linear optimization in applications -...

LINEAR OPTIMIZATION IN ApPLICATIONS

LINEAR OPTIMIZATION IN ApPLICATIONS

S.L.TANG

香港付出版社

HONG KONG UNIVERSITY PRESS

Hong Kong University Press

141F Hing Wai Centre

7 Tin Wan Praya Road

Aberdeen, Hong Kong

。 Hong Kong University Press 1999

First Published 1999

Reprinted 2004

ISBN 962 209 483 X

All rights reserved. No portion of this publication may

be reproduced or transmitted in any form or by any

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Printed and bound by Liang Yu Printing Factory in Hong Kong, China

CONTENTS

Preface

C hapl er I : Inlroduclion ' . 1 Formulation of a linear programming problem 1.2 Solving a linear programming problem

1.2.1 Graphical method 1.2.2 Simplex method 1.2.3 Revised si mplex method

Chapler 2 : Primal and Dual Models 2.1 Sharlow price I opportunity cost 2.2 The dual model 2.3 Comparing primnlnnd dual 2.4 Algebraic .... '3y to find shadow prices 2.5 A worked example

Chapler 3 : Formulating linear Optimization Problems 3.1 Transportation problem 3.2 Transportation problem with distributors 3.3 Trans-shipment problem 3.4 Earth moving optimization 3.5 Production schedule optimization 3.6 Aggregate blending problem 3.7 Liquid blending problem 3.8 Wastewater treatment optimization 3.9 Critical path ora precedence network 3. 10 Time-cost optimization ofa project network

C hapter 4 : Transport'siion Problem and Algorithm 4. 1 The general form of a transportation problem 4,2 The algorithm 4.3 A furthe r example 4.4 More applications of Iran sport al i on algorithm

4.4. 1 Tmns-shipment problem 4.4.2 Earth moving problem 4.4.3 Product schedule problem

4.5 An interesti ng example using transportation algorithm

vii

I 2 3 4 6

9

• 9

II 13 15

I. 19 22 24 27 30 32 34 37 40 42

51 51 53 59 65 65 68 6. 71

VI Linear Optimization In Applications

Chapter 5 : Integer Programming Formulation 5.1 An integer programming example 5.2 Use of zero-one variables 5.3 Transportation problem with warehouse renting 5.4 Transportation problem with additional distributor 5.5 Assignment problem 5.6 Knapsack problem 5.7 Set-covering problem 5.8 Set-packing problem 5.9 Either-or constraint (resource scheduling problem) 5.10 Project scheduling problem 5.11 Travelling salesman problem

557914790258 『I

司/呵/吋

/OOOOOOOOAYAVJAYAVJ

Chapter 6 : Integer Programming Solution 6.1 An example of integer linear programming solutioning 6.2 Solutioning for models with zero-one variables

105 105 112

Chapter 7 : Goal Programming Formulation 7. 1 Linear programming versus goal programming 7.2 Multiple goal problems 7.3 Additivity of deviation variables 7.4 Integer goal programming

117 117 120 122 127

Chapter 8 : Goal Programming Solution 8.1 The revised simplex method as a tool for solving goal

programming models 8.2 A further example 8.3 Solving goal programming models using linear

programming software packages

131 131

137 140

AppendixA Appendix B Appendix C

Appendix D

Examples on Simplex Method Examples on Revised Simplex Method Use of Slack Variables, Artifical Variables and Big-M Examples of Special Cases

145 153 161

162

PREFACE

This b∞k is 001 wriUen 10 disεuss the malhematicsof linear programming. lt isdesigncd

10 illustrate, with praclical cxamples, thc applications orlîncaroptimi且tion techniques

The simplcx mcthod and the rcvised s implcx mcthod. thcre forc , 3re includcd as

叩pcndiccs only in thc book

啊lC book is wrillcn in simple 3nd easy 10 understand languagc. 11 has put logelhcr

a very useful and comprehcnsivc sct ofworked examplcs. bascd 011 fcallifc problcms

using lin閏 r progmmming and ÎIS c .Klcnsions including intcgcr programming and goal

programming. The examplcs 3rc uscd 10 explain how lincar optimi甜tion can be app1ied

in cnginceri ng and busin閏s/managcmcnl problems. Thc stcps shown in cach w可orkcd

cxamplc 3re clear and 間sy 10 rcad

Thc book is likcly 10 be uscd by tcachers. taught course students and 悶search

studcnts orboth engincering :md businesslmanagcment disciplincs. l t 惱， howevcr, not

suitablc for studcnlS or purc rnalhcmalics becausεthc conlentS or Ihe book cmphasizc

叩抖ications ralher than theories

INTRODUCTION

1 .1 Formu lation of a Li near Programming Problem

Linear programming îs a powe血1 mathematical 1001 for the optimization of an

objcctive under a numbcr of constraints in any given situation. Its application

C曲 be in maximizing profils or minimizing costs whlle making the best use of

the limited resources avaîlable. B目ause it is a malhematical t∞1 ， it is best

explained using a prac!ical example

Example 1.1 A pipe manufacluring company produces tWQ types of pipes, type I 叩d Iype n

The storage space, raw material requirement 胡d production rale are given as

below:

且==目 b且i 工站直且 (&固paoy Ay晶ilallili!)!

Storage space 5m月p'pe 3 m2/pipe 750 m2

Raw materials 6 kglpipe 4 kglpipe 8∞ kglday

Production rate 30 pipesthour 20 pipeslhour 8 hourslday

SJHO

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tctnu 阱呵

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叫自問宙間血

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仲叫叫間耐

iy

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memcdm ddhM3a ihk-Rm H

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叫咱們

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hkmv

Solution 1. 1 LetZ =

X) =

100al profit

number of Iype I pipes produced each day

number of type n pipcs pr'吋uced each day '‘, =

Sînce our obj且llve IS lO maxlmlze pro缸， we write 曲。bj配tive function,

equation (0)、 whîch wîll calculale lhe IOlal profil

MaximizeZ= 10xI +8X2 (0)

2 Linear Optimization 的 Applications

Xl and X2 in equation (0) are called decision variables.

