej are obtained at the

point where Ej maximize the difference between the revenue and the cost

of export activities. Let

2:
A
m
V
II

n
h. . - .. P. + . + . .
J Pf EJ (E a1J 1 wJ Pw k3) EJ
(1.53)
.. . + . + . .
aU‘P.I wJ Pw kg) EJ

.+ . . .. .-
(1J OJ E3) PT EJ (

d-M:

where (Z aij Pi + wj Pw + kj) is the unit cost of exporting one unit
.i
of projuct j.
Since Pf > 0, R3" (Ej) = 2 Dj Pf < 0. Hence, the unrestricted
maximum of Rj (Ej) is attained at
+ . P + k. - . P
wa 3Y3

f) (1.54)

for Rj' (rj) = 0. Hence the maximum of Rj (Ej) for Ej 3_O is attained

Chenery, Kretschmer, p. 377.

28

at

[11)
II
r-‘-\

(1.55)

Next, the authors want to determine the optimal proportion of

domestic production d. needed to compute ii and M1. .For this, the

J
total cost of producing the fraction dj (dj = Xi ) domestically

X, + Mi

and importing the fraction (1 - dj) is

1 - dj) (1.56)

n
T. (d.) = (T a.. Pi + mj Pw + Cj) dj + gj Pf (

n
where (E a . P. + w. Pw + Cj) is the cost of producing domestically
1

131.1
one unit of product j. Using Function (1.47), Equation (1.56) becomes

n
. . = .. . + . + . + . . . - . .
TJ (d3) (1; a13 P1 1.3 PW or) 8de + gJ P. (1 dJ) (1 57)

Since Tj" (dj) = 2 Bj > 0, therefore the unrestricted minimum is

attained at

81f

D
—_l_ _ _ -
tj - 2 j (g. P E a.j Pi wj Pw a.) (1.58)

for Tj' (tj) = 0, hence the minimum value of Equation (1.57) in the

interval 0 5-dj §_1 is attained at

O tj §_0
dj = {tj O E-tj f_l (1.59)
1 t. 3_l

J

29

From the knowledge of dj and Ej, X1 and fii follow directly from
Equation (1.48). If c1j = o it implies that )1, = 0. However, if SJ. >

0, then a, the ratio of imports to increase in production of sector

1 x
A Mi
(m. = we) will be:
1 .
X.
1

It also follows:

M1 = mi Xi (1.61)

Let us substitute Equality (1.61) into Equation (1.48).

A A A A n A
X. + m. X. — Z a.. X. - Z a.. E. = Y., i e S (1.62)
1 1 1 jeS1 13 j j 13 j 1 l

where S1 is the set of i such that di > 0. There are as many equations

as unknowns (i.e. Xi) 50, X1 are uniquely determined. Now, in order to

find Mi’ one must use Equality (1.61) for i in S], but, for those 1 not

in 5], Mi are given by

m)

M. (1.63)

A n
= . + .. . + .
1 Y1 E a X E a

jeS1

As the given set of shadow prices has the above properties, it

n A A
follows that 2 (Ci X1 + ki Ei) is the minimum value of Function

1—1
(1.52) subject to Constraints (1.48), (1.50) and (1.51). Moreover,
if pw > 0, there will be a strict equality in Constraint (1.50) and

if Pf > 0, there will be a strict equality in Constraint (1.51) too.

 

30

Hence, the problem is reduced to finding a set of shadow prices sat—
isfying the above properties.

The next step for the authors is to derive a set of shadow
prices. For this, they proved the following results.1

"Theorem 1: For each PW and Pf there exists one and only one

solution to the following system of equations.

. P , . < O .
93 1: SJ —- (164)
P = {g P — ——l—- s 2 O < s < 28 j = 1 n
f f 2 j ’ — j— J
48.
J
. P - . + , .
QJ f sJ BJ sJ :28J
where Sj = 93 Pf - g aij Pi - “j Pw - aj j = 1 n

Lemma 1: The following iterative process converges to the

solution of (1.64). Assume initial values of s], 52, . . . sn, say
(0) (0) (0)
s1 , $2 . . . sn . Set
(0)
gj Pf , sj :_0 (1.65)
(1) _ 1 <0) 2 (0)
P — P - — , < . < . =
J gJ f 48j (sJ ) 0 __sJ __ZBJ,J l n
(O) (O)
. P — . + -. , . >
9.1 f SJ 33 SJ —ZBJ
Determine subsequent values in the order sn(k), Pn(k+1),
s (k), P (k+1) . . . s (k), P (k+1) from the recursive formulas
n—1 n-1 1 l
(k) 3 (k) “ (kn)
s. =.P-Z ..P. -2 ..P, -.P-.
.1 93 f (:1 an 1 15341813 2 (”.1 w 0‘3

1Chenery, Kretschmer, pp. 379-380.

31

(k)
gj Pf , Sj : O (1.66)
(k+l) _ 1_ (k 2 (k) =
1’1 ' 1911f 483- (5111’°<Sj 32811 1 '1
_ (k) (k)

Theorem 2: If Pw and Pf are nonnegative, then the Pj given by

(1.64) satisfy

min [wj Pw + aj; gj Pf] 5-Pj §_gj Pf (1.67)

Also, for a fixed nonegative value of PW, the Pj satisfying (1.64)

are nondecreasing functions of Pf.

Theorem 3: Suppose PW+ is nonnegative, Pf+ is positive.

+ + +

P. = i Z a.. P. + . P + . + . d. d. + . P + l-d.
J on1<1111 ,J , wJ w a3 SJ J) 3 9J f ( J)1
_J—
j=1...n (1.68)
and the E1, X1, and M1 say Ei+’ X1.+ and M1.+ are determined in the
manner described above for the optimal set of prices.
Then if L+ = z (w. x.+ + w. E.+) (1.69)
i 1 1 1 1
+ + +
and D — Z (91 Mi - hi E1 )

1
the minimum value of (1.52) subject to (1.48), (1.50) and (1.51) with
+ + + H
1 1'

L* = L+ in (1.50) and 0* = D in (1.51), is 2 (Ci x1 + k1 5.
.i

This completes the method of derivation of the shadow prices.

The authors applied this model and its solution to Southern

Italy where an economic development program.was carried out.
All of the models looked at can be used for the formulation of-

economic plans. The first model is concerned with the maximization of

32

the growth rate of net output subject to investment constraints. The
second one is a general model which can be used with any objective
function as long as it suits the model. The third one minimizes the
total investment required to satisfy the final demand subject to some
resource limitations.

The nodel discussed in this research borrowed elements from the
previous models. 1

Lange's formulation provided the objective function. Eventhough
the growth rate of net output is not explicitly maximized in this re-
search, the same results are obtained with maximizing the value added
(this research objective function). Recall that the net output and
the value added of the whole economy are equal.1 In the growth rate
formula (AY/I), investment I is constant. Maximizing growth rate im-
plies maximizing net output. Lange's model and the model of this re-
search seek the same thing but by different means. His decision var—

iables are the a. '5 (fractions of total investment made in sector

jst
j in period t and having a cycle of 5 years). Production levels

(x. ) are the decision variables in this research.

1t
Wagner's model provided the framework for capacity and demand
constraints. His capacity constraints were modified to suit this
study purpose. Increase of sector capacity is thought in terms of
projects implementation in this research.2 Demand constraints are
slightly different from Wagner's ones. Input-output coefficients
matrix A and capital coefficients matrix B are not distinguished

1cf. Equality (1.4). However, net output and value added of a
sector are not equal.

2cf. Equality (3.3) in section 3.1, Chapter 3.

33

as is the case in Wagner's model. Imports and exports variables are
added to Wagner‘s demand constraints to take into account the import—
ance of foreign trade, as in Chenery and Kretschmer's model.

Funds available for investments are a limited resource like
labor and foreign exchange are in Chenery and Kretschmer's formula-
tion.

More details on the model are given in the next section which

presents the purpose of this research.

1.5 Objectives of This Research

Leontief nodels are very helpful in designing programs of econ-
omic development. In the last two decades nany underdeveloped count-
ries implemented such programs. This research proposes a model based
on Leontief input-output coefficients that would help in allocating
resources to achieve some economic goal.

In general, a plan of economic development must achieve both of
the following objectives:

- First, state the goals to be accomplished by the end of the
plan. These may be increasing the real Gross National Output or
National Income by, say 8 percent per year, developing some sectors more
than others, and so on. Certain goals can be quantified while others,
such as quality of education, cannot be.

- Second, provide means for such a program. These are resources
that must be allocated within the plan. Investment and manpower are
examples of resources. The only resource dealt within this model is
investment. By investment it is meant all funds available for spend-

ing in projects of any types (industrial and agricultural). This is

34

a major constraint underdeveloped countries face because of its short-
age.

Assume economic planners want to maximize the value added in the
whole economy over the planning period. The maximization of value
added as the economic planners' objective is chosen for the following
reasons:

- First, value added is a quantitative economic criterion.

— Second, value added is also the origin of all types of in-
comes: wages, profits and so on. So maximizing value added is also
maximizing National Income.

— Third, increasing the value added created within the whole
economy contributes to reduce unemployment which is a big problem
underdeveloped countries face.

Now the economic planners' problem can be stated in the follow-
ing question. How can investment be allocated in a plan of economic
development aimed at maximizing value added created by all sectors of
the economy?

In the next chapter, a nodel is proposed that can help answer
this question.

A mathematical program is used to formulate the economic plan-
ners' problem in a Leontief type of framework. It maximizes value
added subject to three types of constraints:

1) constraints of demand imposed on sectors

2) constraints of capacity of sectors

3) constraints of investments.

There is an interrelationship between capacity and investment.

When an investment is made, the corresponding capacity will increase.

35

Moreover, when there is already enough capacity in a sector to satisfy
demand, there will be obviously no need to make any investments. Both
of these elements are taken into account in the mathematical formula-
tion. But, among competing investments in different sectors, the al-
gorithm ranks them according to their value added.

The solution of this mathematical program leads to an optimal
allocation of investment among the sectors of the economy. The plan-
ners have an answer to the triple question:

- how much noney must be invested

- where the investment must be made

- when it must be invested.

Once these answers are provided, the economic planners will know the
sequence of projects that must be carried out to maximize the value
added by the end of the plan.

Determining the investments per sector and per year is the pur-
pose of this research. Since investment is in short supply in underde—
veloped countries there must be a Wise way to use it, in order to get
the most out of it. For instance, if a country has a given amount of
money to be invested, this investment can be made in any sector or
group of sectors with different effects on the total value added
created in the whole economy. There is also the problem of investing
whether in downstream or upstream industries. There are many econom-
ists who argue that investing in upstream industries generates more
benefits for the whole economy than in downstream sectors. They
claim: investments made in upstream activities provide capacity of
producing goods that will be used by downstream industries and thus,

favoring their development. For instance, the chemical industries

36

are examples of upstream activities; their productions constitute in-
puts to many downstream sectors such as agriculture, food industries
and so on. This theory recommends investments in these upstream in-
dustries. Underdeveloped countries are heavily dependent on developed
countries for equipment and machinery needed to implement industrial
projects. It would be logical to develop upstream sectors that would
provide other sectors with these inputs. The Algerian plan of economic
development has been neinly based on this theory: priority is given
to heavy industries. When the model is applied to the Algerian case
it will be possible to check the claim of this theory. Is it correct
to state that upstream industries lead to the maximization of value
added?

The model, presented in the next chapter, uses two types of
Leontief input-output coefficients. These are domestic and foreign
coefficients. This distinction is made because underdeveloped count-
ries must import many of the inputs and HDSt of the equipment neces-
sary for production pruposes. These coefficients would change from
one year to the other. Let us suppose, for instance, an investment
made in the steel industry results in excess capacity; in the future
the different industries importing steel will reduce their foreign
purchases to buy more domestic steel. This situation might result in
a change of domestic and foreign input coefficients related to the
steel sector. Domestic input coefficients might increase, showing
the decrease in foreign purchases of steel. Hence, in the next per—
iod, new input coefficients must be used, reflecting the change of
situation. This situation occurs each time there exists an opportun-

ity of purchasing more domestic inputs.

37

Both types of input coefficients are used in the model to take
into account the dynamics of the situation.

Data that must be used to compute both of these coefficients
are not available. The model of the next chapter is presented from a
theoretical point of view. It is also used to solve an example to

show the method of solution.

CHAPTER II

THE DYNAMIC PROGRAMMING SOLUTION TO THE DYNAMIC
LEONTIEF MODEL FOR AN INVESTMENT ALLOCATION

In this chapter the problem defined previously is formulated in
a mathematical program using a Leontief type of framework. The model
concerns underdeveloped countries deciding to implement programs of
economic development.

The key question it must answer is how investment can be allo-
cated in a plan of economic development in order to maximize value
added. As indicated previously, three types of constraints are used,
namely, final demand, capacity and investment. Once the model is
defined, the method of solution by dynamic programming will be pre-

sented and applied to a numerical example.

2.1 Definition of the Model

Suppose there are m sectors in the economy.1
Let us define the variables and parameters of the model.
The variables are expressed in monetary units (m u) and the

subscript t denotes the relevant time period t.

Xit = production of sector i.in period t.

f

it final demand for goods of sector i in period t.

1It can be assumed: a sector comprises a group of similar pro-
ducts. See Hollis Chenery and Kenneth Kretschner, ”Resource alloca-
tion for economic development," Econometrica, volume 24, No. 4, 1956.

38

39

sit = inventory of goods in sector 1 at the end of period t.

eit = exports of goods by sector 1 in period t.

zit = imports of final goods 1 by the whole economy in period
t.

cit = capacity of sector i in period t.

”it = slack capacity in sector i in period t.

git = investment required to build an average capacity in sec-

tor i period t.
The parameters are:

Vit = value added per unit produced in sector i in period t ex-

pressed in m u.

aijt = ratio of the value of domestic input 1 consumed by sector

j to the output of sector j.

dijt = ratio of the value of foreign input i consumed by sector
j to the output of sector j.
0% = coefficient of conversion of investment into capacity in

sector 1.

An investment of g. in sector i will increase capacity by(x.g. .
1t 1 1t

8 = number of projects built in sector 1 in period t, Bit is

it
an integer.

The value of sector j output is equal to the sum of the value of

inputs (domestic and foreign) consumed by sector j and the value added

of sector j.1

Ma

d (2.0)

m
For 3 = 1’ th = 1:1 aiktxkt 1 1 1 iktxkt 1 thxkt’

The sum of sector i output, imports of inputs 1, and imports of

1cf. section 1.2.

\—

 

40

final goods 1 must equal the sum of final demand of goods 1, inter-
mediate demand of inputs 1, exports of final goods 1, and change of
inventories in sector 1.

The balance equations of the model are:

m m
Xit 1 1‘51 dijtxjt 1 zit = 1111'. 1;, aijtxjt 1 eit 1 sit
#1
- S1(t—l)’ 1 — 1 m
Xit + uit = Cit’ 1 = l m
m

In the first set of balance equations, 2 d.. x. , is explained

i=1 1jt jt

J'ri

. x. is alread incl d d in . as an be seen
1t 1t y u e x1t C

from Equation (2.0). To avoid double counting, jfi appears under the

by the fact that d1

summation sign.

After rearranging, the balance equations are:

Suppose a set of values for fit’ eit’ sit and Cit 1s prov1ded.
Now everything that is on the right hand side of the balance equations
15 known. The model has 3 m var1ables, namely, Xit’ Zit’ uit’ even

though zit and ”it are residuals, they can be determined right away

41

once a set of values for Xit is found. Moreover, the model has 2 m
constraints.

In this model, the balance equations can be considered as the
constraints of a linear program once an appropriate objective function
is chosen. The economic planners are interested in maximizing the value
added by the whole economy. The value added of sector 1 is the dif—
ference between the values of its output (Xi) and of its consumed in—
puts ( g xij)' Therefore, for a given year, the value added by the

i=1
whole economy is:

111 m m m
2 V,X_ = Z X_ - Z Z X..

1 1 i=1j=l '3
This is also the total net output.1 So, maximizing the total value
added is the same thing as maximizing the net output.

The economic planners' objective, over a T year plan can be ex—

pressed by:

T m

max 2 z V- X-
t=l i=1 '1 '1

Assume the planners have N m u available for capacity building
during a T year plan. Of these N, n1 m u are available for invest-
ment in period 1, 112 in period 2, and so on.

In other terms:

1cf. Equation (1.4) in section 1.1, Chapter 1.

 

 

 

 

   

42

The planner's problem can be expressed as:

k m
max Z = Z E v. x.
t=1 i=1 1111
k m k
2 2 8. g. f_ z n , k = l, 2, . . . T, (2.3)
t=1 i=1 1t 1t t=1 t
(LPk)
m m
xit 1 ji1dijtxjt 11:1 aijtxjt 1 zit = 1111’. 1 eit 1 $11; ‘ S1(t-l),
jfi
1 = 1 m, t - l k
Xit + uit = Cit’ 1 = l m, t = l k
Xit 3_O, ”it 3 O, zit 3_O, i = l . . . m, t = l . . . k

The difficulty in solving this mathematical program stems from

the following reasons.

—- cit’ i.e., the capacity of sector 1 in period to is unknown

for t = 2, . . ., T, a_priori. However, Cit depends on the
t
level of investments made previously, namely 2 Bip g1p
p=l '

-- An investment 9. may change the value of the parameters
1t

and dij in subsequent periods. This will be dis-

Vit’ aijt t
cussed in section 2.2.

-— For each value of k, Inequality (2.3) contains k constraints.
The solution of the problem (LPk) does not guarantee that the
set Bi1911, . . ., Bikgik’ 1 = 1 . . . m,is optimal for
1LPk+l)‘ A dynamic programming approach will be used to

solve this problem in section 2.3.

43

2.2. Assumptions and Dynamics of the Model:

2.2.1 Assumptions:

The model presented in section 2.1 will be used to find an
optimal allocation of investment among the sectors, such that the
value added is maximized over the planning period in a non—integrated
(underdeveloped) economy.
ijt’ dijt and mi must
be known in year 1 of the planning period; Vit’ aijt’ dijt and dijt

may change in subsequent Periods whereas oi stays constant. The con-

The initial values of the parameters Vit’ a

stancy of ai stems from the fact that technology is assumed constant
during the planning period. This is a realistic assumption for a model
in which T = 4 or 5 years as is the case in this research.

For convenience, inflation has been ignored: the cost of build-
ing the same capacity in a given sector is constant each year of the
plan.

The parameters v and di are given for an integrated

it’ aijt
economy. An integrated economy can be defined as an economy able to
produce any kind of goods, and in which imports and exports as a per-
centage of gross national product are less than 10 percent. These

parameters are denoted as V: and 51.

1 13 respectively and are known for

a given year; only values for one year are used.

For an integrated economy the following equality holds:

m— .—
2 a.. + v. = l for any j

44

For a non-integrated economy the relation 15:1

_ (aijt + dijt) + Vj = l for any j

1

"MS

1

Finally, the value of “i is computed for an integrated economy

and is used for the non—integrated economy.

2.2.2 Dynamics of the Model

In period 1, the model (LPk) becomes:

"I
max Z = 2 v. x.
i=1 11 11
m
1:, B11911 5'11
(L13)
m T11
X11 13-51 dijlle 1:1 aijlle 1 7-11 ' 1111 1 '5‘11 1 $11 ‘ 510’
iii
1 = l m
x1111'11‘C11’1'1 1“
Xil 3_O, ui] 3_0, Zil 3_O, 1 = l . . . m

Suppose k sectors out of m decide to increase their capacities.

Let K be the set of these k sectors. For i e K, sector 1 invest Bilgil

m __
1 Z aij f_l, aij 3_0 is used by Hollis Chenery and Kenneth
i=1

Kretschner, “Resource allocation for economic development," Econo-
metrica, volume 24, no. 4, 1956.