There are three constraints which govem the number of type 1 and type 11 pipes

produced. These constraints are: (1) the availability of storage space, (2) the raw

materials available, and (3) the working hours of labourers. Constraints (1), (2)

and (3) are written as below:

Storage space: 5Xl + 3X2 三 750 、‘'，

.E.A /，.‘、、

、

Raw material 6x 1 + 4X2 三 800 一一一一一一一一一一一一一一-一一-一一 (2)

x x W orking hours :一土+一土豆 8 一一一一一一一一一一一一一… (3)

30 20

When constraint (3) is multiplied by 60, the unit of hours will be changed to the

unit of minutes (ie. 8 hours to 480 minutes). Constraint (3) can be written as :

2Xl + 3X2 三 480 (3)

Lastly, there are two more constraints which are not numbered. They are Xl 三 O

and X2 三 0， simply because the quantities Xl and X2 cannot be negative.

We can now summarize the problem as a linear programming model as

follows:

勻，旬

x nxu + xm nununuAU E』《

Jnunδ

=弓

IOOA

『

Z<

一歪歪

咒。

222

utxxx n1343 .m﹒戶+++

a-b!lll AUXXX EWSζJfO

司4

、‘.，/

nU J'.‘、、

(1)

(2)

(3)

Xl ~O

X2~ 0

1.2 Solving a Linear Programming Problem

There are two methods in solving linear prograrnming models, namely, the

graphical method and the simplex method. The graphical method can only

solve linear programming problems with two decision variables, while the

simplex method can solve problems with any number of decision variables.

Since this book will only concentrate on the applications of linear prograrnming,

Introduction 3

the rnathernatical details for solving the rnodels will not be thoroughly treated.

In this section, the graphical method and the simplex method will only be briefly

described.

1.2.1 Graphical Method

Let us look at the linear prograrnrning rnodel for Exarnple 1.1:

Max Z = 10Xl + 8X2 “…一

subject to

5Xl + 3X2 至 750..._..... 的自…"“…“…叮叮………“凹，叫“_........ (1)

、‘，/

AU /EE‘、

6Xl + 4X2 至 800 ...叮叮叮叮“ “一 …......................_.......呵呵 ω 叮叮叮叮一 "曰 “ … (2)

2Xl + 3X2 歪 480 山_..……"叮叮叮………… 一。" “ _.......“…… 一 " “ … (3)

Xl ~O

X2~ 0

The area bounded by (1) : 5Xl + 3X2 = 750, (2) : 6Xl + 4X2 = 800, (3) : 2Xl + 3X2

= 480, (4) : Xl = 0 and (5) : X2 = 0 is called the feasible space, which is the

shaded area shown in Fig. 1.1. Any point that lies within this feasible space wi11

satisfy all the constraints and is called a feasible solution.

X2

Note: the x) and X2 axes are not drawn on the same scale.

X)

Fig. 1.1 Graphical Method

The optirnal solution is a feasible solution which, on top of satisfying all

constraints, also optirnizes the objective function , that is, rnaxirnizes profit in

this case. By using the slope of the objective function , -(10/8) in our case, a line

can be drawn with such a slope which touches a point within the feasible space


and is as f;缸 away as possible from the point of origin O. This point is

represented by A in Fig. 1.1 and is the optima1 solution. From the graph, it can

be seen that at optimum,

Xl = 48 (type 1 pipes)

也= 128 (type 11 pipes)

max Z = 1504 (profit in $), ca1culated from 10(48) + 8(128)

From Fig. 1.1 , one can also see whether or not the resources (i.e. storage space,

raw materials, working time) are fully utilized.

Consider the storage space constraint (1). The optima1 point A does not lie on

line (1) and therefore does not satisfy the equation 5Xl + 3X2 = 750. If we substitute Xl = 48 and X2 = 128 into this equation, we obtain:

5(48) + 3(1 28) = 624 < 750

Therefore, at optimum, only 624 m2 of storage space are used and 126 m2 (i.e.

750 - 624) are not used.

By similar reasoning, we can see that the other two resources (raw materia1s and

working time) are fully utilized.

If constraint (1) of the above problem is changed to 5Xl + 3X2 三 624， that is, the

available storage space is 624 m2 instead of 750 m2, then line (1) will a1so touch

the feasible space at point A. In this case, lines (1), (2) and (3) are concurrent at

point A and a11 the three resources are fully utilized when the maximum profit is

attained. There is a technical term ca11ed “optima1 degenerate solution" used for

such a situation.

1.2.2 Simplex Method

When there are three or more decision variables in a linear programming model,

the graphica1 method is no more suitable for solving the model. Instead of the

graphical method, the simplex method will be used.

5 Introduction

As mentioned earlier, the main theme of this book is applications of linear

Therefore, programming, not mathematical theory behind linear programming.

no detail description of the mathematics of linear programming will be

presented here. There are many well developed computer programs available in

the market for solving linear programming models using the simplex method.

One of them is QSB+ (Quantitative Systems for Business Plus) written by Y.L.

Chang and R.S. Sullivan and can be obtained in any large bookshop world-wide.

The author will use the QSB+ software to solve all the problems contained in the

later chapters of this book.

Examples of the techniques employed in the simplex method wi1l be illustrated

in Appendix A at the end of this book. Some salient points of the method are

summarized below.

First of all we introduce slack variables SI，也 and S3 (S 1, S2, S3 ~ 0) for Example

1.1 to change the constraints from inequalities to equalities such that the model

becomes:

(Oa) Z - 10Xl - 8X2 = 0

subject to

(1 a) 5Xl + 3X2 + SI = 750

(2a) 6Xl + 4X2 + S2 = 800

(3a) 2Xl + 3X2 + S3 = 480

The initial tableau of the simplex method is shown in Table 1.1. It is in fact a

rewrite of equations (Oa), (la) , (2a) and (3a) in a tableau format.

品一oool

已一oolo

趴一0100

也-A343

h﹒一心562

Z一1000

Basic Variable (Oa) Z (1a) SI (2a) S2 (3a) S3

Initial Simplex Tableau for Example 1.1 Table 1.1

Linear Optimization 的 Applications6

After two iterations (see Appendix A), the final tableau will be obtained and is

shown in Table 1.2 below.

Basic Variable (Oc) Z (1c) Sl (2c) Xl (3c) X2

3-82A6 s-aooa 2-4932 s-loan吋

趴一0100

也一oool

叫一oolo

z-1000

Table 1.2 Final Simplex Tableau for Example 1.1

To obtain a solution from a simplex tableau, the basic variables are equal to the

values in the RHS column. The non-basic variables (i.e. the decision variables

or slack variables which are not in the basic variable column) are assigned the

Therefore, from the final tableau, we can see that the optimal value zero.

solution is:

Z = 1504

Xl = 48

X2 = 128

Sl = 126 S2 = 0

} non仔.枷i比cvar缸ri昀岫a油bl臼叫a缸re叩a叫l ω

It can be seen that this result is the same as that found by the graphica1 method.

S 1 here is 126, which means that the slack variable for storage space is 126 and

therefore 126 m2 of storage space is not utilized. Sl and S2 are slack variables

S3 = 0

for the other two resources and are equa1 to O. This means that the raw materia1s

and the working time are fully utilized.

Revised Simplex Method

The revised simplex method is a1so ca11ed the modified simplex method. In

this method, the objective function is usually written in the last row instead of

1.2.3

the first. Ex剖nples of the technique are illustrated in Appendix B. QSB+ uses

the revised simplex method in solving linear programming models. The initial

tableau for Example 1.1 is shown in Table 1.3.

Introduct的n

Basic X) X2 S) S2 S3

Variable Q 10 8 。。。 RHS S) 。 5 3 。。 750 S2 。 6 4 。。 800 S3 。 2 3 。。 480

L 。。。。。。Cj -Zj 10 8 。。。

Table 1.3 Initial Simplex Tableau for Example 1.1 (Revised Simplex Method)

After two iterations (see Appendix B), the final tableau will be obtained. It is

shown in Table 1.4.

Basic x) X2 S) S2 S3 Variable Q 10 8 。。。 RHS

S) 。。。 -0.9 0.2 126 x) 10 。。 0.3 -0.4 48 X2 8 。。 -0.2 0.6 128

L 10 8 。 1.4 0.8 1504 Cj -Zj 。。。 -1.4 -0.8

Table 1.4 Fínal Símplex Tableau for Example 1.1 (Revísed Símplex Method)

7

PRIMAL AND DUAL MODELS

2.1 Shadow Price I Opportunity C。到

The shadow price (or called opportunity cost) of a resource is defined as the

economic value (increase in profil) of an exlra unit of resource at the optimal

point. For example, the raw material available in Example LI of Chapter I is

8曲旬; the shadow price of it means the increase in profit (or the increase in Z,

the objective fu nclîon) if the raw material is incre越ed by onc un叭， 10801 kg

Now, I叭叭= shadow pricc of storage space($Im2)

Y2 = shadow price of raw material ($lkg)

Y3 = shadow price of working time($/minute)

This means that one additionaJ m2 of storage space available (i.e. 751 m2 is

available inslead of 750 m2) 圳11 increase Z by YI dollars; one additional kg of

raw malerials available wi l1 increase Z by y2 dollars; and one additional minute

of working time available will increase Z by Y3 dollars

8ased on the defini tion of shadow price, we can fonnu lale anolher Iinear

programming mode1 for Example 1.1. T >is new model is called the dual model

2.2 The Dual Model

Si nce thc productîon of a type I pipe requires 5 m2 of storage space, 6 kg of raw

material 朋d 2 minutcs of working time, the shadow price of producing one extra

type I pipe will be 句1 + 6 y2 + 2Y3. This means that the increase in profit (i .e. Z)

due to producing an additionallYpe I pipe is 5Yl + 6 y2 + 2狗， wh.ich should be

greater th曲。r at least equal 10 $ 10, Ihe profit level of selling one type I pi阱， m

order to juslify the extra production. Hence, we can writc thc constraint that

5Yl + 6Y2 + 2Y3 ;::: 10 (1)

A similar argument app l i臨 10 type II pi阱. and we can write another constraint

that

3Yl +4Y2+3Y3;::: 8 (2)

10 L的earOptl的7ization 的 Applications

It is impossible to have a decrease in profit due to an extra input of any

resources. Therefore the shadow prices cannot have negative values. So, we

can also write:

yl 這 O

y2 這 O

y3~ 0

We can also inte中ret the shadow price as the amount of money that the pipe

company can afford to pay for one additional unit of 自source so that he can just

break even on the use of that resource. In other words, the company can afford,

to pay, s旬， y2 for one extra kg of raw material. If it pays less than y2 from the

market to buy the raw material it will make a profit, and vice versa.

The objective this time is to minimize cost. The total price, P, of the total

resources employed in producing pipes is equal to 750Yl + 800Y2 + 480Y3. In

order to minimize P, the objective function is written as:

Minimize P = 750Yl + 800Y2 + 480狗一…一一一一一………一 (0)

We can now summarize the dual model as follows:

Min P = 750Yl + 800Y2 + 480Y3 subject to

一一一一一一……一一 (0)

5Yl + 6Y2 + 2Y3 ~ 10

3Yl + 4Y2 + 3Y3 ~ 8

yl ~O

y2 ~ 0

y3 ~ 0

The solution of this dual model (see Appendix A or Appendix B) is:

min P = 1504

yl= 0

y2 = 1.4

y3 = 0.8

(1)

(2)

We can observe that the optimal value of P is equal to the optimal value of Z

found in Chapter 1. The shadow price of storage space, yl , is equal to O. This

means that one additional m2 of storage space will result in no increase in profit.