 

45

m u in period 1. It should be noted that

,EK 811911i n1

An investment of 9,] does not always result in a capacity in-
crease of digi, in the year of investment. This is only possible if
100 percent completion is assumed. This means that if an investment
is decided at the beginning of year 1 it will be completed during
year 1, so that capacity increases to c1.1 + “igil during year 1. In
the following, a method will be proposed to bring to a 100 percent
completion in the year of investment investments whose completion
times are actually less than a 100 percent. The reason for this will
become clear in section 2.3.

Suppose an investment is completed at h% each year. So it will
take %%-years to reach a capacity level of c1] + ”1911 in sector i for
an investment made in period 1. Let the life time of an investment
91] be 8 years, c Z_l%a During this life time it will be assumed that
sector 1 operates at full capacity. During these e years the value

added per unit in sector i is vi]. The capacity of sector 1 is

oil + h% x digi] in year 1

c1] + 2h% x digi1 in year 2

c1] + kh% x digi1 in year k k 5_l%

c1] + digi] in year %%-+ l, %%-+ 2, . . . e

The value added generated by investment 91.1 is

Vil (h% x o1911) for year 1

46

o , 1_
Vil (th x @1911) for year k, k 5-h%
1

Vil (“1911) for years 621+ l to e

The total value added generated by investment 91] in a years is

Vil (h% x digi1) + . . . + v1.1 (kh% x digi1) + v].1 (6%911) + ...
V11 (011911)

01"
l_
11%
Z
k=l

P ’ V111 1‘ 111011911 +19 ' W) 011911]

Investment 911 can be compared to an investment wi] with 100 percent
completion in year 1 and with the same value added Vil and the same
life time 8. So under the same conditions investment wi] generates the

following value added each year for a years.
V11 (0111111)

For 5 years the total value added is
Q ‘ 1:(Vn X “1W111

The method proposed consists of setting P = Q and solving wi].

 

 

1
v [1? kh‘Xog +(e-1—ag )1
P 11 H °1 11 11% 1 11
W11 €"110‘11~ EV110‘1
1
1111 1
911%; kill; + (E ' 11%)]

"11' e

+

47

 

 

1 1
117 kh%=h7m k=h°’[1—x1(1+1)]=1—(l—+1)
E °Z /°2 11% 11% 2h%

k 1 k=1

The last two equations give:1

 

1 1 1
91112166711) 1 E ‘ 11%3
€

“’11

1 1
9111 ‘2+ 8 ' 2h%)
w].1 e (2.4)

 

In Formula (2.4) 0 < h% 5_l. The case where h% = O is ruled out. Also

note:

 

This 1mpl1es wi] S 91].
Formula (2.4) does not hold if h% is not constant.

In summary, investing 911 in a project completed at h% each year
will be equivalent to investing wi] with 100 percent completion the
first year 1f 91] and wi1 have the same a, v.1 and ail'

1

For any i e K, such that E 81191] g_n], the capacity of sector
16K

1, during year 1, becomes ci] + disilwi]. Set K is not unique, there

are many. Let us label them:

KS, 5 = l . . . 51

Now, for a given set KS, the LP to solve is

1Murray R. Spiegel, Mathematical Handbook (New York: McGraw-Hill
Book Comapny, 1968).

48

m
Max Z - 2 v x.
1:1 11 1l
m m
X11 1.21 dijlxal E1 a13'1’31 + 211 ” fn + E11 + $11 ' 510’
#1
1 = l m
(LPKS) x11 + u11 - C11, 1 ‘ l m
X11 + u11 = C11 + a1311w11, i = l . . . m and i 5 KS
X11 3_O, Z 1 z_0, u 1 3_0, 1 ' l m

Let x11, 211a and U11 be an optimal solution to (LPK ). The increase
5
of sector i capacity, i 6 KS, might change the structure of the economy

and the parameters v. and d. This change will occur if some

it’ aijt i3 t

u11 3_0. Recall u11 = C11 + a1811w 11 - x11, i. e. , unused capacity.

Let

0K = {i, u11 > 0} denote the set of sectors having slack
5
capacity.

The planners would like to use U11 > 0 to reduce imports of goods i,

i e 0KS , by the whole economy. Each sector j, j = l . . . m, used to
* ' * '

buy from sector 1, i e 0KS , a1j1xj. The quantity u11, 1 3 OK , can

be shared by the different buyers of goods i on a proportional basis.

Let y1j be the increase of sector j purchase of a given i c 0y :

‘s
u* (a. x* )
= il ijl jl . . =
Yij m , 1 e OKs, J l m (2.5)
, 1
1.21 aiJlXJl

Yij must be computed for each i a 0K ,

y1j represents the decrease of purchase of foreign goods 1 by

49

sector j.

The new coefficients aijt that will be used in period 2 are:

a.. xt + Y.“
= 111 Jl 13 . = . .
a112 x1 , 1 a 0K , J l . . . m, but Jfl (2.6)
Jl s

The new coefficients dijt are

d112 = a1j1 + dijl - a112, 1 e OKs, J = l . . . m, in (2.7)

The reason for Formula (2.7) is that technology is assumed constant in

the model or in other terms:

+ dijt+l = aijt + dijt’

a1.‘1.t+1 i e OKs, j = l . . . m but jfi.

The validity of Formulas (2.6) and (2.7) is explained as follows. The

different sectors j=l . . . m use the coefficients a111 and (11.1.1 in
period l which are those given in period zero. At the end of period

1 when (LPK ) is solved, the planners have an optimal solution:
5
x11, 211 and u11. If some u11 > 0, they will revise their estimates of

aijt and dijt and Will use Formula (2.6) to get a1:12 and Formula (2.7)
to get d112.
In fiormulas (2.6) and (2.7) it was pointed out that j#i. y.. is

13
the decrease of foreign purchase of goods i by sector j; this does not

have any effect on the value added by sector j, unless j=l. In this
case, the decrease of purchase of foreign goods i by sector i contri-
butes to increase the value added by sector 1. y11 will be shared by

the two elements on a proportional basis.
-- The first element contributes to increase the value added of

sector 1. This element is p11.

50

-- The second element contributes to increase the purchase of

domestic good i by sector i. This is p21.

1.111 = V1. X Y11, 16 0K5 (2.8)

v1 is the value added coefficient of an integrated economy, so

v1 Y11 is the increase of value added in sector i due to a decrease

of purchase of foreign good i by sector i in an amount of y11.

1121: Y11' 1111-: 18 0K5 (2.9)

Now the parameters of sector i, i a 0K , become:
5

a.. X"! + p .
111 11 21 .
a.. = ~————————-——— , 1 e 0 (2.10)
112 x11 KS
V. x"! + p .
11 11 11 .
v. = ————————————— , 1 e 0 (2.11)
12 x11 KS
d112 = am + dm + v11' am ' V12’ ‘5 0K5 (2'12)

Formula (2.12) is justified by the assumption of constant technology.

In summary: for a given set Ks’ s l . . . 51, such that Z 8

. 11911
18KS

f_n1, the planners solve LPK and use the unused capacities U11, where
S
1 e OKs, to find a112 v1a Formulas (2.5), (2.6) and (2.10), (11.11.2 v1a

(2.7) and (2.12) and v12 via (2.11) for each i 9 0K and repeat the
5
previous steps for all values of s, s = 1, . . . , $1.
The next section will show the solution of LPk by Dynamic Pro-

gramming.

51

2.3 The Dynamic Programming Solution

The Dynamic Programming solution to the planner's problem re-

quires the following steps:
-- definition of states and stages
—- use of a recursion formula
-— satisfaction of the optimality principle.

The planner's problem LPk was formulated as

k m
max Z = tEl iEl Vitxit
k m k
tEl 151 Bitg” £t§1 nt k =1’ ' ’ T
LPk

m m

xit +151 dijtxjt '151 aijtxjt + zit = fit + eit + Sit ‘ Si(t-l)
#1

Xit + ”it = Cit

x1t 2,0, u1t 31o, z1t 3_0, i = l, . . . , m, t = l, . . . , k

2.3.1 Stages and States of the Problem

The number of stages in LPk is mT. There are m stages for each
year in a T year plan. Each one corresponds to a given sector i,
i = l . . . m, in a given year of the T year plan. Let (ik) denote
the stage corresponding to sector i in year k, i = 1, . . . , m and
k = 1 , . . . , T.

The state of the system in stage (ik) depends on the planners'
decision to invest, in sector i period k, an amount equal to y m u

when r m u are available for investment in stage (ik). Note y15 r.

52

This implies that r — y was invested optinnlly in the previous

m(k-l) + i-l stages.

2.3.2 Recursion Formula
Let P1k(y) denote the immediate value added generated by invest-
ing y m u in stage (ik).

Let Qi-1,k (r—y) be the optimal value added obtained when r-y

m u was invested in the previous m(k-l) + i-l stages.

The recursion formula is:1

k
0- (r) = max [P- (y) + Q- (r-y)l. r < z n (2.13)
1k YE? 1k 1-l,k _'t=l t

Equation (2.13) is true for any ifl. Inequality (2.3), i.e.

k m k
z z a. g. g z
t=1 i=1 “5 ‘t t=1

k = l, 2, . . . , T

m
becomes 121 811911 5_n1 for k = 1,

and for k = 2 it corresponds to two constraints

m 2 m

2 8. g. 5_n , and, Z Z 6. g. 5_n + n .

1:1 11 11 1 t=1 1:1 1t 1t 1 2
Both of these constraints imply that the maximum invested in the econe
omy in year 2 is n1 + n2 and the least is n2. If an amount less than
n2 were invested in year 2, the following would occur:

1The subscript k to Wagner's recursion formula was added.

Harvey M. Nagner, Princi les of 0 erations Research (2d ed.; Engle—
wood Cliffs, N.J.: Prentice-Hall, Inc., 19755, p. 264.

53

m m m
811911 + .2 B12912 in1+ n2 1 .2 812912 < "2

i=1 i=1 1=l

this implies

m m
1.51 8119115 n1 + "2 '131 812912
In
since 112 — 2 8.29.2 > 0 which includes the possibility of
i=1 ‘ l
m n

131 81191] > n1 because iil B11911 5_n1 + something p051t1ve; this

possibility would not happen if 112 was the least amount that could be
invested in year 2. This includes the case where zero is invested in

year 2 in all sectors i except i = 1. So the recursion formula for

i = l is:
Q]k(r) = max [P]k(y) + Qm,k-l (P1101 (2J4)
5’3”
k k-l
where r :_ 2 nt , and , r-y < 2 nt.

t—i _ t=1

2.3.3 Principle of Optimality:

Bellman1 stated the so-called ”principle of optimality“ encount—
ered in any recursion formula of a dynamic progranming problem as
follows:

An optimal policy has the property that whatever the initial
state and initial decision are, the remaining decisions must

constitute an optimal policy with regard to the state resulting
from the first decision.

1Richard E. Bellman, Dynamic Programming (Princeton: Princeton
University Press, 1957), p. 83. Stuart E. Dreyfus and Averill M. Law,

The art and theor of d namic ro rammin (New York: Academic Press,
19775.

54

The decisions in this research are the investment decisions y.
The principle of optimality as stated above applied to the backward

type solution to dynamic programming problems.

The recursion formulae (2.13) and (2.14) are of the forward
types.1 The model is known only at the beginning of period 1 of the
T year plan. In subsequent years, the state of the model will depend
on the previous investment decisions r—y for a given available amount
r in these subsequent years.

The principle of optimality for a forward type solution can be
restated as:

An optimal policy has the property that whatever the present
state and present decision are, the previous decisions must constitute
an optinal policy with regard to the state resulting from these de—
cisions.

Formulae (2.13) and (2.14) satisfy this statement.

For a given available investment r in stage (ik), a decision to
invest y in this stage determines both the state of the system in
stage (ik) and the decisions concerning the previous m(k-l) + i-l
stages. These previous decisions are an optimal policy for entering

this state at stage (ik).

2.3.4 The Dynamic Programming Approach

Next, the planners' problem (LPk) is solved by dynamic program-

ming.

1George F. Hadley, Nonlinear and D namic Pro rammin (Reading,
Mass.: Addison-Wesley Pub. Co., 1964). Harvey H. Wagner, Principles
of 0 erations Research (2d ed.; Englewood Cliffs, N.J.: Prentice-Hall,
Inc., 1975).

55

For a T year plan, there are mT stages. The states in any stage
depend on the combination of r and y.

Starting with the first year (k=1), the corresponding stages are
(l,l) to (m,l).

For i=1, the recursion formula (2.14) becomes
011(r) = P11(r) for r 5_n1 (2.15)

because by definition Qm o(r-r) = 0
P11(r) is obtained by solving the following LP for a given set

of values of f11, e11, s11.and s10,and for r = 611911, 811 = O, 1, 2,

P11(r) = max Z
m
max Z = E v. x.
1:1 11 11
m m
X11 + 121 dijlle 1:1 aijlle + 211 = f11 + e11 + 511 ' 510
j#1
1 = l m
x11 + u11 = C11 for ifl
x11 + u11 = C11 + o‘1811W11
H1Z°’%13°’%130

This LP has to be solved for each value of r, obtained for a
given value of 811. The computation of P11(r) requires solving £11 =
max [811, integer such that 811911 5 n1] LP's.

Now, for i=2, . . . , m, the recursion formula (2.13) is applied

011(r) = max [P11(y) + QH 1(r-y)l. r 5_n1 (2.16)
xfir ’

56

The computations are carried first for i=2 and so on. For each i, the
change in the capacity of the first i sectors, for a given pair of

(r,y), must be accounted for.

.- + .. = _ + . . . ' = .
xJ1 uJ1 cJ1 aJBJ1wJ1 J l 1
x11 + u11 = c11 for j > 1
also
B11911 ‘ y
and
i-l
Z 8. g. - r-y
1:1 31 31

So the LP that must be solved, to get [P11(y) + Q1_1 1(r—y) for

a given pair (r,y), is

m
max Z = 2 v. x.
1:1 11 11
811911 = y
i-l
2 8. g. - r-y
j=1 11 31
m m
x11 + 121 d1j1XJ1 ' 1:1 aijlle + Z11 " TC11 I e11 + 511 ' 510’
jfi
1 = l m
x11 + u11 = cJ1 + ijj1w11, J - 1 1
x11 + u11 = cj1 for j > 1
x11 :_0, 211 3_O, uj1 3_0, j = l . . . m

For each i one LP must be solved for each value of y. The number of

57

LP% that must be solved, for each i, and for a given value of r, is
811 where 811 = max [811, 1nteger such that B11911 = y 5_r].

For the case where i = 2, . . . , m and y=0, there is an inter—
esting property of the model that can be used to make the dynamic
programming solution appealing. For example, if i=2 and y=0 the pre-

vious LP becomes

1'1 ' 11 + e11+ S11 ‘ 510’

X11+ u11 = C11+ o‘1811‘111

C21
— le for j > 2

X
+
C

I

x. > 0, 211 3_0, u11 3_O, for j = l, . . . , m

It can be seen that this LP has been solved (see (2.15)).

Max Z = Q1_1 1(r) = Q1 1(r) when i=2 or

[P2] (y=0) + 02-],l(r)] = Q],](r)-

This implies that P21(y=0) = 0.

lbreover, for i 3_2 and y=o, P11 (y=0) = o. This is generalized
to any k, i.e., P1.k (y=0) = 0, for i 3_2.

The case of P1k(y=0) when i=l will be discussed shortly. For

k=2 the model deals with year 2. First, there is a new set of values

58

for f12, 212, 51.2 and 511. Second, some parameters such as v12, a112
and dijz might have changed because of investment decisions in year 1.
Both of these situations must be accounted for.

The case of i=1 is treated separately. For i=1, the recursion

formula (2.14) gives

0120‘) = max [P12(y) + 0m111r-y11 (2.17)
YEP
with
2
r < Z n and r-y §_n .
_t=l t 1’

First of all, Qm 1(r-y) is known by Formula (2216) for any pair (r,y)
satisfying the three Conditions above. This will determine the values
of v12, a1jz and d1J2 if they are different from those of year 1. Also,
the new capacities, resulting from investing r-y in year 1 are known.

For y=0 and any r, r 5_n1, P12(y=0) is obtained by solving the following

LP.
m
max Z = Z v. x.
1:1 12 12
m m
X12 +121 dij2xj2 ‘ 1.51 aij2xj2 + Z12 — fi2 + e12 + 512 ‘ S11:
in
1 = 1 m
X12 + u12 = C12 1 = 1 m
x12 3_0, U12 3_0, 21.2 3_0, 1 = l . . . m
The optimal solution to this LP is:
m
= = * * . . .
P12(y 0) 151 v12x12, where x12 15 an optimal product1on level

in sector i, i = l, . . . , m.

59

This implies that P12(y=0) 3_0
For y > O and a given r, i.e., for 812 = l, 2, . . . such that
y = 812 912 §_r, P12(y) is obtained by solving this LP with a new con—
straint reflecting the change in capacity of sector 1. More precisely,

x12 + u12 = C12 + o‘1812W12'

For i 3_2, the recursion formula (2.13) is applied.

Q- (r) = max [P- (Y) + Q- (r-y)] , r < n + n
12 YET 12 1—l,2 —- l 2

Again P12(y=0) = O. For example i=2, y > 0 and a given r, i.e., for
822 = l, 2, . . . such that y = 822922 5hr, P22(y) is obtained by
solving the LP

m
max Z = 1:] v1.2x12
m m
X12 + 121 dithjZ ' 151 aij2xj2 + 212 ' f12 + e12 ‘ 511 + 512’
in
1 = l, , m
x12 + U12 = C12 for i > 2
x12 + u12 = c12 + O‘1512W12
X22 + u22 = C22 I O‘2522W22
x1.2 3_0, u1.2 3_0, 21.2 3_0 i = l, . . . , m

The value of the right hand side of capacity constraints i, if2, is

determined by the investment decisions associated with Q1 2(r—y). The

same approach is used for i=3, 4, . . . , m. Furthermore, the case of
k=3, . . . , T and i=1, . . . , m is similar to the one just discussed
k=2.

It is now possible to give the very reason for having adopted

60

the concept of w1t:

tion in the year the money is spent for an investment g1t. This rea—

the equivalent investment with 100 percent comple-

son satisfies the principle of optinality in the model of concern. For
a given stage (ik) and a state represented by the pair (r,y), in the
quantity [P1k(y) + Q1_1’k(r-y)], the principle of optimality says:
Q1_1’k(r-y) is the optimal value added generated in stages m(k-l) + i—l
when r-y m u were invested in these stages to reach state (r,y) at
stage (ik). The main difference between w. ‘n-

1t t1
tegrates two things: the immediate value added (generated in the year

and g1t is that w1

of investment) and the future value added (generated by the completion
of the project); whereas g1t takes into account only the immediate value
added. If g1.t were used instead of w1t, it would favor investments
having the best immediate value added which would prove to be non—
optimal at some stage (ik).

A simple example will carify things. An amount y=2 is invested
in any of the 2 sectors, assume 100 percent completion in year 1 for
sector 1 and 50 percent completion for sector 2. Let both investments
have the same life time, 5 years. Let the value added by each sector
be v1 = .2 and v2 = .3.

Assume for convenience a1 = a2 = 1, Bit = BZt = l, and that both
sectors are operating at full capacity, C11 = 20, c21 = 15. The total
added value is .2 x 20 + .3 x 15 = 9.5.

Assume that with investment y=2, they will be still operating at
full capacity.

If g1t is used, the value added, generated by y=2 in sector 1,
will be .2 x 2 = .4 and in sector 2 it will be .3 x l = .3.

Investment in sector 1 will be favored by the recursion formulae.

61

(2.15) gives

011(0) = 9.5
011(2) = 9.5 + .4 = 9.9
(2.16) gives
021(0) = 9-5
02112) = max1P21101+ 011(2) . P2112) + 011(0)]

wax [9.9 , 9.8] = 9.9

In year 2 it will be found out that investment in sector 2 was
actually the best. Under the same Conditions as year 1 (f12, e12,
$12, 51.1 are those for period 1 even though the sum tends to increase
each year), the total value added in year 1 and year 2, if the invest-
ment in sector 1 was carried out, would be: 9.9 + 9.9 = 19.8.