Primal and Dual Models 11

This is reasonable because there has a1ready been unutilized storage space. The

shadow price of raw material，抖， is equa1 to 1.4. This means that one additiona1

kg of raw materia1 will increase the profit level by $1 .4. The shadow price of

working time，豹， is equal to 0.8. This means that one additiona1 minute of

working time will increase the profit level by $0.8. It can a1so be interpreted

that $0.8/minute is the amount which the company can afford to pay for the

extra working time. If the company pays less than $0.8/minute for the workers

it will make a profit, and vice versa.

2.3 Comparing Primal and Dual

The linear programming model given in Chapter 1 is referred to as a primal

model. Its dual form has been discussed in Section 2.2. These two models are

reproduced below for easy reference.

位也al 2屆l

Min P = 750Yl + 800Y2 + 480Y3

subject to

5Yl + 6Y2 + 2Y3 ~ 10

3Yl + 4Y2 + 3Y3 ~ 8

?但

x oo +OOO 1508 x784 的歪歪歪

=0222 FJtxxx 21343 LC+++ 、f﹒必刊J

111ll A

叫

xxx

hnsζJrO

呵，L yl 這 O

Xl ~O y2 這 O

y3 ~ 0 X2 ~ 0

It can be observed that :

(a) the coefficients of the objective function in the prima1 model are equal to the

RHS constants of the constraints in the dua1 model,

(b) the RHS constants of the constraints of the prima1 model are the coefficient

of the objective function of the dua1 model, and

(c) the coefficients of yt , y2 and 豹， when read row by row , for the two

constraints of the dua1 model are equa1 to those of Xl and X2, when read

column by column, in the primal model. In other words, the dual is the

transpose of the primal if the coefficients of the constraints are imagined as a

matnx.

12 Linear Optimization in Applications

The general fonn of a primal model is :

Max Z = C1X1 + C2X2 + .‘+ CnXn subject to

a11x1 + a12x2 + ., ... + a1nXn 三 b l

~月 I + ~2X2 + ..... + ~nXn 三 b2

丸llX I +丸泣X2 + ..... + ~Xn 三 bmall xj 主 O

The general fonn of the dual model wil1 be :

瓦1in P = b tYl + b2Y2 + .. ... + bmYm

subject to

allYI + ~tY2 + ., •.. +丸llYm 主 c 1

al2YI + ~2Y2 + ., •.. +丸。Ym 主 c2

a1nYI + ~nY2 + .. '" +丸mYm 這 cn

all Yi 這 O

The two models are related by :

(a) maximum Z = minimum P, and

、‘，/

ku /'.、

ð. Z Yi= 一一一

ð. bi

where Yi stand for the shadow price of the resource i and bi is the amount of the

ith resource available.

It should be pointed out that it is not necessary to solve the dual model in order

to find Yi' In fact, Yi can be seen 企om the final simplex tableau of the primal

model. Let us examine the final tableau of Example 1.1 :

13

地一oool

叫一oolo

z-1000

Primal and Dual Models

Basic Variable Z Sl XI X2

We can see that the coefficient of Sl , slack variable for storage space, in the first

row (the row of the objective function Z) is equal to Yh the shadow price of

storage space in $/rn2. The coefficient of 鈍， slack variable for raw rnaterial, in

the first row is equa1 to Y2, the shadow price of raw rnaterial in $/kg. Again, the

coefficient of 缸， slack variable for working tirne, in the first row is equal to 豹，

the shadow price of working tirne in $/rninute.

Therefore, it is not necess缸Y to solve the dua1 rnodel to find the values of the

decision variables (ie. YI，但 and Y3). They can be found frorn the prirnal rnodel.

The sarne occurs in the revised sirnplex rnethod. The final tableau of the revised

rnethod for Exarnple 1.1 is :

Basic Xl X2 Sl S2 S3

Variable q 10 8 。。。 RHS Sl 。。。 126 XI 10 。。 48 X2 8 。。 128

2日 10 8 1504 Cj -Zj 。。

y3 Y2 Yl

In a sirnilar w呵， the values of yl , Y2 個d Y3 can be seen frorn the row of Zj .

Algebraic Way to Find Shadow Prices

There is a sirnple algebraic way to find the shadow price without ernploying the

2.4

sirnplex rnethod. Let us use the sarne ex缸nple， Exarnple 1.1 , again to illustrate

how this can be done. The linear prograrnrning rnodel is reproduced hereunder

、‘E/

AU /'.‘、、

for easy reference:

Z=10Xl +8X2 M位


AUAUAU P、JAUOO

司IOOA

且可

<一<一<一

0222 txxx d343 戶+++

:ll1 叫

XXX

S

《JfO

司4

(1)

(2)

(3)

As can be seen 企om the graphical method, storage space has no effect on the

optimal solution (see Section 1.2.1). Line (1) : 5x) + 3x2 750 therefore does

not pass through the optimal point. Since raw material and working time both

define the optimal point, the solution is therefore the intersection point of line

(2) : 6x) + 4x2 = 800 and line (3) : 2x) + 3x2 = 480.

Assuming that the raw material available is increased by ~L kg, the optimal

solution will then be obtained by solving:

6x) + 4x2 = 800 + ~L

and 2x) + 3x2 = 480

Solving, we obtain :

l=48+L aL 10

and x2 = 1孔令 ~L

Substituting x) and x2 into the objective function, we have:

z + ~Z =叫48 + 1~f\ ~L) +叩28 - +~L) 3 10 ~~I -，~-- 5

Simplifying, we get

z + ~Z = 10附 +8(128)+÷aL

Since Z = 10(48) + 8(128)

az=?aL

i.e-A三1.4 (i.e. shadow price ofraw material in $/kg) ~L

Primal and Dual Models 15

Similarly, if we assume that the working time available is increased by AM

minutes, we can, by the same method, obtain that :

各= 0.8 (i.e. s叫p心

2.5 A Worked Example

Let us now see a practical example of the application of shadow prices.