If the same investment was made in sector 2, the total value
added would be 9.8 + 10.1 = 19.9.

Using w21 Formula (2.4) gives

1 1
= 921 (2 + e ' _2h%)

W21 6
e = 5, h%= 5
l
21 5 5 ' '

Now the value added generated by investing 2 m u in sector 2, is
.3 x 1.8 = .54 so Formula (2.16) gives

021(2) = max [9.9, .54 + 9.5] = 10.04, i.e., investment in
sector 2 is favored.

This concludes the remarks made on w.

1t and g1t. The problem

62

faced by the planners is very tedious. For any stage (ik), the plan-

ners have to solve Bik LP's where

Bik = max [Bik 1nteger such that Bikgik = y §_r]

T m A
The total number of LP‘s is E Z Bik‘ This quantity can be very

k=1 i=1
large.
Next a method will be proposed to reduce considerably the number

of LP's to solve.

2.4 Insights in the Model

2.4.1 Reduction of Computations by Marginal Analysis

Max 2 = VX such that AX = b and X 3_0

V is a 3 m—component vector of constants

X is a 3 m-decision variable vector

A is a 2m x 3m matrix

b is a 2m—component vector of constants

Once this LP problem is solved, the following information can be
obtained from the final tableau.

J = set of 2m basic variables

8'] = inverse of the basis matrix B; B is a 2m x 2m matrix

b = vector of 2m optimal right hand side constraints

n = vector of 2m dual variables.
and V

Let X denote respectively the vector of 2m optimal basic

B B
variables and the vector of 2m basic coefficients of the objective

function associated with XB.

63

Linear programming theorems give:1

—1

B b = E
-1 _
VB B — n
VBXB = Max Z = nb

The right hand side ranging method (abbreviated as r.h.s.r.m.)

provides the tool for achieving the reduction of computations.

th

Let the right hand side of the i constraint increase by y.

Let 3(y) be a 2m column vector where all components are zero except the

ith one equal to y.

For y=l, the new vector of optimal right hand side becomes

-1 th 1

B" = B (b + 3(1)) = E'+ i column of B-

the new objective function is
max Z = n(b + 3(1)) = nb + ith component of n.

1

Now, for any y, so long as B' (b + 3(y)) = 5” remains non negative,

i.e., Eh 3_0, for j=l , . . . , 2m, the basic set remains optimal
(and n the optimal dual vector). If y is large enough, 5" may becone

negative in which case the old basic set is no longer feasible.

-1 .
II = b D + B I I = , O I I , I
J J 11 y j 1 2m (2 18)
-l - -l . .th .th
where 811 15 the element of B corresponding to the 3 row and 1

column.
53 will become zero for some value of y if B31 is negative.

For the set J to remain basic, the maximum value of y should be

1George B. Dantzig, Linear Programmin and Extensions (Princeton,
N.J.: Princeton University Press, 19631. Leon S. Lasdon, Optimization
Theor for Lar e S stems (New York: Macmillan, 1972). Allan w. Spivey
and Robert M. Thrall, Linear 0 timization (New York: Holt, Rinehart,

and Winston, Inc., 1970).

64

—1

11 < 0] (2.19)

A E-
= y = min [ —%T- , B

This means that the right hand side of the ith constraint can increase

by y leaVing the set J basic and n the optimal dual vector.

For 0 < y §_y the value of the objective function is

th

Max Z = n(b + 3(y)) = nb + 3(y) (i component of n) (2.20)

Equation (2.20) will be used to reduce the number of LP's to solve.

2.4.2 Further Insights in the Model
The nodel of concern has the nice following property. Once the
value of y is reached, increasing the right hand side of the ith con-
straint will not improve the value of the objective function. In
other words for y 3_§, Max Z = nb + 8(y) x n1, where n1 is the ith
component of n. This is so because of the following reasons.

V is a 3m component vector where all components are zero except
the first m ones, those corresponding to Xit' Since there are 2m
constraints, m for demand constraints (f1t, e1t, s1t, Si(t-1))’ and,
m for capacity constraints, it is expected all x.

1t
tive level in the optimal solution unless some constraints have a

will be at a posi-

zero right hand side. At the optimal solution for any x1t, one of

two cases must happen:

1) 21t 3_0 and u1t = 0, or,
* =
2) u1t 3_0 and 21t 0.

The first case states that the sector i is operating at full
capacity. Increasing the capacity of sector 1 by y §_y will improve

the value of the objective function by 8(y)n1 and increase the optimal

65

value of Xit' Now, for y > y the result is a surplus capacity and the
second case occurs. Z1t will leave the basic set J, u1t will enter
it and x1t will keep its value. Since u1t has a zero coefficient in

vector V, its contribution to the objective function is zero.

This nice property will further reduce the number of LP's to
solve, since, beyond y the objective function remains the same even
though the basic set changes.

The basic matrix has a special characteristic if one is able to
know what the basic variables are and consequently whether or not the

slack capacity variables u1t are basic.

If none of the u1.t are basic one does not have to solve an LP

to get the optimal solution, i.e., Z* and E} but this is not enough;

one must calculate B'] to perform the r.h.s.r.m. computations. In the

following, a method will be proposed to find 8']. In the basic set

J,there are 2 m basic variables m for z1t and m for x11: since the

case where none of the slack capacity variables u1t are basic is
assumed. The method consists in arranging the set J starting with

the import variables 2. occupying the first m places and the last m

it

places being occupied by the x1t's. J has the following format:

lt’ ZZt’ ' ' ' ’ th’ Xlt" X2t’ ‘ ' ‘ ' xmt}'

The basic matrix B is also arranged by columns according to the
order in set J. Its inverse or B—1 is obtained very easily: it is
B but where all figures switch sign except those equal to 1. Note that

the one's are on the diagonal.

Example: J = {3,4,l,2} i.e., m = 2

03
II
COO-4
OO—JO
...
OOO—'

This method can be used only when one knows what the basic se-

quence is and if none of the u. are basic.

1t

There is one instance where this is the case. This is when Vit’

a and dijt at stage (k,l) (where k1: 2) are recomputed. This is so

ijt
because of the following reasons. The parameters are recomputed in
state (k,l) when one or more sectors are operating below capacity in

stage (k-l,m), i.e., the last stage of the previous year; the slack

capacity is used to reduce d1. and increase v. and a. Once the

jt 1t 1jt’

new parameters are obtained, the sectors that had U1t

jt 3_0. The case of

d.. < 0 occurs when the slack capacity u1 more than offsets the

> 0 will operate

at full capacity with the new parameters if d1

1jt m t
imports jg] dij(t-l)xj(t-l)’
jfi

which leads to dijt < 0 in the Formulae (2.7) and (2.12)

dijt = aij(t-1) + dij(t-]) ' aijt’ for in,
and
As long as d 0 in the previous formulae, its slack capacity is

ijt 3
reduced to zero, which implies it cannot be basic. So, one will have

to solve an LP only if dijt becomes negative implying that dijt 1s
zero and slack capacity exists. Note also, if one or more slack cap-
acity variables are basic and the basic set J is known, an LP need not

be solved. Instead, the inverse basic matrix B.1 must be found and

67

2*, b and W are computed respectively, which can be easier than solv-
ing an LP.
The previous section in conjunction with the right hand side

ranging method will make the dynamic programming approach efficient.

2.5 Dynamic Programming Solution in a New Light

In this section, a synthesis is made from both of the last sec-
tions to arrive at the final format of the dynamic programming solu—
tion.

Year 1: Year 1 corresponds to the first m stages.

 

*Stage (1,1): recursion formula (2.14) gives

011(r) = P11(r) r < n1 (2.21)

For r=0, the following LP must be solved:

"1
max Z - iil v11x11
m m
X11 + 131 d131XJ1 151 a1j1xj1 + 211 = f11 + e11 + S 1 ' 510:
Jfl
1 = l m
X11 + ”11 ‘ C11: ‘=1 m

N130’fi130’fi13m

The solution gives P11(0) = 011(0), and the basic set J. For r > 0,
an annunt r = 811911 is invested in sector 1, resulting in a capacity '
increase of G1811W11 in sector 1. Using the r.h.s.r.m., the maximum
increase in capacity of sector 1, for the set J to remain basic can

A

be obtained. Let R be such a capacity increase.

68

R = max {a1811w11} (2.22)

Since 01 and w11 are fixed, R depends only on 811

 

. 13
B = (2.23)

For any r = 811911 such that 811 §_B11 (811 integer) it follows

Q11(r) I P11“) I “1811w11"m+1 (2'24)

where n corresponds to the dual associated with constraint m+1

m+1

(the first capacity constraint).

For any r = 811911 such that 811 > 811 (511 integer) the maximum

value added is

Q11(r) I P11”) I O“11311“‘11'Im+1 (2'25)

Note that 811 is not restricted to be integer. At state (1,1) the
values of Q11(r) = P11(r) for r §_n1 and the optimal basic set J and

output levels B'are recorded in Table 2.1.

Tab1e 2.1 Stage (1,1)

 

69

Recursion formula (2.13) gives

Q11(r) = 51:);[131-10’) + Q1_'|,‘|(r‘.Y)]a Y‘ in] (216)

Table 2.2 for stage (1,1) has the following format:

Table 2.2 Stage (i,l)

 

The shaded area corresponds to y > r.

In this table, the only boxes filled are those satisfying y 5_r.
In other words, the computations of Formula (2.16) are carried out up
to the point where r=y in each row of the table.

There is an interesting property in the dynamic programming re—
cursion formula (2.16). Q1_1 1(rsy) is constant for (r—y) fixed.
The quantity (r-y) assumes values from zero to r. For each value of
(r-y), there is one line crossing the boxes in a diagonal manner; this
line crosses only one box in each row and each column. It is along

this line that the r.h s.r.m. is applied. For example, in the case of

70

line r-y=0, y takes on the value 0, g11, 2911, etc. P11(y) + Q1_1,1(r-y)
is the quantity that must be computed for each box along this line.
Since Q1_1 1(r-y) is constant for all these boxes, the only difference
between them is the value of y. The value of y only affects the ith
capacity constraint. The application of the r.h.s.r.m. is made pos-
sible, because P11(y=0) and Q1_1;1(r-y), for a given (r-y), have the
same basic set J.

Recall that P11(y=0) = O for l=2, . . . , m, as was shown pre-
viously. Once the r.h.s.r.m. is carried out for all lines (r-y), one
is able to find Q11(r) along each row of the table, to record the
value of y, the basic set J and 51

There is a case where one has to be very cautious. The value

of Q11(r) which is equal to max [P11(y) + Q1_1 1(r-y)] is also the
ysr ’

solution of a linear program where y is invested in sector i, r-y
has been optimally invested among the other i-l sectors and zero will
be invested in the other i+l to m sectors. There is one case where
this relationship does not hold. In other words, Q11(r) can be less
than or equal to the 2* solution of the corresponding LP. This
happens if both of the following conditions hold.

lst condition: Q1_1’1(r-y), i.e., the value added generated by
optimally investing (r-y) among the (i—l) first sectors, is obtained
with

A

P = 0 + a.

1-1,1 1-1 X 81-1,1 X w1—1,1 X 1Tm+1-1

P1_1 1 is the maximum generated value added along its line in stage

(i-l,l) because the optimal investment decision leads to slack

71

capacity in sector i-l.
2nd condition: The value added obtained by investing y in
sector i also results in a slack capacity in sector i, i.e., P11(y) =

O + a1 x 811 x w1.1 x nm+1. In both of these cond1t1ons, the optimal

decision has resulted in slack capacity in two adjoining sectors lead—
ing to 811 and B1_1 1 which are not necessarily integer.

This situation is due to the non-independence of the sectors,
situation which is not taken into account by the r.h.s.r.m. This
weakness of the r.h.s.r.m. does not affect the recursion in future
stages, namely i+l to m, and, can be remediated by solving the LP
corresponding to Q11(r) and thus getting Z*.1

Year k: k=2, . . . , T

 

All that follows applies for any k=2, . . . , T.

In stage (l,k) P1k(y=0) is not equal to zero, contrary to the
others.

*Stage (l,k):

Recursion formula (2.14) gives

Q1k(r) = 11111:: [P1k(y) + Qm,k_1(r-y)l

where k-l

Z n

t=1 t

r 5. nt , and, r-y 5

The table in stage (l,k) has the following format.

1In the example solved in section 2.6, this situation occurs in
stage (2,2) for r=28 along line r-y = 20. '

72

Table 2.3 Stage (l,k)

 

The upper shaded area of the table is y > r.
k—l
The lower one is r-y > Z n

t-l 1'.

73

For P1k(y=O) and a given value of r, an LP must be solved. This

LP has two characteristics:

-- First of all, it has a new right hand side for demand con—

straints: fik’ e1k, 51k, Si(k-l)
-- Second, the parameters Vik’ aijk and dijk might have changed
for some i's, because of investments in the previous k-l years.
*Stage (i,k): i=2, . . . , m
The same steps used in stage (1,1) are applicable here.

An example will be used to show how the dynamic programming

approach works.

2.6 An Example

Assume there are two sectors, m=2. The following data pertain

to year 1.
a.. i ..
j 1 1 ijl 2 v. j 1 131 2
______._______________;L
l 2 3 .2 l l 2
2 3 2 .3 2 l 1

Assume there is a two year plan with the following demands, and

also assume that e1t = s1.t = 0.

74

t
i 1 2
f1t 15 18
f2t 20 22

The initial capacities are C11 = 20, c21 = 24.

The coefficients of conversion are

011=1
dz = .9

The percentage of completion of project in sector 1 is 100 per-
cent and in sector 2, 50 percent.
The life times of the projects are:
e = 5 years in sector 1
e = 10 years in sector 2.
The investment available in each year is 14 m u, i.e.,
n1 = 112 = 14.
The costs of a project in sector 1 and sector 2 are, respectively,

2 and 4. The equvalent investments w1 obtained by using (2.14) are:

t
2 (15.15%)
W1t - _—-——_TT-__——_ — 2, and
1
4 (§-+ 10 - l)
.21 = —T—= 3-8

The solution by dynamic programming will be given in 4 stages,
2 for each year.

2
Max Z = Z vjx .

2 2
X21 I E d23'1"J'1 E a2jlle I Z21 I 20
j—l j-l
if?
x11 + u11 = 20
x21 + u21 = 24
or
Max Z = 2x11 + .3x21
x11 + .lx21 - .2x11 = .3x21 * 211 = 15
x21 + .2x11 - .3x11 - .2x21 + 221 = 20
x11 + u11 ' 20
x21 + u21 - 24
or
Max 2 = 2x11 + .3x21
8x11 — .2x21 + 211 = 15
-.lx11 + .8x21 + 221 = 20
X11 I u11 I 20
= 24
x21 I u21
“1:0,Q130,%130,#L2

The solution of this LP gives the following information.
2* = 11.2 E: (3.8, 2.8, 20, 24)
J = {211, 221, x11, x21} , n = (0, O, .2,.3)

3'1 =

O—J—‘m
I
#OOJN

COO-d
OOé‘O

From now on, the basic variables in the set J will be denoted
by their column number, i.e.,

J = {211, 221, x11, x21} = {3, 4, l, 2}

76

If it was known that both of the input variables 211 and 221
were basic, 8'1 could have been found by the method illustrated in
section 2.5.2 without having to solve an LP.

Year 1

 

* Stage (1,1)
P11 (r) = Q11(r) r 5_ 1 l4
(r=O) = 11.2

3
II

P
11
For r > 0, the r.h.s.r.m. will be applied.

Firs, none of the slack capacity variables in are basic, this
implies that P11(r) > 11.2, for r > 0.

Sector 1 capacity corresponds to the third constraint, i.e.,

i=3, so Formula (2.19) gives

- 5; -1 3 8
y=minI—ET, Bj3<0]=—'§=4.75
J -Bj3 I

Formula (2.19) is identical to Formula (2.22)

y = R = 4.75

13 4.75

From Formula (2.21), 611 = 317—: —2—= 2.375
1 11

r = 811911, 811 integer. If 811 > 2, sector 1 will operate
under capacity.
The discussion following Formulas (2.22) and (2.23) gives
r = 2 011(2) = P11 (0) + 1 x l x 2 n3
= 11.2 + l x l x 2 x .2
= 11.6

1
I
4:

011(4) = 11.2 + l x 2 X 2 x .2
= 12

77
For 811 > 2 or r > 4 Formula (2.24) gives
011(6) = 11.2 + 1 x 2.375 x 2 x .2 = 011(8) = 011(10) = 011(12)
= 12.15

Table 2.4 contains data on the level of investment r, the

corresponding value added (Q11(r)), and the optimal basic set J and

output levels 51

Table 2.4 Results of Stage (1,1)

m:

l" Q11(r) J b-
O 11.2 3, 4, l, 2 (3.8, 2.8, 20, 24)
2 11.6 3, 4, l, 2 (2.2, 3, 22, 24)
4 12 3, 4, l, 2 (.4, 3.2, 24, 24)
6 12.15 5, 4, l, 2 (1.25, 3.275, 24.75, 24)
8 12.15 5, 4, l, 2 (3.25, 3.275, 24.75, 24)
10 12.15 5, 4, 1, 2 (5.25, 3.275, 24.75, 24)
12 12.15 5, 4, l, 2 (7.25, 3.275, 24.75, 24)
14 12.15 5, 4, l, 2 (9.25, 3.275, 24.75, 24)
The above table requires some explanations.
For r=0, J and E are those obtained by the LP solution found
previously.

For r=2 and r=4, the r.h.s.r.m. gives the same basic set J as

for r=0. To obtain 5} for r=2 and r=4, Formula (2.18) is used, i.e.,

— — -1
. = . + .. '
b1 b1 811 r,

+
N
II
ON

sor for r=2' E“ =

..-
NNMN
#N- -

The case of r=4 is obtained similarly.

78

For r=6, by the results of the r.h.s.r.m. J changes, because
sector 1 is operating below capacity. This implies that the import
variable 211 (or 3) leaves the basis and slack capacity u11 (or 5)
will enter it. The new basic set is J = {5, 4, l, 2}. However, an LP
need not be solved, since the new basic set is known, the inverse

basis matrix associated with it is obtained by the method of section

2.4.2.
—1.25 o 1 - .25 1.25
B—1 _ .125 1 o - .775 B'— 3.275
1.25 0 o .25 24.75
0 o o 1 24

= (.25, 0, 0, .35).

For r=6, b'is computed as follows
—- -1 -1

b = B b = B (15, 20, 26, 24) = (1.25, 3.275, 24.75, 24)
For r=8 5‘: B4 (15, 20, 28, 24) = (3 25, 3.275, 24 75, 24)
For r=lO 5': B4 (15, 20, 3o, 24) = (5.25, 3 275, 24.75, 24)
For r=12b B'I (15, 20, 32, 24) = (7.25, 3.275, 24.75, 24)

For r=l4 EI= =B-I (15, 20, 34, 24) = (9.25, 3.275, 24.75, 24)

* Stage (2,1)

= P 14
021(r) 3:: I 2161) + Q11(r- -y)l r <

P21(y=0) = 0
a2 = .9
3.8

"21

92t 4

In each box of Table 2.5 corresponding to stages (2,1) two values
will be recorded: P21(y) and Q11(r-y).

Remember the r.h.s.r.m. is used along the lines r-y.

79

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AN5.2N.o.mN.me..e.Nv N.P.e.m a mm.m_ ~.__+mo._ N_+N._ m_.NF+N._ m2.m_+o N_
Ame.em.m.mw.mo..e.v N._.e.m 5 mm.m_ o._F+m_.F m_.N_+N._ m_.N_+o o_
Ame.mm.em.oe..mo._v N._.e.m e mo.m_ N.__+mo._ Np+mo._ m_.N_+o w
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m a » ALV_No NF w e o »x

 

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A—.Nv macaw mo mp—zmmm m.m mFQmH

80

Sector 2 capacity corresponds to the fourth constraint, i.e.,
i=4.
1) line r-y = O Q11(r-y = 0) = 11.2 with corresponding
j = {3, 4, l, 2} and 5': (3.8, 2.8, 20, 24)

A = - . J -] = c2—.—8—=
y R min E-B'I , 814 < 0] .8 3.5
j4
A _ 3.5 3 5
821 I 9x3._8 I 3 42 ‘ I-0234
P21(y=0) - O
P21IYI4I I P21II’I0I I 0‘2 X 821 x w21 X I4
= 0 + .9 x l x 3.8 x .3
= 1.026 = 1.03
P21(y=8) = P21(y=0) + .9 x 1.0234 x 3.8 x .3 = P21(y=12)
= 1.05
2) line r-y = 2 Q11(r—y = 2) = 11.6 with corresponding
J = {3, 4, 1, 2} and EI= (2.2, 3, 22, 24)
37: 2: 133—: 3.75
is . £5.
21 3.2 ; I'0965
P21(.y=0)
P21(y=4) = O + .9 x l x 3.8 x .3
= 1.03
P21(y=8) = O + .9 x 1.0965 x 3.8 x .3 = P21(y=12)
= 1.125 = 1.13
3) line r-y=4 Q11(r-y = 4) = 12 with

J = {3, 4, 1, 2} and 5'= (.4, 3.2, 24, 24)

§=§'=%=4

81

A - 4 _
82] — 3.42 _ 1:]696

P21(y=0) = O
P21(y=4) = O + .9 x l x 3.8 x .3 = 1.03
P21(y=8) = O + .9 x 1.1696 x 3.8 x .3 = 1.2
4) line r-y = 6 Q11(r-y = 6) = 12.15 with
J = {5, 4, 1, 2} and BI= (1.25, 3.275, 24.75, 24)

 

 

 

 

y = R = min [1.2: , —3'§;g] = 4.2258

A _ 4.2258 _

82] — 3.42 _ 1.2356

P21<y=0) = 0

P21(y=4) = O + 82 x 821 x w21 x n4 = O + .9 x l x 3.8 x .35

= 1.197 2 1.2

P21(y=8) = 0 + .9 x 1.2356 x 3.8 x .35 = 1.48
5) line r—y=8 Q11(r-y=8) = 12.15 with
J = {5, 4, 1, 2} and EI= (3.25, 3.275, 24.75, 24)

y = R = 1111111363 , ————3'§;I5’1 = 4.2258

A _ 4.2258 _

82] _ 3.42 ‘ 1.2356

P21(y=0) = 0

P21(y=4) = 0 + .9 X 1 X 3.8 X .35
= 1.2
6) line r-y=lO Q11(r-y=lO) = 12.15 with
J = {5, 4, 1, 2} and EI= (5.25, 3.275, 24.75, 24)

 

94:11.52, , 4,—1.4.2258

‘4
UI

 

 

 

82

P21(y=0) =0
P21(y=4) = o + .9 x1 x 3.8 x .35 =1.2

tocompleteTable2.50neproceedsasfollows. Alongeachrow,the

value or 021(r) = $3: [P21(y) + Q11(r-y)] and the corresponding value
of y are recorded. “10 get J and Shane proceeds as follows. The
value of y is zero for the first two rows, i.e., r=0 and r=2, this
implies that J and 5 remain the same as in stage (1,1).

For r=4, Q21(r=4) = 12.23 is obtained when 4 are invested in
stage (2,1) leading to 1 project, i.e., 821 = 1. Since 821 = l (821

= 1.0234, along line r=y=0, J is the same as the one found for

Q11(r-y=0); E'is obtained by using (2.18).

O)
H
LO

X
(A)
be
ll
UL)
.b
N

3.8 .2 4.48
—-_ 2.8 -.8 _ .06
b — 20 + 0 x 3.42 — 20

24 l 27.42

For r=6, using the same reasoning, it follows

2.2 .2 2.88
_ 3 -.8 _ .26
24 l 27.42
For r=8,
.4 .2 1.08
— _ 3.2 -.8 _ .46
24 l 27.42

83

For r=lO,
1.25 -.25 .40
-_ 3.275 - 775 _ .63
b ‘ 24.75 I .25 X 3'42 ' 25.6
24 1 27.42,
For r=12,
3.25 ' -.25 2.4
-— _ 3.275 - 775 _ .63
b ‘ 24 75 I .25 X 3'42 ‘ 25.6
24 1 27 42

For r=l4, the best thing to do is to invest 8 in sector 2 which means
that 2 projects will be built. However, along line r-y=6, the maximum
immediate value added of 1.48 corresponds to 1.2356 projects. This
means by investing in two projects sector 2 will operate below capa-

city. Therefore, the basic set J is {5, 6, l, 2}

-l.2908 - 3226 1 o
. -1 _ -.l6l3 -1.2903 0 1 _
”It“ 3 ‘ 1.2903 .3226 0 0 ’ I I I 2’ '3’ 0’ 0)
.1613 1.2903 0 o

5': (15, 20, 26, 30.84) B.I = (.19, 2.61, 25.81, 28.23)

Year 2

 

In stage (l,k), k 3.2, parameters value might change, due to
excess capacity in year 1.
* Stage (1,2)

012(r) = $3? [P12(y) + 021(r-y)l

r < n1 + r12 = 14 + 14 = 28

r - y g_l4

 

 

84

To get the value of P12(y=0) in stage (1,2) an LP must be solved
because the right hand side of demand constraints has changed, and, so
did the capacity constraints because of previous investments.

1) line r-y=0 Q21(r-y=0) = 11.2 with
J = {3,4, l, 2} and 5': (3.8, 2.8, 20, 24)

There is no need to solve an LP because none of the slack ca—
pacity variables are basic. The increase in the right hand side of
demand constraints will not improve the value of 2*, but the import
variables 212 and 222 (i.e., 3 and 4 in J) will increase, respecively,
by the same amount as demand, namely, 3 and 2.

P12(y=0) = 11.2 and 5': (6.8, 4.8, 20, 24)

Let us apply the r.h.s.r.m. along line r-y=0. Sector 1 capacity

corresponds to the third constraint, i.e., i=3.
y=R=—8—=8.5

82] :1); = 4.25
P12(y=2) = 11.2 + 1 x 1 x 2 x .2 = 11.6
P12(y=4) = 11.2 + .8 = 12

P12 y=6) = 11.2 + 1.2 = 12.4

(
P12(y=8) = 11.2 + 1.6 = 12.8
(

P12 y=10) = 11.2 + 1 x 4.25 x 2 x .2 = 12.9 = P12(ye12) = , , .
= P] 2(y=28)
2) line r-y=2 021(r—y=2) = 11.6 with

J = {3, 4, 1, 2} and 5-: (2.2, 3, 22, 24)

Following the same reasoning as above, it follows

P12(y=0) = 11.6 and E= (5.2, 5, 22, 24)
§ = E = §4§ = 6.4

85

B12 = —5— = 3 25

P12(y=2) = 11 6 + 4 = 12

P12(y=4) = 12.4

P12(y=6) = 12.8

P12(y=8) = 11.6 + 1 x 3.25 x 2 x .2 = 12.9 = . . .
= P12(y=26)

For lines r—y=4 up to r-y=8, the computation will not be carried out
because they follow the same procedure as above since J does not con-
tain any slack capacity variables. The figures are in Table 2.6.

3) line r—y=lO 021(r-y=l0) = 13.35 with

J = {5, 4, l, 2} and 5'= (.4, .63, 25.6, 27.42)

Here, since slack capacity is present in sector 1, the parameters

v12, a1112 and d112 w1ll change. The slack capac1ty u1 .4 is used
to reduce imports of goods 1 by the whole economy.

Equation (2.5) gives

Y11 .2 x 25.6 + .3 x 27 42 13.346

_ .4 x .3 x 27.42 =

Y12 I 13.346 '2455

Equations (2.6) and (2.7) give

_ .3 x 27 42 + .2465 =
a122 I I'I27742 III '309

d122 = .3 + .1 - .309 = .091

Equations (2.8) to (2.12) result in

p11 = .3 x .1535 = .04605

u21 .1535 - .04605 = .10745

 

86

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mm.mp+pm.¢p mm.m_+N¢.m— m©.mp+m~.mp mm
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87

A55.55.5.55.55.5.5.55
A55.55.5.55.55.5.5.55
A55.55.5.55.55.5.5.55
A55.55.5.55.55.5.5.5v
A55.55.5.55.55.5.5._5

A55.55.55.55.5.55.55
A55.55.55.55.5.55.55
A55.55.55.55.5.55.55
A55.55 .55.55.5.55.55
A55.55.55.55.5.55.55
“55.55.55.55.5.55.5V
A55.55.55.55.5.55.55

 

 

5._.5.5 5_ 5.55 5.5—+5.55 5.__+5.5_ 55.5_+5.5_ 55.5_+5.5_ 55.5_+5.5_ 55.5_+_5.5_
5.5.5.5 55 5.55 5.__+5.5_ 5.__+5.5_ 55.5_+_.5_ 55.5_+5.5_ 55.5_+ 5.5.
5.5.5.5 5_ 5.55 5.__+5.5_ 5.5_+5.5_ 55.5_+_.5_ 55.55+ _5_
5.5.5.5 5 5.55 5.__+5.5_ 5.__+5.5P 55.5_+ _.55
5._.5.5 5 5.55 5.5—+5.55 5.5_+ 5.5_
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5.5.5.5 5 55.55
5.5.5.5 5 55.55
5._.5.5 5 55.55.
5.5.5.5 5 55.55
5._.5.5 5 55.55
5.5.5.5 5 55.55
5._.5.5 5 55.55
555.55.5.5.5v 5.5.5.5 5 5.55
A55.55.5.5.5.5v 5._.5.5 5 5.55
5 5 5 A555_5 55 55 55 55 55 5_

 

 

A555555=555 5.5 55555

a112 25.6

= .2 x 25.6 + .04605 =
v12 25 6IIIIIIII
d - .2 + .1 + .2 -

112 _

88

= .2 x 25.6 + .10745 =

.204 —

.204

.202

.202 = .094

The LP that must be solved with the new parameters has the following

form in period 2.

Max 2 = .202 x12 + .3 x22
x12 + .091 x22 - .204 x12 - .309 x22 + 212
x22 + .2 x12 - .3 x12 - .2 x22 + 222 - 22
X12 I u12 I 26
x22 + u22 = 27.42
or
Max Z = .202 x12 + .3x22
.796 x12 - .218x22 + :12
-.l x12 + .8x22 + 222
X12 I U12
X22
X12 3_O, z1.2 3_0, U12 3_O.
From the

18
22
26
27.42

discussion of section 2.2.2, whenever the parameters

for a given sector are recomputed, this sector will operate again at

full capacity.

This is the case for sector 1; since sector 2 was also

operating at full capacity there is no need to solve this LP since the

basis will comprise 212 and 222, and, B'I can be computed as shown in

section 2.5.2.

J = {3, 4, 1, 2}

89

E'= BIIb = BII (18, 22, 26, 27 42) = (3.28, 2.66, 26, 27 42)
2* = 13.48
P12 = 13.48

The r.h.s.r.m. application provides the following value

3.

N
00

y=R=.—7—9—6=4.1206

512 = 5=l§9§-= 2.0603

P12(y=2) = 13.48 + 1 x l x 2 x .202
= 13.884 2 13.88

P12(y=4) = 13.48 + .808
2 14.29

P12(y=6) = 13.48 + .1 x 2.0603 X 2 X .202 = P12(y=8) = . . .
= P12(y=l8) = 14.31

4) line r-y=l2 021(r-y=12) = 13.35 with

J = {5, 4, 1, 2} and b = (2.4, .63, 25.6, 27.42)

 

Since u11 = 2.4, v12, a1‘12 and d1‘1.2 must be recomputed.
_ 2.4 x .2 x 25.6 _ 12.288 _
Y11 I .2 {I25T5I1IT3IXI27742'I 13.346 I '9206
_ 2.4x .3 X 27.42 _
Y12 I IIIITEififiTIIIIII I 4794

_ .3 x 27.42 + 1.4794 =

a122 I 27.42 '354
d122 = .3 + .1 - .354 = .046
611 = .3 x .9206 = .276l8
021 = .64442
_ .2 x 25.6 + .64442 -
a112 I IIIIII2576 I '225
- .2 x 25.6 + .27618 = _211

v12 I 25.6

d112 = .2 + .1 + .2 -

The LP to be solved i

Max Z = .211 x12 +
.775 x12

-.1

fizio’fizimum

Here again the basic set is J = {3, 4, l, 2}, for the

vious reasons.

A _ A _ 4. 5 =
y — R - :775 6 129
812 I 64%gg_= 3.065
P12(y=2) ' 14.13 + l x l x 2 x .211 = 14.52
P12(y=4) = 14.97
P12(y=6) = 15.37
P12(y=8) = 14.13 + 1 X 3.065 x 2 X .211
= P12(y=16) = 15.42
5) line r-y=l4 021(r—y=l4) = 13.63 with

S

-.308 x

90

.225 - .211 = .064

.3 x22

+2

22 12

> 0

J = {5, 6, 1, 2} and EI= (.19, 2.51, 25.81, 28.23)

18
22

28
27.42

same pre-

27.42)

Since u11 and u21 are basic, all parameters must be recomputed.

D)
V
C
._a
._a
ll

91

.19

.19 x .3 x 28.23 =

_ .3 x 28.23 + .33018 =

v22 I 28.23 '3I2

.19 x .2 x 25.81 = .98078
Y11 22 iI253fiIIIT37F28.23 13.631

 

Y12 I 13.631 'II8
_ .3 x 28.23 + .118 _
a122 I IIIIITZIZEIIIIIIII I '304
d122 = .3 + .1 - .304 = .096
011 = .3 x .072 = .0216
1,121 = .0504
_ .2 x 25.81 + .0504 _
a112 I 25.81 I ‘202
_ .2 x 25.81 + .0216 _
V12 I IIIIII25I81IIIIII' I 'ZOI
d112 = .2 + .1 + .2 — .202 - .201 = .097
b) u21 = 2.61
= 2.61 x .3 x 25.81 : 20.20923 =
Y21 .3'x 25T§TIIIZTGF28I23' I127289
_ 2.61 x .2 x 28.23 =
Y22 I IIIIIIfiiifififIIIII I I006
_ .3 x 25.81 + 1.5094 =
a212 I IIIIIIIifiiifiIIIIIIII '359
d212 = .3 + .2 - .359 = .141
612 = .3 x 1.1006 = .33018
022 = .77042
_ .2 x 28.23 + .77042 _
a222 I IIIIIIII28I23TIIIIIII I ‘277

.072

1.5094

92

d222 = .2 + .3 + .1 - .227 - .312 = .061

The above values are summarized in the following tables.

d..

a..
. 1 132 . 1 132
j l 2 v12 j l 2 <
l .202 .359 .201 l .097 .141
2 .304 .227 .312 2 .096 .061

Once the parameters new values are included the LP is:

Max Z = .201 x12 + .312 x22
.798 x12 - .208 x22 + 212 = 18
-.218 x12 + .773 x22 + 222 = 22
X12 I u12 I 26
x22 + u22 = 30.84

The basic set is J = {3, 4, l, 2}, for the same previous rea-

sons; and, hence

g 0 - 798 .208
—1 _ 1 .218 -.773 _
B — 0 0 1 0 , n — (o, 0, .201, .312)
0 0 0 1
5': BIIb = BII (18, 22, 26, 30.84) = (3.67, 3.83, 26, 30.84)

2* = 14.85 = P12(y=0)

; = 8 = §§g%-= 4.599

A _ 4.599 _
812 II _2—— - 2.299

P12(y=2) = 14.85 + l x l x 2 x .201 = 15.25
P12(y=4) = 15.65
P12(y=6) = 14.85 + 1 x 2.299 X 2 X .201 = . . . = P12(y=14)

15.77

93

To complete Table 2.6 one must record the values of Q12(r),

y, J and b. For r=O to r=l4, y=0; J and 5 are those obtained when the
different LP's used to get P12(y=O) were solved.

For r=l6 and r=l8, the value of Q12(r) was obtained with

J = {3, 4, 1, 2} which means that 5, for those values of r, can be

found using Formula (2.18).

b = (3.67, 3.83, 26, 30 84) + (3rd column of BII)a1812w12

For r=20 up to r=28, Q12(r) attains a maximum value because sec-

tor 1 will be operating below capacity. So, the slack variable u12

will enter the basis and 212 (or 3) will leave the basic set. The

new basic set is J = {5, 4, l, 2} and remains the same from r=20 to

r=28. B is given below.

10 o .798 -.208 -1.253 0 1 -.261
B-l = 0 1 -.218 .773 .8—1 _ .273 1 0 -.716

1 0 1 0 _ 1.253 0 0 .261

0 0 0 1 0 0 0 1
I I VB BII = (o, 0, .201, .312) BII = (.25, 0, 0, .365)

For r=20 53=B_I (18, 22, 32, 30.84) = (1.4, 4.83, 30.6, 30.84)
r=22 53=B-I (18, 22, 34, 30.84) = (3.4, 4.83, 30.6, 30.84)
r=24 5= (5.4, 4.83, 30.6, 30.84)

r=26 b= (7.4, 4.83, 30.6, 30.84)

r=28 b= (9.4, 4.83, 30.6, 30.84)

* Stage (2,2)
It is the last stage of this example.

0220“) = max [P22(.Y) III Q12(Y“Y)] 1" i 28

v<r

94

Sector 2 capacity corresponds to the fourth constraint, i.e.,
i=4.
1) line r-y=O Q12(r-y=0) = 22.4 with
J = {3,4, l, 2} and 5'= (6.8, 4.8, 20, 24)

The r.h.s.r.m. provides the following information

bj

y = 8 = m1n'[——:T-, 831 < 0] = 55§ = 6

A " ——6_ - .__.6_ :

822 I .9 x 3.8 I 3.42 I 7544

P22(y=0) = 0

P22(y=4) = 0 + .9 x 1 x 3.8 x .3 = 1.03

P22(y=8) = 0 + .9 x 1.7544 x 3.8 x .3 = 1.8 = ... = P22(y=28)

For lines r=y=2 to r—y=18 the results are obtained in a similar way
and are shown in Table 2.7.