Example2.1

A company which manufactures table lamps has developed three models

denoted the “Standard",“Special" and “Deluxe". The financial retums from the

three models are $30, $40 and $50 respectively per unit produced and sold. The

resource requirements per unit manufactured and the total capacity of resources

available are given below:

扎1achining Assembly (hours) (hours)

Standard 3 2

Special 4 2

Deluxe 4 3

A vailable Capacity 20,000 10,000

σ。、』'，

n3 ••• A

•• A

eELEa-mm 劃

h

DAit

2

3

6,000

(a) Find the number of units of each type of lamp that should be produced such

that the total financial retum is maximized. (Assume all units produced 缸e

also sold.)

(b) At the optimal product mix, which resource is under-utilized?

(c) If the painting-hours resource can be increased to 6,500, what will be the

effect on the total financial retum?

Solution 2. 1

(a) The problem can be represented by the following linear programming

model:

Max Z = 30Xl + 40X2 + 50月


subject to

3x1 + 4x2 + 4x3 至 20，000

2x1 + 2x2 + 3x3 至 10 ，000

x1 + 2x2 + 3x3 至 6 ，000

x,. x~. x, ~ 0 l' L'-2' .l"1t.. 3

where X 1 = number of Standard lamps produced

X2 = number of Special lamps produced

X3 = number of Deluxe lamps produced

Introduce slack variables Sl' S2 and S3 such that:

3x1 + 4x2 + 4x3 + Sl 20,000

2x1 + 2x2 + 3x3 + S2 10,000

x1 + 2x2 + 3x3 + S3 6,000

Using the simplex method to solve the model, the final tableau is:

Basic Variable Z

Sl x1 x

shadow price of painting hour

The final tableau of the revised simplex method is :

Basic x1 x2

Variable Cj 30 40

Sl 。。。x1 30 。x2 40 。

Zj 30 40 Cj -Zj 。。

x3 Sl

50 。-2

。。l.5 。60 。-10 。

S2 S3

。。

-0.5 10 |T -10

shadow price of painting hour

RHS

4,000 4,000 1,000

160,000

From the final tableau of either method, we obtain the following optimal solution:

Primal and Dual Models

Basic varìable

S} = 4,000

x} = 4,000

X2 = 1,000

Z = 160,000

17

Non-basic variable

X3 = 0

S2 =0

S3 =0

Therefore, the company should manufacture 4,000 Standard lamps, 1,000

Special lamps and no Deluxe lamps. The maximum financial retum ìs

$160,000.

(b) Machìnìng ìs under-utilized. Since SI = 4,000, therefore 4,000 machining

hours are not used.

(c) The shadow price of painting-hours resource can be read from either

tableau and is equal to $10/hour. This means that the financial retum wi1l

increase by $10 if the painting-hour resource is increased by 1 hour. If the

painting hour is increased by 500 (from 6,000 hours to 6,500 hours) , then

the total increase in financial retum is $10 x 500 = $5 ,000, and the overall

fìnancial retum will be $160,000 + $5 ,000 = $165 ,000.

It should be noted that the shadow price $10/per hour may not be valid for

infinite increase of painting hours. In this particular problem, it is valid until

the painting hour is increased to 10,000. This involves post optimality

analysis which is outside the scope of this book. The software QSB,

however, shows the user the range of validity for each and every resource in

its sensitivity analysis function.

It is also worthwhile to note that while the 2月 values of the final revised

simplex tableau represent shadow prices of the corresponding resources, the

Cj - 2月 values 自P時sent the reduced costs. The reduced cost of a non-basic

decision variable is the reduction in profit of the objective function due to

one unit increase of that non咱basic decision variable. In Example 2.1 , the

non-basic decision variable is 肉， and so X3 is equa1 to 0 when the total profit

is maxirnized to $160,000. If we want to produce one Deluxe lamp in the


product mix (i.e. X3 = 1) under the origina1 available resources, then the total

profit will be reduced to $159,990 (i.e. 160,000 - 10) because the reduced

cost of X3 is -10 (the Cj - Zj va1ue under column X3).

Exercise

1. A precast concrete subcontractor makes three types of panels. In the production

the quantities of cement, co缸se aggregates and fines aggregates required are as

follows:

Cement Course aggregates Fines aggregates (m3/panel) (m3/panel) (m3/panel)

Panel 1 3 2

Panel II 2 3

Panel m 2 3 4

The subcontractor has the following quantities of cement, coarse aggregates and

fines aggregates per week :

Cement : 300 m3

Coarse aggregates : 500 m3

Fines aggregates : 620 m3

The financia1 retum for panel types 1, II and m 缸e 20, 18 and 25 respectively.

Find the number of panels of each type that should be made so that the total

financial retum is maximized. Which resource is under-utilized and why? If the

subcontractor can obtain some extra coarse aggregates, what is the maximum

cost per m3 the subcontractor can afford to pay for it?

In Chapter 1, we have seen how a Iinear programming model for maximizing

profit under limited resources is formulated. In Chapter 2, we have also seen

how its dual model is formulated. This chapter contains more examples on the

formulation of linear programming problems.

3.1 Transportation Problem

Example3.1

Goods have to be transported from three warehouses 1, 2 and 3 to two customers

1 and 2. The three warehouses have the following quantities of stock to be

transported per week.

warehouse Ouantitv of stock (tonnes)

2

3

勻，MQVAU

，且呵

JH43

The requirements of the two customers per week 訂e:

Customer O~reauired (tonnes)

2

29

33

The costs of transporting one tonne of stock from each warehouse to each

customer is given in Table 3.1.