2) line r—y=20 Q12(r-y=20) = 29.4 with

J = {5, 4, 1, 2} and 5': (1.4, 4.83, 30.6, 30.84)

Here, if the r.h.s.r.m., is applied a smaller P22(y) will be
obtained when y > 0, because sector 1 is already operating under
capacity (U12 = 1.4). The r.h.s.r.m. does not take into consideration
the fact that after building in sector 2, 822 projects where 822 > 822,
sector 1 might operate at full capacity, because of the interdepend-
ence between both of these sectors. It was pointed out 822 > 822,
i.e., the case where the marginal value added generated by investing
in sector 2, 822922 does not comprise the marginal value added gen—
erated by secoor l (which was operating at full capacity) resulting
from the investment in sector 2. So to get the whole marginal value

added due to an investment of 2922 = 8 in sector 2, the following LP

 

 

 

95

 

Am¢.~m.~n.m—..-.v ~.—.o.n a 2.5.5.33..— otémzu. oo.o~+~m. 3.3.5..— mmdNKT— v.m~+em.~ ¢.m~+m~.— 5.3.5 ow

 

.55.55.55.5...5.... 5...5.5 5 5...5 5.55+55.. 55.55+55. 55.55.. 55.55+55.. ,55.55+ 5.. 5.55+55.. 5.55.5 55
.55.55.55..5.55.5..5.. 5...5.5 5 55.55 5.55.5.. 55.554... 55.55+55. 55..5+.5.. 55.5555... 5.55+55.. 5.55.5 55
.55.55.55.55.5.5.... 5.....5 5 55.55 5.55+55.. 55.55+55. 55.55. . 55.55.55.. 55.554.5.. 5.55+5 55
.55.55.55.55...5..5. 5...5.5 5 55.55 5.55+ 5.. 55.55+... 55.55+55. 55..5+.5.. 55.55+.5.. 5.55.5 55
.55.55.55.5....55.55 5...5.5 5 55.55 5.55+55.. 55.55+55. 55.55+ . 55.55+.5.. 55.5545 5.
.55.55.55..5.5..5.5. 5...5.5 5 55.55 5.55+ 5.. 55.55+... 55.55.55. 55..5+55.. 55.55+5 5.
.55.55.55.55.5..5.5. 5...5.5 5 55.55 5.55455.. 55555555. 55.55+ . 55.5545 5.
.55..5.55.55.5.5..5. 5...5.5 5 55.55 5.55+_5.. 55.55+... 55.55+55. 55..5+5 5.
.55..5.55.55.5.55.5. 5...5.5 5 55.55 5.55+55.. 55.55.55. 55.55+5 5.
.55..5.55.55.5.55.5. 5...5.5 5 55.55 5.55. 5.. 55.55+.5. 55.55+5 5
.55..5.55.55.5.55.5. 5...5.5 5 55.55 . 5.55.55.. 55555+5 5
.55..5.55.55.5.55... 5...5.5 5 55.55 5.55+55.. 55.55+5 5
.55.55.5.5.5. 5...5.5 5 5.55 . 5.55.5 5
.55.55.5.5.5.55 5...5.5 5 5.55 . 5.55.5 5

5 5 5 ...555 55 55 55 5. 5. 5 5 5 ».

 

.5.5. 55555 55 55.5555 ..5 5.5..

 

96

must be so1ved according to the discussion in section 2.5.

Max Z = .201 x12 + .312 x

22
.798 x12 +—.208 x22 + z12 = 18
-.218 x12 + .773 x22 + 222 = 22
X12 + ”12 = 32
x22 + u22 = 37.68
X12 3 0’ Z1'2 30’ “i2 30

Its so1ution gives

Z* = 18.13

J = {3, 6, 1, 2} b = (.27, .19, 32, 37.49)

1 .269 —.739 O
B—1 _ 0 -1.294 -.282 1
0 0 1 0
0 1.294 .282 0

The margina1 va1ue added due to an investment of 2922 = 8 is
18.13 - 15.77 = 2.36.

The quantity of 15.77 corresponds to the va1ue added due to in-
vesting 6,in stage (1,2) when 20 were avai1ab1e in stage (1,2).

Tab1e 2.8 summarizes the resu1ting investment decisions.

The tota1 va1ue added in the 2 year p1an is 31.76 m u.

The optima1 so1ution requires that 14 m u be invested in each
year.

At the end of year 2 the capacities are:

C12 = 32

c22 = 37.68

The optima1 production schedu1e in year 1 is:
X11 = 25,81, x2] = 28.23, U11 = .19, ”21 = 2.61

97

Tab1e 2.8 Investment per Year and Sector Unit: m. u.

Investment Year 1 Year 2
Sector 1 6 6
Sector 2 8 8
Tota1 14 14
in year 2:

x12 = 32, x22 = 37.49, 212 = .27, u22 = .19

The optima1 va1ue added in year 1 is 13.63 and in year 2 =
31.76 - 13.63 = 18.13.

The app1ication of dynamic programming to the examp1e, presented
above, shows the tediousness of computations, carried out to get the
optima1 so1ution. A computer program cou1d be designed to perform
a11 of the steps invo1ved in the so1ution as indicated in section 2.5.
This is not the purpose of this research. A more important reason 1ed
the author not to design this computer program. It was the non-
app1icabi1ity to the A1gerian case. Data used to compute both of the
input coefficients (domestic and foreign) are not avai1ab1e in A1geria.
Even though imports are known by sector, there is no way of knowing
from which foreign sectors they come from. Moreover, import figures
contain both intermediate and fina1 demands.

That is why, the node1, presented in the next chapter, wi11 be
used for the A1gerian app1ication. It uses data as given in the
A1gerian input-output tab1e. The so1ution of the economic p1anners'

prob1em is obtained by mixed integer programming.

 

CHAPTER III

THE MIXED INTEGER PROGRAMMING SOLUTION TO THE DYNAMIC
LEONTIEFF MODEL FOR AN INVESTMENT ALLOCATION

In this chapter, a mixed integer programming so1ution wi11 be
presented. However, the previous Leontieff mode1 wi11 have to be
modified to make its so1ution by MIP (a shorter notation for mixed

integer progranming), possib1e.

3.1 The MIP Formu1ation

The MIP requires that a11 coefficients of the mode1 are known.

and the input coefficients a.. must

The va1ue added coefficients v. th

3t
be known.

The variab1es and parameters of the mode1, where the subscript

t denotes the re1evant time period are defined.

Let:
Xit = production of sector i in period t expressed in m u.
Bit = number of projects bui1t in sector i in period t. Bit
is integer.

fit = demand of fina1 goods i in period t expressed in m u.
eit = export of fina1 goods i in period t in m u.
zit = imports of fina1 goods i in period t in m u.

it = inventory of fina1 goods i in m u at the end of period t.
9, = cost of a project in sector i expressed in m u, g, is

98

 

99

constant.
Cit = capacity of sector i in period t in m u.
”it = s1ack capacity of sector i in period t in m u.
Vi = va1ue added per unit produced in sector i, V, is known.
aij = ratio of the va1ue of input i consumed by sector j to
the output of sector j. aij is known.
a = coefficient of conversion of investment into capacity in

sector i.
An investment of 9i in sector i wi11 increase capacity by aigi'

The mode1 has the fo110wing ba1ance equations:

= f. + s

aijxjt + zit 1t ), i=1 . . . m

it ‘ Si(t-1

The difference between this set of ba1ance equations and those
m

presented in the previous chapter is that Z d
i=1
J7”

ijtxjt is ignored in
the present case. The input-output tab1es pub1ished to date provide
data concerning on1y aij and not dij for a given year. This is the
main reason this mode1 wi11 be used in the A1gerian app1ication.

In the fo110wing, the MIP formu1ation of the prob1em wi11 be
given. The concept of equiva1ent investment w.

it
is constant.

is used again: the

subscript t is dropped because wit

In period zero, the capacity of sector i is known, i.e., Cio'

In period 1, an investment of Bi1gi wi11 increase capacity by

Bi1aiwi' So the fo110wing re1ationships ho1d:

100

Xi1

l
O
+
"(D

C11 ‘ io
Fina11y,

. + . = . + . . .
X11 u11 Cio B110‘1wi ’ or,

C .

10 Xi1 I ”1'1 ' Bi1o‘iwi (3'1)

Equa1ity (3.0) can be genera1ized to any t
t

c. = x. + u. — 2 B
r=1

iraiwi’ i=1, . . . m, t=1, ...T (3.2)
It is a1so assumed that n1 m u are avai1ab1e for investment in
period 1, n2 in period 2 and so on. Let N be the amount avai1ab1e

for a11 T periods. Investments of period 1 must satisfy

Inequa1ity (3.3) is genera1ized for any t as fo11ows:

t m

2 £8.g.<

p=1 i=1 1p1_ n t=1, . . . , T (3.4)

P

"Md"

p 1

It is a1so assumed the p1anners want to maximize the va1ue
added over a T year p1an.

The formu1ation of the p1anners' prob1em in MIP form is:

= f. + e. + s. - Si(t-1)’ i=1 ... m

 

101

t
xit + ”it ' DE] Bipaiwi = cio’ 1:] m
t m t
z 2 8- g. 5_ z n , t=1 T
p=1 i=1 1" 1 p=1 P
Xit 3_O, zit :_0, ”it 3_0, Sip-3 0 and integer.
T
Note a1so that Z n = N.
p=1 ‘0

The previous prob1em can be presented in a matrix form. This
is done in Tab1e 3.1, where capita1 1etters X, U. Z, B, V, F, E, AS,
CO, G stand for m-component vectors, (I-A) for an mxm matrix, I for
the mxm identity matrix, W for an mxm diagona1 matrix. The variab1es
are X, U, Z and B where B must be integer. The right hand sides are

t
F, E, AS, Co and 2 hp. The coefficients of the objective function

p-1

are V. AS is the net inventory in period t.

t
Mixted integer programming wi11 be used to find the optima1

so1ution for the investment a11ocation in the A1gerian mode1. Before

attempting to do that, the parametens of the mode1 must be estimated.

3.2 Estimation of the Parameters and Data

E1even sectors are considered in this app1ication. Sectors
which are not primary producers, name1y services, trade, and transpor-
tation, are exc1uded.

Each of these sectors is 1abe1ed from 01 to 11.

Agricu1ture 01

Food industries 02

Petro1eum O3

The MIP Formu1ation

Tab1e 3.1
obj. fct.

102

[IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII IIII
N
A
=5 ‘F
C
:IIIIIIIIIIIIIIIIIIIIIIIIIIIII!IIIIIIIIIIIIII a. ‘3
A
> :1:
IIIIIIIIIIIIIIIIIIIIIIIIHIIIIIIIIIII
IIIIIIIIIIII IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
A
>- :F
IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII
=’ IIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII

 

 

103

Uti1ities 04
Mining 05
Stone and c1ay products 06
Mechanica1, e1ectrica1 industries 07
Chemica1 products 08
Texti1e and 1eather 09
Misce11aneous manufacturing 10
Construction industries 11

In what fo11ows, methods are presented to estimate each param—
eter or constant of the mode1. Such estimations concern the vi's,

aij's, right hand sides of constraints, gi's, ai's and nt's.

3.2.1 Estimation of the vi's and aij's

The 1969 Input-Output tab1e of the A1gerian economy provides
the necessary data to estimate these parameters since the present
study concerns the first four year p1an which started January 1st,
1970.1

The estimation of the va1ue added coefficients Vi is obtained
by dividing the va1ue added of sector i by its production. The es-
timates are found in the bottom row of Tab1e 3.2.

The input coefficients aij were defined as the ratio of the
va1ue of goods i consumed by sector j to the output of sector j.
Their estimates are obtained by dividing the amount purchased by
sector j from sector i by sector j production. The aij's are a1so
in Tab1e 3.2.

1The Input-Output Tab1e is at the end of the study: Tab1e A1.
The data was obtained from : Secretariat d'Etat au P1an, Tab1eaux

de 1'Economie A1gerienne (A1ger: SNED, 1973, pp. 272—273).

 

104

0000. 5005. 5500.

5505. 0555. 5505. 5005. 50N5. 05—5. 005N. 0555.

 

5.

 

0000. 0500. 0000. 0000. 0000. 0000. 0000. 0N0o. 0050. 5500. 0500. 55 .pmcoo
00N0. 000N. 0500. 0550. 05—0. 00N0. 00N0. 0500. 0000. 0050. 5000. 0— .055:
0000. N500. 5000. 0500. N500. 0500. 0500. 00 5055pxwh
0550. 0000. 0N00. 5000. @050. 0500. 0050. 5500. 0000. N000. 00 m5wuwewzo
00N5. 50N0. ¢000. 00No. 0005. 0000. 500—. 0500. 0000. 0500. 5000. 50 .0050 .00:
0000. 0000. N000. 0000. 05¢N. NNoo. 5N50. 0N00. 0000. 00 0:000
Nooo. ¢5N0. 0000. 00 0:505:
5N00. 55—0. 0N00. 5000. 0000. 0550. 5500. 0005. 0000. N500. 0050. 00 wwwuw5wpo
0500. 0000. 5000. 5000. 0050. 5000. 0000. 0000. 5000. 5000. 0050. 00 500500505
0000. vN—o. 0000. 0005. 0050. N0 0005
0050. 5000. N000. N550. 50 0550<
55 05 00 00 5o 00 00 #0 00 N0 50 0 5
5 .5.

5..> 0:5. 5.

5 55 555555.55 5.5 5.555

105

3.2.2 The Estimation of the Right Hand Side
of Demand Constraints

The right hand side of demand constraints comprehends four
elements: demand of final goods fit’ exports of final goods eit’
initial and ending inventory of final goods Si(t-l)’ and Sit'

The 1969 Input-Output Table gives the demand, the exports and
the change in inventory of final goods. To get an estimate of the
demand and exports of final goods for each year of the four year
plan, a constant rate of increase for both of them is assumed.

The value of demand and exports as projected in the four year
plan is known for the ending year.1 The estimate of the growth rate
9 is obtained from the following formula
Value73 = Value69 (l + g)4
Table 3.3 contains the growth rates of demand for food products

and industrial goods.

Table 3.3 Rate of Increase of the Different Types of Consumption

 

Increase 1969-731 Annual Increase

Goods % %
Agriculture and Food Products 23.5 5.42
Industrial Goods 30 6.78
Overall Mean 24 5.53

1Source: Ministry of Information and Culture, The four year
plan (Algiers: SNED, l970, p. 70).

1Ministry of Information and Culture; The four year plan (Algiers:
SNED, l970).

106

Table 3.4 shows the increase rates of exports for four categor-
ies: l) agriculture and food products, 2) mining products, 3) Pet-

roleum products, and 4) industrial goods.

Table 3.4 Value of Exports in 1969 and 1973 (as Projected). Unit:
Millions of DA

 

Goods 19691 19731 Annua1%1ncrease
Agriculture and Food Products 860 950 2.6
Mining Products 160 250 11.8
Petroleum Products 3250 5130 12.1
Industrial Goods 180 450 25.74

1Source: Ministry of Information and Culture, The four year
plan (Algiers: SNED, 1970, p. 33).

Furthermore, it is assumed that the growth rate of the net
change in inventory is equal to the one used for the demand of final
goods. This assumption is necessary since there is no data available
to make a better estimate.

The right hand side of demand constraints is presented in Table
3.5. Table 3.5 requires some explanations. First, the following
growth rates are used in computing the demand and net change in inven-
tory of final goods (see Table 3.3).

-- 5.42% for agriculture and food sectors

-— 5.53% for stone and clay products and construction indus-

tries sectors

—- 6.78% for all other sectors

107

Table 3.5 The Right Hand Side of Demand Constraints. Unit:
Millions of DA

Sector 1969 1970 1971 1972 1973
01 Agriculture 2040 2145 2255 2371 2494
02 Food ind. 4791 5030 5280 5543 5821
O3 Petroleum 3642 4055 4516 5031' 5606
04 Utilities 202 216 230 246 263
05 Mining 175 194 215 239 265
06 Stone and Clay 118 124 131 139 146
07 Mech., Elec. 629 690 760 841 936
08 Chemicals 945 1014 ' 1089 1171 1260
O9 Texti1e-1eather 2100 2251 2415 2593 2787
10 Misc. mfc. 709 765 828 898 976
11 Const. ind. 1852 1954 2063 2177 2297

Source: Secretariat d'Etat au P1an, Tab1eaux de 1'Economie
Algerienne (Alger: SNED, 1973, p. 273).

108

Second, the exports are multiplied by the growth rates shown in
Table 3.4:

-- 2.6% for the agriculture and food sectors

—- 12.1% for the petroleum sector

-- 11.8% for the mining sector

—— 25.74% for mechanical and electrical, chemicals, textile-

1eather and miscellaneous manufacturing sectors.1

3.2.3 Estimation of the Right Hand Side of
Capacity Constraints

The estimation of initial capacity is based on the Gross Na—'
tional Product of 1969, and on the assumption that the economy oper-
ated at full capacity in 1969. There are no data available to pro-
vide a better estimate for initial capacity. The main reason for
doing so, is, that the data found in the Input—Output Tab1e show al-
most every sectors imported goods. Table 3.6 presents the initial

production capacity for the different sectors of the economy.

3.2.4 Estimation of projects costs 91's

The cost of a project is understood as the average investment
made in a given sector to acquire the necessary fixed assets to achieve
an average production.

The estimation of g1 is based on American data. (The main reason
for this choice is that Algeria has to import most of its capital
goods from western countries. So, American data can provide a fair

estimate of a project cost.

1The stone and clay products sector will not export anything
during the four year plan because a strong demand will arise during
this period. The exports were very low in 1969 for this sector
1 million of DA.

109

Table 3.6 The Right Hand Side of Capacity Constraints. Unit:
Millions of DA

Sectors Value
01 Agriculture 3312
02 Food ind. 3565
03 Petroleum 4679
04 Utilities 369
05 Mining 185
06 Stone and clay 310
07 Mech., e1ec. 1256
08 Chemicals 493
09 Texti1e-1eather 1443
10 Misc. mfc 498
11 Const. ind. 2043

Source: Secretariat d'Etat au P1an, Tab1eaux de 1'Economie
Algerienne (Alger: SNED, 1973, pp. 272-273).

110

The data used in this estimation is in Table A2 in the Appen—
dix. The data concerns the period 1969-1972 according to the differ-

ent sectors used in this nodel.

3.2.4.1 General Assumptions
Three assumptions are made:
*1st Assumption.
It is assumed that the life-time of a project is
-- ten years in agriculture, food products, textile-leather and
miscellaneous manufacturing. =
-- fifteen years in construction industries.
-— twenty years in mining and stone—clay products sectors.
-- thirty years in utilities, petroleum, mechanical-e1ectrica1
and chemical sectors.
The first assumption is justified because the lifetime of a
project increases with the degree of capital intensity.
*2nd Assumption.
The completion time of a project is also assumed to vary with
the degree of capital intensity. It is
-- one year in agriculture, food products, textile-leather,
miscellaneous manufacturing and construction industries
sectors.
-- two years in mining and stone-clay products sectors.
—- three years for all other sectors.
*3rd Assumption.
It is assumed that the fixed assets available in the correspond-

ing American sectors have been in use for half their lifetime. It is

lll

necessary to make this assumption to avoid a low estimate of the

value of fixed assets.

3.2.4.2 Method of Estimation
Data obtained for all sectors, with the exception of agriculture
and mining, measure the net book value of fixed assets.

To obtain a fair replacement value of fixed assets, two assump—
tions are made.

*lst Assumption.

It is made in conncection with the depreciation method. Among
the three methods generally used: straight line, sum of the year's
digits and the double declining balance, the sum of the year's digits

will be used.1 Table 3.7 contains the values of the sum.

Table 3.7 Sum of Years' Digits

3

§ t = n (n2+ 1

n Sum
5 15
7 28
10 55
15 120
20 210
30 465

1This assumption is ignored for fixed assets having a groos book
value figure, namely agriculture and mining.

112

*2nd Assumption.

To get a fair replacement value of fixed assets, inflation is
taken into consideration. It is necessary to make such an assumption
to avoid a low replacement value of fixed assets whose life-time is
quite high. For instance, a fixed asset with a thirty year life-time,
in use for fifteen years, will certainly have a high replacement value
compared to its acquisition cost, because of the inflation in the cap-
ita1 goods area.

The rate of inflation for capital goods was in the vicinity of
4 to 7% in the sixties.1 The rate of inflation chosen is 5%.

The method of estimation proceeds along the following steps.

-- First, the net book value (NBV), of fixed assets per estab-

lishment is found."

—- Second, the acquisition cost of the fixed assets is computed.
An example will clarify the method of obtaining the acquisition cost.
Let us suppose, the net book value is NBV, the average life—time of
the fixed'assets is ALT and they have been in use for T years:

T = 5%I-, since the fixed assets age is assumed equal to half their

life time. The sum of years digits method of depreciation gives:

ALT
NBV x z t
t=1
acquisition cost = T ; t=1, . . ., T, . . . , ALT
Z t

t=1

-— Third, to obtain the replacement value of the fixed assets,

1U. S. Department of Labor, Bureau of Labor Statistics, Chartbook

nprices, wages and productivit (Editors James McCall and John
Tschetter, Vol. 2, No. ,May 1976, p. 25).