Cost per stock

Warehouse

Table 3.1 Cost matrix of the transp。前ation problem

This means that it costs $9 to transport one tonne of stock from warehouse 1 to

customer 1, $10 to customer 2, and so on.


The problem is to determine how many tonnes of stock to transport from each

warehouse to each customer per week in order to minimize the overall

tr缸的portatíon cost.

Solution 3. 1 Let xij be the decision variables so that xij is the number of tonnes of stock

transported from warehouse i to customer j per week as follows:

warehouse Customer GGG

It is best to represent the problem in the form of a transportation tableau (Fig.

3.1).

Customer 2 Supply

之j l旦lX11 X12 12

丘l 斗W訂e- 2 X21 X22 20

house

3 主j ~

30 X31 X32

Demand 29 33

Fig. 3.1 A transpo叫ation tableau

The objective function is to minimize overall transportation cost:

Minimize P = 9X 11 + 10x12 + 6X21 + 7X22 + 8X31 + l1x32

There are three constraints on the amount of stock transported from each

warehouse; the amount must not exceed the supply available:

Formulating Linear Optimization Problems 21

Warehouse 1 : x lI + X12 $; 12 ………一…一………………一…………(1)

Warehouse 2: X21 + X22 三 20 … (2)

Warehouse 3: X31 + X32 三 30 …………一………一…………一…一… (3)

There are two constraints on the amount of stock demanded by the customers:

Customer 1 : X11

+ X21

+ X31

= 29 一…一………戶。"一……_._..._...……… (4)

Customer 2: X12

+ X22

+ X32

= 33 …… … "的“呵呵 "“一………_._...-帥_.......... (5)

So, the linear programming model for this problem is summarized as follows:

Min P = 9X 11 + 10x12 + 6X21 + 7X22 + 8X31 + l1x32

subject to

XII + X12 ~三 12

X 21 + X 22 ~三 20

X31 + X32 ~三 30

(1)

呵呵 (2)

一 (3)

向"一 (4)X11 + X21 + X31 = 29

X12 + X22 + X32 = 33 一一…………一…………一……… (5)

all xij 主 o (i = 1, 2, 3 and j = 1, 2)

The solution for this model is:

Overall minimum transportation cost = 503

XII = 0

X 21 = 0

X 31 = 29

X 12 = 12

X 22 = 20

X 32 = 1

The above example is usually called a transportation problem. A

transportation problem can be solved by the simplex method of course, or by

another algorithm technique which will be discussed in Chapter 4.


3.2 Transportation Problem With Distributors

Example3.2

This example is related to but a little more complicated than Example 3. 1.

Assume that the three warehouses ( i = 1, 2, 3) are transporting their stocks to

the two customers (k = 1, 2) directly and/or through two distributors G = 1, 2). Distributorships are offered at the discretion of the warehouses only when

needed. Distributors 1 and 2 have weekly capacities to store and distribute 40

and 35 tonnes of stock respectively. The system can be represented by Fig. 3.2.

warehouse Distributors Customers

Supply = 12

Demand= 29

Supply = 20

Demand= 33

Supply = 30

Fig. 3.2 A transportation problem with distributors

The costs Cik of transporting a tonne of stock from i to k on route Zik (i.e. direct

delivery without through the distributors) have already been given in Example

3. 1. The cost of transporting a tonne of stock from each warehouse to each

distributor (i.e. i to j) and from each distributor to each customer (i.e. j to k) are

as follows:


Route i - i Unit cost C Route i - k Unit cost C

1-1 4 1-1 3

1-2 6 1-2 5

2-1 5 2-1 6

2-2 4 2-2 4

3-1 5

3-2 3

How should the stocks be distributed to the customers such that the overal1

distribution cost is minimized?

Solution 3.2 Objective function:

Minimize P = LCij xij + LCjk Yjk + L Cik Zik

= 4x"

+ 6XI2 + 5x21 + 4X22 + 5x31 + 3X32

+ 3y" + 5YI2 + 6Y21 + 4Y22

+ 9z11 + 10z'2 + 6z21 + 7Z22 + 8z31 + l1z32

Constraints:

The warehouse supply constraints:

LXii + LZik 三 Supply of warehouse for i = 1, 2, 3 j=1 - k=1

i.e. X11 + X12 + Zll + ZI2 三 12 一……………………一個……._--_._-_...……(1)

X21 + X22 + Z21 + Z22 ~ 20- (2)

X31 + X32 + Z31 + Z32 三 3。一一…一一…一…………一……._.-叫“一……… (3)

The distributor capacity constraints:

LXij 歪 Capacity of distributor for j = 1, 2

i.e. X11 + X21 + X31 三 40 ……一一一一……一一一……一一---_....._.-一 (4)

X12 + X22 + X32 三 35 一一一一一…一一一一一……一一一一…一… (5)

24 Linear Optimization 的 Applicat.的ns

The distributor input-output constraints:

~Xii ~ ~Yik for j = 1, 2 i=1 - k=1

i.e. xlI + X21 + X 31 ~ YII + Y I2 …一一一一一…一一一一一一一一一一一 (6)

X I2 + X22 + X32 ~ Y21 + Y22 -: (7)

The customer demand constraints:

~Zik +抖= Demand of customer for k = 1, 2

i.e. ZII + Z21 + Z31 + YII + Y21 = 29 …………一…一一一一----~.~…一一… (8)

ZI2 + Z22 + Z32 + Y I2 + Y22 = 33 …………一一--_._----………--一一一 (9)

The final set of constraints are:

all xij ~三 0， Yjk ~ 0, Zik 去。


Minimum overall distribution cost = 417 X 11 = 12 YII = 12 ZII = 0

X I2 = 0 YI2 =0 Z I2 = 0

X21 = 0 Y21 = 0 Z21 = 17

X 22 = 0 Y22 = 30 Z22 = 3

X 31 =0 Z31 = 0

X32 = 30 Z32 = 0

3.3 Trans-shipment Problem

In Example 3.1 , we have seen what a transportation problem is. A

transportation problem involves a set of source locations and another set of

destinations. We have also seen an extension of the transportation problem in

Example 3.2 with the introduction of a set of distributors in between the sources

and the destinations. Now, we shall see another extension of the transportation

problem, called the trans-shipment problem, in which stock can flow into and

Formulatíng Línear Optímízatíon Problems 25

out of intermediate points in a network, depending upon where they are needed

most. Let us now see a trans-shipment example.

Example3.3

Fig. 3.3 A trans-shipment problem

In Fig. 3.3, the warehouses 1 and 4 have positive numbers, which means that 18

and 12 tonnes of stock respectively are to be transported out from these two

nodes in the network. Demand stations 3 and 6 have negative numbers, which

means that 14 and 16 tonnes of stock respectively are required by these two

nodes. Nodes 2 and 5 are intermediate nodes which neither supply nor demand

any items of stock. These two nodes are called trans-shipment nodes.

Sometimes, nodes 3 and 4 缸'e also called trans-shipment points because stock

can be both in and out in these nodes. The costs Cij are unit costs of

transportation, similar to those of Examples 3.1 and 3.2. Note that delivery is

possible in both directions between nodes 2 and 5. The unit costs C25 and C52

may be the same or may be different (in this example, they are different). Nodes

4 and 5 also allow travelling in both directions; C45 and C54 m旬， again , be the

same or different (in this ex缸nple， they are the same). The requirement is to

relocate the stocks at minimum cost.

Solution 3.3 Let xij be the tonnes of stock transported from i to j (i ~ j). The objective

function is:


Minimize P = C 12 XI2 + C23 X23 + C34 X34 + C25 X25 + C52 X52

subject to

+ C53 X53 + C45 X45 + C54 X54 + C46 X46 + C56 X56

10xI2 + 20X23 + 15x34 + 22x25 + 18x52

+ 12x53 + llx45 + llx54 + 13x46 + 16x56

At node 1 : x 12 歪 18 (1)

At node 2 : - x12 - X52 + X23 + x鈞、= 0 ……一一一一 (2)

Atnode3:-X23 - X53 + X34 = -14 一…一一一一 (3)At node 4 : - X34 - X54 + X45 + X46 至 12m…一……一 (4)

At node 5 : - X25 - X45 + X52 + X53 + X54 + X56 = 0 (5)

At node 6 : - X46 - X56 = -16 一一一一一一一一一一 (6)

all xij 主 O

The optimal solution of the above linear programming model can never have X25

and X52 both positive, nor X45 and X54 both positive.

The solution of this model is given below:

Total minimum trans-shipment cost = 768

X I2 = 18

X 23 = 14

X 25 = 4

X 46 = 12

X 56 = 4

other xij = 0

There are two modifications to a trans-shipment problem. The first modification

is the addition of capacity constraints to the nodes. For example, if node 2 can

only process the trans-shipment of a maximum of R2 tonnes of stock, node 3 can

only process R3 tonnes, and node 5 can only process R5 tonnes, then we have

three additional constraints:

Formulating Linear Optimization Problems

At node 2, x12 + XS2 三 R2 (7)

At node 3, x23 + X S3 三 R3 ..-“一……………一“一………………一一…叮叮叮曰“一 …… (8)

At node 5, X2S + X4S 三 RS …………一…………………一…………………一… (9)

27

The second modification is the addition of capacity constraints to the routes.

For example, if route 2-3 has only a flow capacity of L23, route 5-2 a flow

capacity of LS2' route 4-6 a flow capacity of L俑， and route 5-4 a flow capacity of

LS4' then we have four additional constraints:

For route 2-3: X23 三 L23

For route 5-2: XS2 三 LS2

For route 4-6: X46 三 L46

For route 5-4: XS4 三 LS4

3.4 Earth Moving Optimization

Example3.4

(1 0)

(1 1)

(12)

(1 3)

A highway contract requires a contractor to alter the terrain of a section of road

work. The work involves cut and fill of earth so that the origina1 profile will be

much flatter. The original profile and the finished profile are shown in solid line

and dotted line respectively in Fig. 3.4.

Section No. 2 3 4 5 6 7 8 9 Cut (m3xl03) 28 32 13

Fill (m3xl03) 10 14 16 10 12 11

Fig. 3.4 Original and finished profiles of cut and fill


The roadway is divided into nine sections. For ex缸nple， 10 x 103 m3 of fill is

required in section 1; 28 x 103 m3 of cut in section 2; and so on. The cost for

cutting including loading is $8 per m3 and that for filling including compaction

is $12 per m3• The unit cost of transporting earth by trucks from one section to

another is $2 per m3 per section.

Our task is to determine how much earth should be moved from where to where

so as to optimize the earth moving operations.

Solution 3.4 Let xij be the quantity of earth in thousand m

3 to be moved from section i to

section j (i *- j).

Therefore, there are 18 (i.e. 3 x 6) decision variables as follows:

Fill section

4 5 6 8 9

2 X 21 X24 X25 X26 X28 X29

Cut Section 3 X 31 X 34 X 35 X36 X 38 X 39

7 X71 X74 X75 X76 X78 X79

Cij can be calculated as follows:

C21 cost per m3 for moving earth from section 2 to section 1

= $8 + $2 X 1 + $12 = $22 •••

Cut Transport Fill

C24 cost per m3 for moving earth from section 2 to section 4

= $8 + $2 X 2 + $12 = $24

••• Cut Transport Fill

Other Cij can be calculated in the same way.