113

the inflation rate is taken into account.
replacement value after T years = acquisition cost x (1 + .05)T
A11 replacement values are computed for 1971, i.e., the second
year of the plan and are used for each of the four years.

Cost of projects (based on American data), gi, are in Table 3.8.

Table 3.8 Cost of Projects: gi (Based on U.S. Data)

,2
seem” ($1,000?000's) (1 ,ooo?ooo's DA)
01 Agriculture .097 .452
02 Food 5.419 25.168
03 Petroleum 67.075 311.496
04 Utilities] 37.592 174.577
05 Mining 2.367 10.991
06 Stone-c1ay 12.999 60.367
07 Mech. Elec. 18.320 85.080
08 Chemicals 25.717 119.432
09 Textile-leather 1.898 8.815
10 Miscellaneous mfc 2.252 10.456
11 Const. industries .179 .830

1U.S. Federal Power Commission, Technical Advisory committee on
finance, National Survey: The financial outlook for the

ower industry: lhe report and recommendations of the Technical Ad-
Visory committee on fiannce, Part II (Washington: U.S. Government
Printing Office, 1974), p. 3.5l.

2$1 = 4.644 DA in United Nations, Statistical Office, Statist-

ical Yearbook (New York: UN, 1975).
3.2.4.3 Estimation of a Project Cost in Agriculture
The cost of an agricultural project is restricted to the value

of machinery and equipment in an average size farm.

114

The market value of equipment is $24.285/acre.1

The average size of a state owned farm in Algeria is 1154 hec-

2 3

tares corresponding to 2851 acres.

The acquisition cost of machinery and equipment in 1969 was
$24.285 x 2851 x (1.05)5 = $88,365.
In 1971 it was: $88,365 x (1.05)2 = $97,422.

Its corresponding value in DA (the Algerian currency unit) is in
column 2 of Table 3.8.
3.2.4.4 Estimation of the Cost of a Project

in the Mining Sector

The gross book value of fixed assets in the mining secotr was
16,432 millions of dollars in 1972.4 The number of establishments was
10,771. The average life-time of fixed assets in this sector is 20
years.

Acquisition cost/establishment, in 1971 = 18’???

 

x (i.05)9

= 2.367 millions of dollars.

1U.S. Bureau of the Census: Census of A riculture 1969 Volume II
General Reports, Chapter 3, Farm management, Farm operations (Washing-
ton D.C.: U.S. Government Printing Office, 1973), p. 153.

2Secretariat d'Etat au Plan, Tab1eaux de 1'Economie Algerienne
(Alger: SNED 1973, p. 121).

31 hectare = 2.471 acres.

4U.S. Bureau of the Census, Census of Mineral Industries 1972

Sub ect Industr and Area Statistics (WaShington, D.C.: U.S. Govern-
ment Printing Office, 1976), pp. 1-89—1-92.

115

3.2.4.5 Estimation of the Cost of a Project in the
Construction Industries Sector.

The 1972 census of construction industries gives the following
data:1
Number of establishments = 437,941
Net value of fixed assets = $12,054 millions
Rental payments for machinery and equipment = $1,972 millions
Now, the cost of a project in this sector can be computed.

Net value of fixed assets/establishment = £37J§25 = $.027 millions

 

To get the acquisition cost, the rental payments for machinery

and equipment are incorporated.

15
.027 x Z
acquisition cost of a project = ——-—7r£:l——-X (1-05)7 + 55%2%%T
Z t

t=1
= $.178 millions

3.2.4.6 Estimation of Projects Costs for
All Other Sectors

The data used to obtain an estimate of the replacement value of
the fixed assets are shown in Table A2 in the Appendix.

However, the non existence of the number of establishments in
1971 made necessary the use of the following assumption. The number
of production workers per establishment is assumed the same for 1971
and 1972. Since the number of production workers and the number of

establishments in 1972, are known, the number of production workers

1U.S. Bureau of Census, Census of Construction Industries, 1972,

Vol. 1, Industr Statistics and S eCial Reports (Washington, D.C.:
U.S. Government Printing Office, 1976), pp. 6—7.

116

per establishment can be obtained. The last figure divided into the
number of production workers for 1971 gives the number of establish—
ments in 1971 for a given sector. Now it is possible to get the net
book value per establishment in 1971. Using the sum of years' digits
method of depreciation1 and the inflation rate, the replacement value
of fixed assets per establishment is (the cost of a project) obtained.

This method is applied to the food sector to show how it is used.

From Table A2 one obtains:

Number of production workers/establishment = liég§j%%Q-= 49.23
Number of establishments in 1971 = li9%%i%%9-= $21,306 millions

Net book value of fixed assets = gf’gg3'4 = $1.158 millions

5
Acquisition cost = 141§§—5—%§—5-1149§1— = $5.419 millions

Table 3.8 indicates the cost of projects in other sectors.

3.2.5 Estimation of the Capacity Increase Due to Investment
An investment 9i will increase capacity of sector 1 by aigi
cone the project is completed. Recall, W1 is an equivalent investment

to gi, being completed the same year it is started

1 1
g' (_+€' a)
wi=—L—g—E——g%— fmmmeMa(ZA)

Multiplying both sides of the previous formula by ai’ it follows

1 1
-0591 (2* E ' m)
(x.W.— —-—-—————
11 8

1cf. Table 3.7.

117

The increase in capacity aiwi must be computed. It can be de-
termined if Gigi is known since 8 and h% are given.

From the data in Table A2 the value of shipments per establish-
ment in 1971 can be obtained (provided, the number of establishments
in 1971 is computed by the method illustrated in 3.2.4.6). However,
the value of shipments does not correspond to full production capaci—
ty; so the following assumption was made--the value of shipments cor-
responds only to a certain percentage of production capacity. Once
the value of shipment is divided by the corresponding percentage of
production capacity, the aigi's, looked for, are obtained. In fact,

there was no need to find the ai's. Table 3.9 contains the values of

aiwi o

The case of the mechanical-electrical sector will illustrate

the application.
The value of shipments per establishment is $5.492 millions.

Since this value corresponds to 80% of production capacity, a project

with fixed assets of $18.320 milliong will increase capacity by

5.432
.80

tained as follows

=$6.865nfi11ions. This figure is precisely a7g7. a7w7 is ob—

1 1 1 3

a7w7 —————————7;—————-—— = 30 = $6.636 millions

Table 3.9 shows the figures for each sector.

1cf. Table 3.8.

118

 

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119

3.2.6 Estimation of the Investment Expenditures

The money available for investment in the four year plan was
defined as N.

The value of N is computed by adding the public expenditures for
investments to the private ones.
The value of N is equal to 20,200 millions of DA.1

Since the data figures available did not specify the anount
to be spent each year, it was decided to divide N into four equal
parts, i.e., 5050 millions of DA per year. Table 3.10 shows the cor-

responding values of E n
t=1

k

t, 1,. -.,4.

Table 3.10 Available Investment in Algeria
Unit: Millions of DA

k
Year t nt
1970, k=1 5,050
1971, k=2 10,100
1972, k=3 15,150
1973, k=4 20,200

Source: Ministry of Information and

Culture, The four year plan (Algiers: SNED,
1970).

1Ministry of Information and Culture, The four year lan (Algiers:
SNED, 1970) Public InVestments: l) allindustries: 12,100 millions of
DA (p. 32); 2) Agriculture: 4100 millions of DA (p. 52); 3) Housing:

1510 millions of DA (p. 84). Private investments only in industrial
sectors: 2190 millions of DA (p. 31).

120

Now, the planners' problem can be solved since all coefficients
and right hand side constraints are known. This is the subject of the

next section.

3.3 Solution by MIP

The planners' problem, which is to maximize the value added over
the four year plan to get the optimal sequence of investments, is
solved by using the mixed integer programming option available in the
APEX package.1

Before going further, a change in the initial tableau, correspond-
ing to the one depicted in Table 3.1 was made. The columns Ut and Zt
were deleted from the tableau; that make the demand and capacity con—

straints inequalities of the 5_type. This avoided punching unneces-

sary cards for these columns.

3.3.1 Introductory Comments

The problem formulation required:

-- 8 m variab1es, i.e., 2 m per year, m for Bit and m for Xit'
Since m = ll,there were 88 variables.

-- 4 (2m + 1) constraints, there are m constraints for demand
and m for capacity, 1 for available investment. There was a total of
92 constraints.

All constraints are of the less than or equal type except those
representing the demand constraints of utilities (sector 04). These
are strict equalities. LeSs than or equal constraints for this sec-
tor demand would mean the possibility of having imports of utilities.

1The APEX package was run on the CDC 6500.

121

This is not very realistic even though it is conceivable.

The listing of the input cards is shown in Table A.3 in the
Appendix. Before interpreting the computer output, some remarks must
be made on the names of variables; The integer variables Bit are rep-
resented by NBRth, number of projects in sector i in year t. The
PRODth's correspond to production levels of sector i in period t,
xit' The demand constraints are denoted by the slack variable asso-
ciated with each as follows:

—- IMPORTth stands for the slack variable zit
—— UNUSEDth stands for the unused capacity of sector 1 in

period t, uit

—- MONEYLEFTt stands for the funds left over after investment

in period t.

The APEX package requires that bounds be set on integer varia-

bles. The bounds were set equal to 16383, maximal value allowed in

this package.1

3.3.2 Analysis of the MIP Results

The MIP results are presented in Table A.4 in the Appendix. The
solution was obtained after 101 seconds of Central Processing Unit
time. The value of the objective function equaled 61369.063 millions
of DA.

The following facts can be observed:

—- First, the funds, available each year, were almost all ex-

hausted, the value of the slack variables associated with the

1In fact, the problem as formulated in section 3.1 is bounded
by the investment constraints.

122

investment constraints are 0.24, 0.109, 4.511 and 1.291 millions of
DA at the end of year 1, 2, 3 and 4 respectively.1
-— Second, the previous remark is complemented by the fact that
Host of the sectors operated at full capacity in the fourth year with
the exception of the following three: agriculture, food industries
and construction industries. These sectors show positive slack for
the capacity constraints (rows 82, 83 and 94 in Table A.4 under CON-
STRAINTS).
-- Third, three sectors show a negative row activity2 in each
year of the four year plan: stone-c1ay products, mechanical-electrical

and chemicals (sectors 06, O7 and 08). To understand this situation

let us refer to the demand constraints.

m
Xit ' J2] aijxjt + zit = fit + eit + Sit ' S1(t-1)
Omitting zit will transform the above equality into a less than or
m
equal constraint. A negative row activity occurs if x.t - Z a .x.
i j=l ij jt

is negative. This implies sector i production is not even able to
satisfy the intermediate demand coming from other sectors, the nega—
tive figure is equal to this deficit. For instance, the right hand
side of the demand constraint for sector 08 products is 1014, the def-
icit is equal to 166.736 (row 9), the sum of both of these figures,
1014 + l66.736= 1180.736, is the value of imports of chemicals for

1
and 93.

2cf. Tab1e A.4 under CONSTRAINTS, column: ROW ACTIVITY, rows 7
to 9, 30 to 32, 53 to 55 and 76 to 78.

cf. Tab1e A.4 under the heading CONSTRAINTS: rows 24, 47, 70

123

the first year, this figure is the value of the slack variable
(row 9).

These three facts support the conclusion that if more funds had
been available the MIP would have used at least part of them. More
funds must have been allocated to the four year plan to achieve the
goals represented by the demand constraints.

The optimal sequence of investments to achieve the maximum sum
of value added over the planning period is what the planners look
for.

This is in Table A.4 under the heading COLUMNS. the values of
the integer variables NBRth are found in column COL ACTIVITY. Table
3.11 presents the value of investments in each sector per year. The
investments are obtained by multiplying the value of NBRth (Bit) by
the corresponding project, 91.

The following facts can be observed in Table 3.11:

-- The petroleum sector invested 27.76% of the funds available
for investment expenditures in the four year plan, no other sector
achieved this percentage. This sector has three characteristics:
first, it has a high value added coeffiéient, .7145; second, it is a
sector that is capital intensive; and, third, the petroleum sector is
highly related to primary activities such as oil and gas extraction.
By the way, the activities usually found in the oil industry were
not very developed in 1969, which is the date for which the value
added coefficient was computed. These activities have a much lower
value added coefficient.

-— The next three highest investments were made in the textile-

1eather, food and miscellaneous manufacturing sectors. They shared

124

Table 3.11 Structure Of Investments in the Four Year Plan. Unit:
Millions of DA

1 2 3 4
1970 1971 1972 1973

Sector Tota1 %

01 Agriculture 1499.284 261.256 221.48 201.592 2183.612 10.81

02 Food 2416.128 377.52 327.184 327.184 3448.016 17.07
03 Petroleum 2180.472 3426.456 5606.928 27.76
04 Utilities 523.731 349.154 349.154 349.154 1571.193 7.78
05 Mining 76.937 43.964 43.964 164.865 0.82
06 Stone-clay 60.367 60.367 0.3

07 Mech. Elect.

08 Chemicals

O9 Texti1e-1eather 511.27 2953.025 379.045 387.86 4231.2 20.95

10 Misc.mfc. 1056.056 1432.472 261.4 2749.928 13.61

11 Const. ind. 22.41 53.12 51.46 55.61 182.6 0.9
Total 5049.76 5050.151 5045.598 5053.22 20198.709 100.0

125

together 20.95 + 17.07 + 13.61 = 51.63% of the total investment.
These activities are in the light industry area, where capital in-
vestments are low compared to other sectors such as chemicals, mech-
anical-electrical, and so forth. The sectors in question are labor
intensive and are heavily dependent on agriculture which is a pri-
mary sector "par excellence.“

-- Agricu1ture had a high level of investment in the first
year, 1500 millions of DA, i.e., 30% of total expenditures of that
year. In the following year there was a sharp drop in spending.
From 1971 to 1973 the investment levels would have seemed to stabil-
ize between 260 and 200 millions of DA. The reason has to do with
production levels. For this the reader should refer to Table A.4
under the heading CONSTRAINTS. Even though there was a huge invest-
ment in the first year, 1500 millions of DA, the slack capacity in
agriculture was only 0.857 millions of DA (row 13). In later years,
investments being much lower (between 260 and 200 millions) the cap-
acity constraints showed a lower slack (rows 36, 59 and 82). One
can conclude that the high investment level in 1970 was due to the
agricultural output capacity having been far behind the demand im-
posed on agriculture. The investments made in 1971 up to 1973,
which were low, were carried out for the sole purpose of keeping up
with the annual increase in demand for the agricultural products.
This conclusion is also corroborated by the fact that demand CONE
straints were bound in 1970, 1972 and 1973 (rows 2, 48 and 71) and

had low imports in 1971 (row 25) of about 4 millions of DA.
-- Next, the sectors generally capital intensive, such as

chemicals and mechanical-e1ectrica1, had a zero investment level by

126

the end of the four year plan, even though they showed relatively
high imports, 1686 and 1184 millions of DA in 1973, respectively,
(rows 78 and 77). Furthermore, in 1970, both sectors had a total of
1180.736 + 717.177 = 1897.913 millions of DA of imports (row 9 and
row 8), sensibly equal to the one obtained for textile—leather and
miscellaneous manufacturing sectors for the same year, 1273.243 +
630.925 = 1904.168 millions of DA (rows 10 and 11); however, in the
following year, sectors 09 and 10 were the best candidates for invest—
ment. 2953.025 + 1056.056 = 4009.081 millions of DA (Table 3.11,
column number 2) alnnst 80% of year 2 spending, even though the value
added coefficients are not that different. The reason for this re-
sides in the high cost of projects in sectors 07 and 08.

—- The food sector followed the same pattern as the agriculture
sector, huge investment in the first year (2416) and stabilization in
the folldwing years, 377 in 1971 and 327 in 1972 and 1973.

-- The investments in the utilities sector were quite steady,
523 millions of DA in 1970 against 349 in the following years. Their
presence was due to the equality of demand constraints for this sec—
tor1 which forced MIP to have strictly positive values for the cor-
responding integer variables.

In summary, the MIP solution favors investments in:

1) sectors of primary activities: petroleum, agriculture and
mining.

2) sectors with low capital investments: food, textile-leather

, and miscellaneous manufacturing.

1The first time MIP was run with less than or equal constraints
for the demand of utilities, no investment took place in any of the
years concerned and we ended up with imports of utilities.

127

These facts are in complete concordance with a certain theory
of economic development that was formulated in the early 1950's and
that is still highly criticized. According to it underdeveloped
countries starting economic plans should invest in sectors with low
capital investments. The reasons behind this theory are:

l) underdeveloped countries are characterized by a high level
of unemployment.

2) a low level of savings that can be invested in different
industrial projects.

So it seems logical that these countries undertake projects
with low capital ratios and in sectors being more labor intensive
than others.1 This theory is confirmed by the investments obtained
in the agriculture, food, textile-leather, miscellaneous manufactur-
ing and construction industries sectors. But, how can the huge in-
vestments in the petroleum sectors of the last two years be explained?
However, Professor Lewis adds,

Capital works with natural resource as well as labor, and a

poor country's natural resource advantage may be such that

it pays to develop some rich but capital intensive natural

resources.2

This explains why large investments are present in the petro-

leum sector. The results obtained by MIP in the four year plan con-

firm this theory. The only way the validity of such a theory, at

1Some of the proponents of this theory are:

Rugmar Nurske: Problems of ca ital formation in underdevelo ed
countries (New York: Oxfora University Press, I953).

Gunnar Myrdal, Economic theor and underdevelo ed re ions (Lon-
don: G. Duckworth, 1957).

2

William Arthur Lewis, Develo ment lannin : The essentials of
economic policy (London: George Allen and Murw1n, Ltd., 1966, p. 55).

128

least in the present case can be tested, is on the time horizon, be-
cause a four year plan may not be long enough to do so.

The opponents of the above theory are the advocates of the so-
called'hnbalanced growth" theory of economic development.1 This theory
claims that economic growth and development can be achieved by invest-
ing in some'Strategic" industries characterized by:

1) a high capital ratio

2) inducing the development of downstream and upstream indus—
tries.

Examples of "strategic“ industries are the steel and the petro-
leum industries. Almost any underdeveléped country wanted to have its
own steel industry, hoping that it would bring economic growth and de-

velopment by some magic formula. As Professor Kindleberger pointed

out,2

In dynamic terms, Bruton3 has suggested that the industries

which embody external economies are frequently capital inten-
sive. These capital intensive investments must be undertaken
before one can take advantages of opportunities for investment

in labor intensive industries.... It is believed that capital
intensive industries are more effective producers of savings

than labor intensive and hence likely to speed development faster
despite a possibly lower static level of output.

To test the validity of these conflicting theories, it was decided to
increase the time span of the plan from four years to eight years.

This is the subject of the next section.

1Albert 0. Hirshman, The strate of economic develo ment (New
Haven: Yale University Press, 1961).

2Charles P. Kindleberger, Economic development (New York: The
MacGraw Hill Bdok Company, 1958).

3H. J. Brutton, "Growth models and underdeveloped countries,"

Journal of Political Economy, August 1955.

 

129

3.3.3 The MIP Solution to the Eight Year Plan
The mathematical formulation of the eight year plan is identical
to that of the four year plan. There are 16 m variables, 8 m for the

integer (Bit) and 8 m for x. There are 8 (2m + 1) constraints; m

1t‘
for demand constraints, m for capacity constraints and 1 for invest-
ment constraints; that is 2m + 1 constraints per year.

All assumptions, stated in the four year plan case, are used
again. The same input-output matrix is used. The rates of increase of

final demand and of exports are used to arrive at the right hand side

values of demand constraints. This is presented in Table 3.12.

Table 3.12 The Right Hand Side of Demand Constraints for 1974—1977

Unit: Millions of DA

Sectors 1974 1975 1976 1977
01 Agriculture 2622 2758 2901 3051
02 Food 6113 6420 6743 7083
O3 Petroleum 6248 6965 7767 8663
04 Utilities 280 299 319 341
05 Mining 294 326 362 401
06 Stone-c1ay 154 163 172 181
O7 Mech.Elect. 1046 1176 1330 1513
08 Chemicals 1357 1465 1584 1715
O9 Textile-leather 2998 3229 3484 3764
10 Misc. mfc. 1064 1163 1276 1406

11 Const. ind. 2424 2558 2699 2849

130

The values of the right hand side of investments constraints are
in Table 3.13. The first part of Table 3.13 is the same as the one
in Table 3.10. The second part, dealing with year 5 up to year 8 of
the plan, requires an explanation.
Table 3.13 Available Investment in the

Eight Year Plan. Unit:
Millions of DA

k

Year til nt
1970, k=1 5050
1971, k=2 10100
1972, k=3 15150
1973, k=4 20200
1974, k=5 33010
1975, k=6 45820
1976, k=7 58630
1977, k=8 71440

Sources: 1970-1973 Ministry of Information
and Culture, The four year plan (Algiers: SNED,
1970); 1974-1977, Secretariat d'Etat au Plan,

Tab1eaux de 1'Economie Al erienne (Alger: SNED,
pp. 280-281).

The second four year plan covering the 1974-1977 period made 64,688
millions of DA available to invest in directly productive sectors.1

This figure is stated in 1974 DA. To get the corresponding value in

1Secretariat d'Etat au P1an, Tab1eaux de 1'Economie Algerienne
(Alger: SNED, 1975, pp. 280-281).

131

1970 DA a discount rate of 6% was used to take into account the de-
crease of purchasing power in 1974 due to inflation. Its rate in

the capital goods area was between around 5 to 7% in the early sev-
enties.1 The corresponding 1970 value is 51,239 millions of DA. Here
again, an equal amount available each year, 12810 millions of DA, is
assumed.

The listing of the input cards is shown in Table A.5. The re—
sults of MIP are given in Table A.6. Both of these tables are in the
Appendix.

Before interpreting the results, it must be pointed out that
the funds available for investment in the 1974-1977 p1an-—51239 mil-
lions of DA-—are more than two and a half (2.5) times the funds for
1970-1973 p1an (20200 millions of DA).

The structure of investments is shown in Tables 3.14 and 3.15.
If the part of the eight year plan covering the first four years is
compared with the investments made in the four year plan, it will be
discovered that they are very close. The percentages of total in-
vestments as shown in the first column of Table 3.15 are almost
identical to the ones of Table 3.11. The only differences are the
non existence of investments in mining and stone-c1ay products sectors
and the presence of investments in mechanical—electrical industries.

Moreover, the distribution of investments in the given years
remained the same: huge investments in agriculture and food indus-

tries in the first year, to fill the big gap between the production

1U.S. Department of Labor, Bureau of Labor Statistics, Chartbook

on prices, wages and productivity (Editors James McCall and John
Tschetter, Vol. 2, No. 11, May 1976, p. 25).

132

 

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133

capacity of these sectors and the demand, smaller amounts invested in
later years for the sole purpose of catching up with the annual in—
creases in the demand for their products, steady investments in the
utilities sector, big investments in the third and fourth year in the
petroleum industry.

Logically, such a similarity of investments distribution, in
both plans for the first four years, could be expected because the
same demands and investments constraints are used. 0n the contrary,
it has been thought that lengthening the planning period from four
to eight years nay have produced completely different results. Such
results could have been the implementation of projects in "strategic"
sectors such as mechanical-electrical, chemicals, stone-clay products
and petroleum in early years of the plan. If this were true, the
proponents of the “unbalanced growth" theory of economic development
would have been completely right in arguing that growth could be
achieved by investing in industries having a huge potential of induc-
ing the creation of upstream and downstream industries.

From the previous remarks one can draw the following important
conclusion--lengthening the time horizon of the plan fron four to
eight years does not have any nnjor effect on the structure of invest-
ments during the time period connnn to both plans.

Now let us analyze the results of the second part of the eight
year plan, that is the l974—l977 period. All the funds made available
between l970 and 1977 were exhausted, 7l439.9l out of 71440 millions
of DA.

The tenth column of Table 3.14 shows the total investments made

between l974 and l977. These figures are computed in percentages in

134

column 2 of Table 3.15. The comparison of the first two columns of
Table 3.l5 shows a pattern of investment totally different between
the l970-l973 and the 1974—l977 periods. With the exception of two
sectors, petroleum and utilities had steady percentages, all others
showed a big decrease or increase. These changes can be grouped in
two categories, labor intensive and capital intensive.

First, for labor intensive sectors such as agriculture, there
is a decrease of their shares in total investment from 10.88% in the
l970-l973 period to l.95% in the subsequent period. The same situa-
tion occurs for fOOd industries, textile—leather, miscellaneous man—
ufacturing and construction industries sectors.

Second, for capital intensive sectors such as chemicals there
is an increase of their share in funds spent from 0% to 22.84%. The
same thing can be said about mining, stone-clay products and mech-
anical-electrical sectors. I

In sumary, investments

— in labor intensive activities decreased from
l0.88 + l6.95 + 2l.04 + l2.32 + .92 = 62.ll% to l.95 + 3.39 + 4 +
4.67+ 0.5 = l4.5l%

— in capital intensive sectors increased from
29.3 + 6.92 + l.68 = 37.9% to 29.79 + 5.11 + l.l8 + 4.59 + 2l.92 +
22.84 = 85.43%

Here again the same reason, used to explain the decline of
investments between l970 and l973 in labor intensive activities, is
given--the projects built after the first year (agriculture and food
industries), the second year (textile-leather) and the fourth year

(miscellaneous manufacturing) served the sole purpose of catching up

135

Table 3.15 Structure of Investments in the Eight Year Plan.

Unit: %
Sum of Years Sum of Years Sum of Years

Sector 1 to 4 5 to 8 1 to 8

(1970-73) (1974-77) (1970-77)
01 Agriculture 10.88 1.95 4.47
02 Food 16.95 3.39 7.22
03 Petroleum 29.3 29.79 29.65
04 Utilities 6.92 5.11 5.62
05 Mining - 1.18 0.85
06 Stone-c1ay - 4.59 3.3
07 Mech. Elect. 1.68 21.92 16.2
08 Chemicals — 22.84 16.38
09 Textile-1eather 21.04 4.0 8.82
10 Misc.mfc. 12.32 4.67 6.84
11 Const. ind. 0.92 0.5 0.66

Tota1 100.0 100.0 100.0

136

With the annual increases in demand for their products. Besides this
reason, the cost of projects in these sectors was rather low compared
to the othersgthese projects created more capacity for each monetary
unit invested. This made these projects more attractive than capital
intensive projects with comparable value added coefficients, when
funds available for investment expenditure were scarce (20200 mil-
lions of DA were available for the 1970—1973 period against 51240
for the subsequent four year period).

The implementation of projects in capital intensive sectors,
delayed by MIP until the fifth or the sixth year, can be explained by
three elements:

First,the low value added coefficients of these sectors made
them unattractive compared to similar projects with low value added
coefficients but with smaller cost and greater capacity created per
monetary unit invested. This explains their delay.

Second, the funds available for investments in the entire per-
iod, extending from 1974 to 1977, were huge compared to the invest-
ments opportunities offered by labor intensive industries, since they
only required enough investment to satisfy the annual increase of

demand for their products.

Third, the situation of non—investment lasting until 1973
created a backlog of unsatisfied demand,for examp1e, rows 77 and 78
(under CONSTRAINTS of Table A.6) shows imports of 1091.86 millions
of DA by the mechanical-electrical sector and 1664.61 millions of DA

by the chemicals sector.

The results of the eight year plan support the theory of econ-

omic development, that is, underdeveloped countries should first

137

implement projects that are labor intensive even though they should
also take advantage of their natural resources requiring more capital
investment. The sole development of capital intensive industries is
not recommended because the economies of Underdeveloped countries are
so disintegrated that investing in these "strategic" activities does
not improve things, in fact, they create idle capital that could have
been employed productively somewhere else. As shown in the eight year
plan, investments in capital intensive industries were only undertaken
when the funds available for investment exceeded considerably the in-
vestments required to take advantage of the opportunities existing in

labor intensive activities.

CHAPTER IV

CONCLUSIONS AND SUGGESTIONS

The purpose of this research was to investigate the problem of
allocating investments in a plan of economic development. The prob-
lem was formulated as a mathematical program where the objective func—
tion was to maximize the value added over the planning period subject
to constraints of demand and capacity. Leontief input-output coef—
ficients were included in demand constraints. The solution of this
mathematical program provided the optimal sequence in which invest-
ments must be carried out to get the maximum value added. The optimal
sequence of investments was defined as the answer to the following
three questions:

- in which sectors should investments be made?

- how much should be invested?

— when should investments be made?

Their answers are important in economic planning because:

- First, economic planners want to know in which sectors invest-
ments must be decided so as to differentiate activities by their con—
tribution to economic development. This contribution is perceived
through increases in value added by the whole economy.

- Second, economic planners do not want to invest more than what

is needed to achieve the goals of the plan because investment is in

138

 

139

short supply (in underdeveloped countries). Furthermore, capacity
and demand constraints of the model set limits to the amount invested.

- Third, an investment made in a given sector this year does
not contribute to economic development in the same way as if made
next year or after. For example, there are more benefits in invest-
ing this year in the mining sector to support the production of
steel next year than in delaying the investment to next year.

To answer these questions two approaches to solving the sane
mathematical program were given:

- the dynamic progranming approach, and

- the mixed integer programming approach.

The first one was presented in Chapter 2 and was applied to a
numerical example. The dynamic programming approach involved two
types of input—output coefficients (domestic and foreign coefficients)
to reflect the importance of foreign trade in underdeveloped countries.
This approach was not used in the application to the Algerian plan
because first, data is lacking to compute both types of input-output
coefficients and second, a computer program must be written to deal
with all details of the solution procedure.

The second approach was presented in Chapter 3 and applied to
the Algerian plan. The MIP solution is different from the dynamic
programming approach in one crucial point: the constancy of value
added and input-output coefficients. MIP cannot be applied if these
coefficients are not constant. In fact, the dynamic programming
solution is more preferable than the MIP one because the assumption
of conatancy is relaxed. MIP answered the triple question of where,

how much,and when investment must be carried out?

140

This approach permitted to test partially the theory of econ-
omic development that reconnmnds to underdeveloped countries to under-
take projects in "strategic" sectors.1 According to this theory, in-
vestments must be made in upstream industries which will induce the
implementation and development of downstream industries. These
“strategic" activities have a high capital ratio, meaning that a sub-
stantial investment outlay must be made to carry out any project. Ex—
amples of these are steel, chemical, petroleum activities and so forth.
Characteristics of underdeVeloped countries such as high level of un-
employment, investments in short supply, high rural population, and
shortage of food contradict this theory. This research showed that
this theory claims were not verified by the MIP results. However, the
MIP solution indicated that investments must be carried out in sectors
such as agriculture, food, etc., in other words in activities with
low capital ratios. The number of jobs created per monetary unit
invested is much greater in activities having low capita1 ratios than
in those with high ones. Investing in these sectors is a way of re»
ducing unemployment. This makes agricu1ture the "strategic” sector
"par excellence,“ around which other activities should develop for
three reasons: first, agriculture is a big user of manpower, second,
underdeveloped countries have a high rural population (between 50 and
80%), and third, they have a problem of food shortage.

The MIP solution also showed that investments were carried out
in labor intensive industries in the early years of the plan to meet

the demand requirements imposed on them. It was only later that high

1

Albert 0. Hirshman, The strate of economic develo ment (New
Haven: Yale University Press, 1961).

141

capital ratio projects were implemented. It was also shown (in sec—
tions 3.3.2 and 3.3.3) that high capital ratio activities could be
developed if there were more funds available for investment than needed
to implement low capital ratio projects (to satisfy final demand). The
funds available for investment in the 1974-1977 period were two and
a half times greater than in the 1970-1973 period. Since labor inten-
sive projects were carried out for the sole purpose of catching up with
the annual increases of final demands, more funds were made available
for possible investments in capital intensive industries. This ex—
plained their implementation after 1974. Labor intensive activities
did not need that many funds to satisfy the increase of demand require-
ments from one year to the other. So the next choice for spending
the funds left over was in sectors with high capital ratios and low
value added coefficients.

The conclusions reached previously were based on the results
obtained with MIP. As it is often said, results are no better than
the information upon which they are based. Problems related to this
kind of situation are looked at. These include:

- errors in data

-sector aggregation.

1) Errors in Data

Errors in data are a very severe limitation to results. They
have different origins.

First,they may come from collecting the data necessary to make
input-output tables. Collection of data for such a purpose is not
always exhaustive for several reasons:

a) Data were not available although they existed. The most

142

important state owned firms were created between 1966 and 1969.1

During that period and the coming years, their main concern was to
get things going such as implementing projects. Moreover, there was
a lack of skilled people who could organize data to get meaningful
information on these firms.

b) Shortage of means and skilled people necessary to collect
data. Both of these affect quality of data.

c) Underevaluation and nonexistence of data are characteristics
of the Algerian private firms.2

Second, errors in data can come from the methods of estimating
final demands. This research took data as given.

The following question can be raised, did these figures repre—
sent estimates obtained with forecasting methods, or targets aimed
at? Both of these may not give the same values of final demands be-
cause targets are of political nature.

Third, cost of projects as computed could be questioned. The
cost of a project was defined as the average investment made to acquire
the necessary fixed assets to achieve an average output level. In
this cost,two elements were missing: first, the cost of building the
project was not taken into account; second, the cost of shipping
equipment3 were ignored. There was no way of determing both of these
costs. Their sum could be quite substantial.

11969 is the year the input—output table refers to.
2When the author worked in a research agency he found out that

interviewees underevaluated or concealed their figures. They feared
interviews because they thought they were for tax purposes.

3Most of the equipment were imported.

 

143

2) Sector aggregation

Eleven sectors were considered in the model. The 1969 input-
output tab1e made this choice compulsory. If the sector aggregation
had been performed at a much lower level, i.e., eleven sectors had
been broken down into a much larger number, the MIP results would have
been more meaningful. For instance, the American input—output tables
are available in two formats: 70 and 450 sectors. This is an import—
ant factor in results analysis, a greater number of sectors will in-
crease the level of homogeneity in each sector. For example, the
American sector corresponding to the mechanical-electrical sector
comprises 25 groups of industries and each of these contains 6 or 7
categories. When a sector like this is broken down into several in-
dustries, the level of homogeneity in each of these will increase
tremendously, because they are very different. Two project costs
can be far apart in two different industries belonging to the mech-
anical-electrical sector.

Other sectors such as agriculture, food and petroleum, are more
or less honngeneous and would not be affected very much by a higher
level of disaggregation. But the mechanical-electrical sector would,
and, this may make a big difference in terms of investments in this
sector; much lower investments would be shown by MIP. Besides having
high capital ratios, the mechanical-electrical industries, to be via-
ble in economic terms, must turn out a high level of output exceeding
demand underdeveloped countries have, twice or more than that. That

is why an increasing number of economists are urging underdeveloped

144

countries to share investments to speed up integration.1

The demand of many industrial goods is so small in underdeveloped
countries, that it is not worth producing them. But if several count-
ries decide to share investments (because the project costs are too
high), their total demand will become so meaningful that it will be
worth implementing such projects. In this way, production capacity
will not remain idle and funds will not be tied up in equipment and
machinery. For instance, to be viable an automobile plant must turn
out an output of at least one hundred thousand cars per year. For an
average population size of an underdeveloped country (twenty million
inhabitants) the demand for cars could be one third of this amount
(this is an optimistic estimate). The amount that could be sold in
foreign markets is very small because of a very strong international
competition.

Although there were problems raised on the results provided by
the MIP solution the following fact should not be overlooked; the
previous analysis gave indications to where investments must be made,
and in what sequence they have to be carried out. Before deciding
where and when investments should be made, economic planners can get
a guideline if they use this model. It might provide them with fur-
ther insights.

In this research, maximization of value added was assumed to
be the economic planners' objective. In section 1.5, three reasons

were mentioned for such a choice: first, value added is a quantitative

criterion; second, it is also the origin of all types of incomes; and

1Such a study was made on the Arab scale. Abdelhamid Brahimi,
Dimensions et perspectives du monde arabe (Paris: Economica, 1977).

 

 

 

 

145

third, it contributes to lessen unemployment. Planners may want to
achieve other objectives.

Improving quality of education or medical care could be objec-
tives planners would like to pursue. It is not easy to assign a fair
quantitative criterion for them.

Being less dependent on foreign countries could be another objec—
tive. For instance, if a country decides to manufacture commodities
previously imported, it may end up importing equipment and machinery
for producing them. So, one form of dependence is replaced by an-
other. Minimizing imports of commodities could express this goal.

In the same line of reasoning, economic planners may want the
trade deficit not to exceed a predetermined level. Adding a new con—
straint (the difference imports-exports should not exceed this trade
deficit) to the model accomodates this requirement.

Political leadership might require that some projects be carried
out in the plan. For example, if a country had natural resources
(like oil,iron, and so forth), downstream activities related to these
could be developed. If the political leadership required their im-
plementation, the economic planners would need to add constraints to
the model. Before doing so, the planners could have used marginal
analysis to find out the consequences of these requirements if they
had already had an optimal solution.

They may have more than one objective to optimize. If this
is the case, MIP cannot handle this situation; and they must turn to
another technique. Goal programming is such a technique that ranks
goals within the objective function optimized. Maximizing value

added, employment, and exports might be an objective function of a

146

model solved by goal programming.

It will be interesting to formulate a Leontief nodel in a goal
progranming format. This could be a subject for further research:
applying goal programming in economic planning is a possibility that

remains to be explored.

APPENDIX

 

 

 

 

147

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148

Table A.2 Data Used to Compute Projects Costs1

 

 

Number of Net Book
. 31251225. 333325?” 3133; axzitsf...
Sector Year ments (1000's). ($1,000,000's) ($1,000.000‘s)_

02 Food 1972 22423 1103.9

1971 1048.9 103348.2 24673.4
03 Petroleum 1972 2017 98

1971 _97 26935.1 16619.2
06 Stone-clay 1972 4616 319.4

1971 305.1 11345.7 9215
07 Mech. Elec. .1972 43138 4114.7

1971 3949.8 227418.8 94171.3
08 Chemicals 1972 8841 413.5 '

1971 404 39937.5 27575.9
09 Textile-leather 1972 31355 2024.9

1971 1958.2 47693.2 12298.5
10 Misc.mfc 1972 76226 1682.2

1971 1612.2 67298.3 35149.2

Source: U.S. Bureau of the Census, Census of Manufacturers, 1972 Industries
series, volumes 1, 2.3 (Washington D.C.: U.S. Government Printing Office, 1975).

1The above data is the aggregate of the data figuring in the 1972 census
of manufacturers. The correspondence between these sectors and the Standard
Industrial Classification (SIC) used by the 1972 census of manufacture is:

Sector 02 ‘ SIC 20 + 21

Sector 03 - SIC 29

Sector 06 : SIC 32

Sector 07 = SIC 33 + 34 + 35 + 36 + 37
Sector 08 a SIC 28
Sector 09 = SIC 22 + 23 + 31

Sector 10 SIC 24 + 25 + 26 + 27

 

 

149

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179

 

400.000 I Own: Dan 00.. 036—30
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181

 

. cooca.oan- u‘u- :Nuon .~V.« o~30£9o xu<4n u... m>sabrouxu ee—
. 999°...qu 5:7 ~u-uoaen ~c-—.mm~u 29“. ma “Eu-:35.— an.
umnmm: cacao-Jon x..—a . ego—...;bN E: 9:. u.‘ mums—«o5; or
em4~a.o 99:34:. uaoeeéow. . saga. new ..x.~a:~_ Cu m>ac~¢o§u pa
sink: :9: ....u.» “...—- . anon—5.75.. 9.39:. m._ m>na§ouz~ we
~00~o.| aaaea.n:o uZu. . cocoe.n:o 2:93“ u; m>~2xcmtu mo
unmz.o oeooo.~mn.~ 5:: . ooaoa.-o~ 2:95. a.. m=3¢naxu :0
. aooou.ec~e~ ..2-1 °u-u .u canning“ an. xo<._m m... J—uwu>wzoz no
. coca—.443 uzuu 99:2... «hummiieu zonaw w; ..=u=wm::: um
mqnonoo sees: .00.. 5:. . 99999.3.- u::=:_ ma c>auoum25 ‘o
. sauna-mi; 9;- ommmn .N u::o~.o::u Xu<um ud :pooowwaza no
aneon .n 99999.nm4 mzun . 923.59. oznoznu w; orcaawmaza so
neon» f cocauowmua mxm: . 9339.33." u:~o=—m mu otaawmaz: no
95096.: uaaae 5: man- . 992:5: uznozue w.— a>oeowm25 so
. .99: $0.. L‘uo o:..m:.o~u «mm¢m .mm xo<4m m4 apmoawmzz: on
amo~wt=o 99399.0: uzuu I ne-ea.mon ozHE—E— ma 31.909942: m.
asuoo: econ—comma: uzul o 9.3.9.20: o:~oz—. u; atnooumaza :-
- cocoa om0mn uzuu wanna; seawefimmn ¥u<4m NJ 1>~eown=za no
. Genoa-Mann uz~u ~c«:n.:n nonmo.:~n xo«..m m4 :>«aewn:=: ~o
m~::m.| .asae.~o- uzuu . eaaaofiowu 6.39:; mg .;:—¢oa:— «a
. 99°27»; mxm. acadeio mauooine xu¢4m w.— 4>ezcomzu as
«owns .- ooeao.~=~ uzac . 03.39.35“ 3:925. ma a>09-xnmru as
. can-a .aouu 5:0 omnuooizu awn—woooel ¥o<aw ma 4>oeucomzn on
. cocoa-o; uz~a 2:: Joan canon-maul xucam m.— 4>~3¢oazu s.
o ocean .ca‘ azuu nomom.e- oamomn:~un ¥o<4m u.- apoguomru an
. oocao.mo~ an: Nit; £m~ omnmJ; XQ¢4w ma arms—«om!» up
{wnon .au cocoa .non 99999.nou . ocean.no~ «.2223. cu :>:o.—«n¢x~ J~
. =0: .03» “.an Duosc.:nu ~en~1u~nm xu<4n ml. :>na~¢oa:~ as
coma»: noao=.-nn uZn. . oooaa.—~om czaz~u a.. tnuohxonxn -
noOuu-t see—3.4T." u‘nu . cacao .¢o:~ uz~o:—u w.— ..»uebcomxu an
. cause .emqmu uzun canon.» oohn~54um~ ¥o<4w u; n—uu4>uzot as
. 999995;: in: o~nuo.o cocoo.cna~ x015 a.. n>uuowmzz= on
miss»: nos-36w: 5:. . cane-.3.. uzuozuu m4 nus—aum-Z: oo
nmaon.| eaaea.n;:u k2”. . aaa°°.nq:u 023.3. m: n>uacwmaz= ~o
uaomnoo aeeaaono: uz—n - 9930.2... czngzm u; n>09auwazz co
coon»: use: .ommw uz~n . aaeac.mm~u 339;. m4 n>~eoum=z= no
9023.: eaeaoénn ...z—u I 99:96:, 259:7. a.. n>Oacwmaza ac
. coo—3.0." uz~u 9%.:on mucoeéo xu<am a.. n>mooumaz= no
:aauaodnl 90909.2: .2~c . cooaoéon czuozum ma n>3aumaz= no
”5‘00: ocean-mun: uz~| . .ooeainca 339:. u._ n-ncuwwaza «a
. =9: .momn “=3: n~¢¢~3~ summnéamw. xo<._m m4 n>~aomm=zs an
. enoaeouqnn 5:: miwsuomu mmn-.~a~n 2915 NJ unaccwmaz: ¢m
000nm: aeoee.-—~ Lzur . coco..:«~ uzmcznu ma n>~ub¢cazu on
. ooeee..oo Lz~u 32:: .«on chancfiua ¥u<4w ma n»3—¢OL:~ um
o gouge-nmmr. u»:- noeso-ow— sno~¢.na:~ xo‘am ma anaeucnmzn om
. ocean‘s: uzu- maunnécmu oo~n~..~nu ¥o<4m m4 n>oepxnmt~ mm
. 3°92..an uz»: munouémm m-aqimua 20¢4m “a n>~3¢omt~ :m
. aeeg .mnu mz~- oiomoonu ooiaméafil xutdm ud area—~33; nu
. cause .ona uz~c oumumivu .3134: 201.5 mg arms—20%; um
com: o: 99999.0»; cacao-oam . neeao.o:~ Baa—in om n>:n~¢emz~ an
a<z~u¢<z cums; ax: 1354 m2! 201“ >h~>-o< 20¢ mah<hu out. 332 zwozaz
no.0 I mxm—.951 coco-a I 39.5%. I BE I 920 I soon anus—2:. I
oooa.— I mzzmsu can... in I fioomm¢ main I 2:- 93sz I 55. huupuwnoo I 500 20:53.3: I ”8:.
nwunm.an~:mu I w>:quno .3 2.5:. V b39240 uwqurou I zenbno :38;
u acct can; nn—uwa¢ w h z a c x h m z o o «n.so.au uzub Duh—as: and:

182

 

 

  

 

 

 

 

 

 

 

 

 

uz~azna ud

 

demanoa ...-..ooou uz—n a ceaaoomoou spuquomxu en—
. one-o.¢-u uz~o 1wu°s.o~ anno~.ma~« ru<4m u; ~hodbzamzu mad
o~¢:o.n ecu-9.4.:n azun . ooaee.:¢4n uz~o=um ma upoe~¢o¢2u ozu
. one.a.oomu mxm. n~e~0.-N« muono.nan xu<4m ma ~>oc~¢on:~ std

one: I aooeo.ann- mxm- . eeaoo.onnu czuozuo ma ~>~e—¢on:— 04¢
«:00» I ooeon.-u mznl o nose-.as“ azuozu. u; ~>oo~¢omx~ m:—
. aesea.~¢n uz~n deems. 000°".uan ¥o<4m m; ~>n-—¢omz~ «an
nua:o c oaeoa.oun nae-9.9«n - ousao.9un oz~o:~a an ~>oohxnmzn at“
mo:no 0 ca .~o~s LZa» . once..~a~n u=uazno ma ~>noh¢oaxn NJ“
JOJO .auoc.no~m Lzuo . oe-eo.n:~o u=~oz~— ma upwnbxoaxu ~:u
anJQQ couscouann mxm: . ca.oo.uea~ c:~o:—. a; ~>unproaxu a:u
. aeo°°.a~oma mama canOquu cahao.oeom: xu<4m ma cpuw4>wzox on-

. eoeoe.»oa~ K‘ua own-«co eucao.nna~ ¥o<4m u; cpuuoumaza can
«eoem.u oaaoa.no4 mz—o . couao.oo: exuo:~_ w; opeuowmaza Kn.
~o~:¢on oe¢o°.—;:u uz~a . 0.9a .n::u uzuaznu ua o>0aoumaza ca—
=:onm.n .aaee.nna uz—u - econ .nw: uz~oz~m ma epooawmaza and
. aocea.dm~¢ uz~o ~«nu:os n-NJm.c:~u zu<4m m4 o>~oawwaz= :nu
ao°o°.aun L2H. n:—o~.: smo—~.men x9<4a ua opossumaz: and

.as .mo— mxm: teouo.: canoo.aou xo<4m ma o>moaum=z= ~nu

ace-a.non uzn. sOeOn.0 anaeo.~on ¥u<.w ma o>:aowm:z: uni

essen.o~ua mxm: m~:0o-m momnu.¢~c: X9<am u; 0>naowmaza an-

ooeeagmamn Lzuo ~0m:o.o« Ou:mn.oamn xuqaw u; epnuowmaza ouu

eoea°.~unn uz~| ne-n. ~o-o.«unn xu<4m u; o>uoawm=z= o~u

ooeeo.oam~ blul o oeooo.omm~ uz«Q:H_ NJ o>uuhuoazu -~

eac.°.nm_u uzuu Quuuo.us "coon-doeu Xudum ma o>°—»¢Otxu an.

. caea..o-n hz—t nanum- ~o~oo.o-n xu<4m ma o>oaurnmz~ mmu

. cocoa.noou mxm. ~m~sooooou ~m~:0.uu~0 xo<4m u; o>oopuomzn auu
Ioueo.| cacao.o~_u Lam: . coca-.ouuu ozuo=~_ u; o>~e—¢oar~ n~u
n:oo~ I oaooo.no‘ nZun . oe-e9.nou oz~a=- ma o>uo_¢omx~ and
mun:o.o eooea.0~n uznn I neoua.o~n uzua:ua u; o>mapxoutu auu
$9930 . naaeoomou 90:09.00" . onauooaou uzwozuw an choc—unatu o~u
Nann9.n page-.mcoo man. . oeoac.momo e=~o=n. a>nu_¢omtn ad.
nonoo.u e.eoa.ou3a Lz~. . coco. 9~40 oz~o=~; opuoncoazu .«u
nun-0.: oneeu om- 52”: I once. mun uz—ozua ma Oran—«omxu ~¢~
. ooae°.ou=nn mxm: a°~.~.ou caoo~.¢oo~n xvsnm ma nuuUJ>uzcx ow“

. ceoeo.naou b.~n -m0uoo ou:po.one~ 20(40 NJ m>auowmaza ma.
:na:m n oneaa.om: mxm: o cage-.as: uzno:n_ ma who¢=wmazs Jud
Neona.l .auoa.n::— mxm. . aaaaeona:u u=H0:—m ma mpoaowmaz: nu“
msoam.o aeeae.nu: K‘s- . aaoao.no: uz~o:~. w; n>oaoumaza nu—
nnnna u ooaeu.am~« man. - no.9..amwu oz~a=~u u; arsooumaz: «a.
«comm.l aaooeoann uznl . ...oo.o«n oznozuu ma m>onouwaza can
. nae-o.m¢. uz—a nunm¢. n:44u.oo— Xu<4m ma m>muauwaza can

. e°°o°.oan Lz~u oossm.~ ~u-:.«on xo<Jm u; m>aoaumzz= on¢

. ooaeooasma uzuo noa°¢.nm mneom.m~o: xo<4m mu m>n°oumazz ~0¢

. aaaaoumOmn Lz~a ~«m:~.au no¢m~.:mmn ¥O<4m ma m>~oaum=za oau

. ..see.munn Lz~n aoe~:. —on~m.«qnn ¥u<4m u; muuoeumaz: mou
anneo.n asaaa.:u:~ uz—a . ooaoo.:~:~ u2uo:uu ma n>uuuxoax~ 49a
. coo-e.aan‘ mxm. an:0«.a¢ o:m=..a~o ¥u<am u; m>ouhxomtn nan

. o-ooconoou h‘uu «GJOI. nannm.~oo~ ¥u<um ma sycopcnttu u.“

. ..o.~nnu Lana ooo~0.w~ou oo.~..o«m| xu<4m m4 n»..~¢onz— «on

4<z_n¢<: awaaa mzx zmxoa ":1 Xucdw >-xuuu< 3oz map<~w -c>h w:¢z syntax
ooae.9 I mzx:umu con-.c I floozunu I ozx I ”:10 I finou wnuzuxct I c—a
.aao.- I mrxmmx ...-..I I snout: maze I :26 ozcvwo I «:~ hum—unfimo I 960 acubcucua< I mzcz

hazpao Upwamzoc I zo~>no bz—It

n-um.an~:mu I U)~—uufino no uzdt>

 

nun-vuud a u z n 1 8 h m z o u s...“ mxm» .~\.n\- u-qa

n wu<k ace.“

183

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as...

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a-eco.a~c:
noeao.m¢mn
cacao.-nn
aaaoe.@4cu
eepoecoa4a
coo-9.405»
aoaacIn_~u
cacao-n—mu
ounce.unu
cocooI-e:
anooau-an
eaooe.nnmo
aeaoa.nou~
ocean —man
oaQQOIanocm
oeaco.naau
aeaaaInn:
aaaooaneau
aneaa.no4
noo- onu—
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xo<4m ma o~uwa>wzoz mam
xu<am Ha o>aaawmzza 4o—
xocdm ma e>nuaum=zs no—
xu<am ma epoaowma:a ~ou
uzuoznu ma o>ceowm=z3 —a«
Xu<4m u; opsocwmaz: so.
xu<4m ma o>oocwmaz= o-
xu<4m w; a>meoumaza cud
xo<4m UA a»:acwm:z: ss—
oz~azun ma e>naoumzzz as“
¥o¢um ma o>~eoum=:: ms“
uzuoznm u; o>aocwm=== a~u
uzua:~— ma n>uauxosxu n~u
Q.ua=~— u; a>°«_zn&:~ us—
cz—ozna m4 o>oehxonzn «~—
XQ<Jm w; opus—xomru e~a
uzwozum w; o>~auroatu on.
uznozna m4 o>oohzoax— cod
aquZHH ma o>map¢aax~ bog
oznmzua as o>:e—¢otxu 00¢
Xu<am w; c>nohuonxu mad
uznaz~_ ma o>~a~annzu aw—
xu<4m ma o>aa~¢amxu no“
xu<4m ma shuwa>wzox no—
xuuam ma u>au°~m:z: «an
cz~o:~_ ma ~>ouawwaza ecu
:014m U4 ~>eecwmaz= own
oz~oznm ma ~>oucwmaz= om—
xo<4m uJ ~>~ocwn=z= nmu
xo<4m U4 sheeowmaz: emu
uzuozuu ma ~>m°cum=za mn-
xo<4m u; s>¢eawmnzz :mu
xo<4m wJ ~>n99mm223 nmu
Xucam ma ~>~oOUnaza um—
xo<4m ma Asqaaumazs «ma
m:—<-w ug>h uzcz wunxzz
I «coo UNuI—x12 I

bum—uufin: I

panhzo abu4azoo I zoubao .z—xn

.a.~a.o— mzuu

.sxanxae unto

can
Sac zonucuoud< I wzuz

184

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sac oma:oI- oeaeaInunmu I
out oooomIa eoaoaInnna« I
me: I sooaoInonuu I
a¢< oumaanm aaeaaInancu I
a¢< I au;oa.non0« I
m¢< :uo~oI eoooaInonou I
I eccenInona— I
a¢< NucnnIa .aoInoncu I
szi conauIn oceaoInonou I
a¢< oumaoIno aoeeuInanan I
o¢< conunInm aaeaaInanou I
ma. ooomhI- sauoaInnnou I
ma: oo-omI: aeeeoInonou I
au‘ conncIoou accenInonau I
c¢¢ --~qua aoaoaInane« I
mad I oes.u.n.nn‘ I
au< :«owo. ace-eInnnoq I
n¢< I eeueaInonuw I
an: monnquu aeaacInono— I
¢¢< :Jsmnqu .aaaeIncno. I
ma. -oo—.u~ aoauaInnnm— I
01‘ ac~n~Ian oeocoInana— I
c¢< ~onooIo aaeeuInnnau I
out ooeomI: uaaaaInano« I
a¢< a::mn.~mn .aoocInonc— I
u¢< mmmnqucu cacaoInnno— I
am: I aoaocInonou I
uu< :«oon acaeaInono- I
I uocaeInnnou I
ax< 04m4:.n« aaeooInonou I
a¢< JanmnIau oaecoInoan I
o¢< mono:I‘u .ueecInoncu I
m¢< comaoII: aoeaoInnna_ I
u¢< Nona... encoaIncnou I
a¢< .oeomI: eeoeuInono— I
ncucwIaom eaeoeInonou I
n¢< o~o=¢.:~u naaooInono— I
mac I noaoaInonm_ I
as: :uouoI ocoeoInono« I
I space-nono« I
auaa-Imu ooaaaIngnon I
m—00:In. eeooaInonou I
n¢¢ acu~4Ioo uaoaoInuno¢ I
a¢< ¢~04~Imm assuaonono_. I
91‘ nanoo . aoeaeIn.nm. I
ex< nanomIa a-eauInanou I
meanqua pageaInon0¢ I
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9:4 uuomnIm‘ oooooInano— I
mu< Juana. aaaaeInnnmu I
4(2_u¢<: yuan: ozn uwxoa aza

aoeaIa I mxxzukc ee=QIa I fiao:umu

aeoeIu I mrzwma aeoqua I Seaman

n-~MI°n~:mu I u>__ousao mo w341>

u uu¢c noqu unnoxuma

 

 
 
 

  

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I 5000 U-z~x<z I «no

unannUfiac I
-n—:n thszoo I zen—mo pz~xa

sac zo-<uc;;< I wr<z

«nIncIOu uzuh csxunxoa wh<o

185

 

 
 
 

I man .
I gym. I
I mxm. I
I II_I I
I 5:". I
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I uv—I I
mm.au.- I2~I .
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n-~n.a»~:m_ I u...uusuo Io wad.»
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a. ~>ooaeaa I,
4; u>meooun no
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4L u>naooan —o
a; «pueoomc av
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hzn opuuxnz on
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bzu oymaxcz ~o
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ham o>~a¢oz On
—:~ o>uoxmz on
_:n spuqumz Is
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~=~ ope—um: mo
uzw 0>Oemmz Jo
hzn a..-um: nw
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nzu ¢>¢u¢mz em
bzn 0>ncwmz on
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Harp ur<x «waxzz
I anon w-z~x<x I cue

fine zo~»10044< I uzIZ

pan—an aruJonu I zeaumo hz~l¢

unIneIo« uzuh ouxenxc. m—15

 

 

 

186

 
  
 

. uz~o . °e«~:. no¢~m.oo~: u>uhu« J: o>~aoomt .mu
. bx". . .«~»4. ~m~ns.oos u>npu< at o>oeooua m:—
. Lz—o . sumo:. canum.uo: m>-o< an o>moaozs cad
. uv—o . cumuuo unuco.o:o u>n~u< 4m o>:.c°wn 54¢
. gru. . om:«~. nuannoummh w>~—u‘ a: ouneaozn o:u
. m2". . °¢o-. Oo~a-.o-~ w>npu¢ a; o>~¢ooma mJu
. mxm. . nouns. ~anuo.~ao~ u>~ho< at cum-comm Jag
. uz~o . oaoom. onuooowcc~ w>unu< 4L m>—_ooxn nJ~
. ur—. a auo::. aoNuc.o«9~ w>~_o< 4a n>o‘oc¢¢ Nan
. uzn. . ouson. eon—~.eo~m u>~hu< a; m>Gecozm «:—
. nan. . oqsoan cacao-no: u>~no¢ a; nuoeooza 9::
. uz~o . ao««:. camnm.onmn u>~pa< 4L m>~°aoum on"
. uz~o . auna:. aa:nm.m~n u>u~o< 4L opaaooxm gnu
. “7.. . aumot. m:ao~.~:n u>u~¢£ a; m>m°covm 5n“
. uzn. . oum-. u—uoo.:~m u>—bu< a; mpaeoous can
. uzu. . om:u~o unsoo.n~«n w>u~u¢ at apnocoua mn‘
. Lz~c . °w¢-. noaau.oans w>unu¢ at upmoooua an.
. uz~o . scum». uenmo.~s:~ w>u~u< 4L m>uocoum nun
. uy—o . cocom. ¢n~_~.mr:~ w>~hu< a; :>«uaozn Nnu
. mzu. . oun::. ae~ma.n~au w>-u¢ 4L cpaucoua «nu
. uz~o . ausgn. u4o«u.~mnm u>~nun A; s>o=ocmm an¢
. Lam. . o«~o:o cause-Hm: w>~hu< 4L a>o°ooun o~u
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