Now, we can write the objective function:

Minimize P = I,Cjj X ij = 22x21 + 24x24 + 26x25 + 28x26 + 32x28 + 34x29

+ 24x31 + 22x34 + 24x35 + 26x36 + 30X38 + 32x39

+ 32x71 + 26x74 + 24x75 + 22x76 + 22x78 + 24x79

Constraints:

The constraints for cuts:

X21

+ X24

+ X25

+ X26

+ X28

+ X29

豆 28"，，，，，，，，，---，，，，呵呵呵呵 …_.....-一 … ………- (1)

X31 + X34 + X35 + X36 + X38 + X39 至 32 -…ω 一………"………一…-，-" (2)

X71 + X74 + X75 + X76 + x78 + X79 三 13

The constraints for fills:

X21 + X31 + X71 = 10

X24 + X34 + X74 = 14 X25 + X35 + X75 = 16

X26 + X36 + X76 = 10 X28 + X38 + X78 = 12

X29 + X39 + X79 = 11

(3)

(4)

(5)

(6)

(7)

一……一… (8)

(9)

all xij 這 o for i = 2, 3, 7 and j = 1, 4, 5, 6, 8, 9

The solution for this model is given below:

Minimum overal1 earth moving cost = 1,816 X 103

X21 = 10

X 24 = 8

X 28 = 10

X 34 6

X 35 16

X 36 = 10

X78 = 2

X79 = 11

other X jj = 0

29

30 Linear Optimi世ation 的 Applications

We can observe that the model formulated for this earth moving problem is

similar to the one for Example 3.1, the transportation problem. In fact, this

example is indeed a transportation problem which can be solved by the simplex

method, or an algorithm which will be discussed in Chapter 4.

3.5 Production Schedule Optimization

Example3.5

A cement manufacturer provides cement for ready-mix concrete companies. In

the next four months its volume of sales, production costs and available worker-

hours are estimated as follows:

Month

2 3 4

Cement required (m3 x 103) 220 300 250 320

Cost of regular time production 2,500 2,600 2,700 2,800 ($ per thousand m3)

Cost of thoovuesratinmd e mp3r) oduction 3,000 3,100 3,200 3,300 ($ per

Regular time worker-hours 2,000 2,000 2,000 2,000

Overtime worker-hours 800 800 800 800

There is no cement in stock initially. It takes 10 hours of production time to

produce one thousand m3 of cement. It costs $150 to store one thousand m3 of

cement from one month to the next.

Now, the cement company wishes to know the optimal production schedule.

Solution 3.5 Let x\ , x2, x3 and x4 be the number of thousand m

3 of cement produced in months

1, 2, 3 and 4 respectively on regul訂 time.

Let y\ , y2' y3 個d Y 4 be the number of thousand m3 of cement produced in months

1, 2, 3 and 4 respectively on overtime.


Let Zl' Z2 and Z3 be the number of thousand m3 of cement in stock at the end of

months 1, 2 and 3 respectively which have to be carried to the next.

After defining the decision variables, the problem can be presented

diagramatica11y as shown in Fig. 3.5.

220 300 250 320

Fig. 3.5 Cement company production model

The objective function is:

Minimize P = 2500x1 + 2600x2 + 2700x3 + 2800x4

+ 3000Yl + 3100Yl + 3200Y3 + 3300Y4

+ 150z1 + 150z2 + 150z3 subject to

Monthly demand constraints:

弘10nth 1 : x1 + Yl - Zl = 220 、‘，/

',A J'

.‘、、

"

Month 2: x2 + Y2 + Zl - Z2 = 300 心 F 叮叮叮叮叮。"一 …。...“白白叫 “………… •. (2)

Month 3: x3 + Y3 + Z2 - Z3 = 250 叮叮叮叮-_....- 叫叫一一_......._.…… 。… (3)

Month 4: x4 + Y4 + Z3 = 320 …一一一一一一一一一一一…… (4)

Monthly regular worker-hours constraints:

10x1 ::; 2000

10x2 三 2000

10x3 三 2000

10x4 三 2000

ω 叮叮叮 (5)

一一一一一一 (6)

"………一一"…"叮叮_..'一 (7)

……………………一……一一… (8)

32 L的ear Optimization in Applications

Monthly overtime worker-hour constraints:

10Yl ~三 800

10Y2 三 800

10Y3 ~ 800

10Y4 ~三 800

一一……一一……… (9)

(10)

一一一一…一一一…(1 1)

一一一一一一一一一一一(12)

all X j ~ 0, yj ~三 o and Zj ~三 O


Total production cost = 2,458,500

X1 = 200 Yl = 50 Zl = 30 X2 = 200 Y2 = 80 Z2 = 10 X3 = 200 Y3 = 80 Z3 = 40 X4 = 200 Y4 = 80

3.6 Aggregate Blending Problem

Example3.6

The contract specifications of a building project require that the course

aggregate grading for concrete mixing must be within the following limits:

Sieve size

63.0mm

37.5 mm

20.0mm

10.0mm

5.0mm

Percenta2:e nassin2: bv wei2:ht

100%

95% to 100%

35% to 70%

10% to 40%

0% to 5%

The contractor has four quarry sites to supply aggregates. These aggregates,

however, do not individually satisfy the above specifications requirement. The

aggregate grading from each of qu缸Ty sites is shown below:


Quarry sites

% by wt. passing each sieve 2 3 4

63.0mm 100% 100% 100% 100%

37.5 mm 90% 95% 100% 100%

20.0mm 50% 的% 80% 100%

10.0 mm 0% 25% 40% 90%

5.0mm 0% 0% 3% 30%

Transportation cost per tonne 12 14 10 11

The contractor needs to prepare 150 tonnes of aggregates for concrete mixing in

the next two days. How should he obtain the aggregates from the quarry sites so

th前 after blending them the aggregates wi11 satisfy the specifications?

Solution 3.6 Let X) , X2, X3 and X4 be the number oftonnes of aggregates supplied from quarry

sites 1, 2, 3 and 4 respectively.

The objective function:

Minimize P 12x) + 14x2 + 10月+ llX4

Constraints:

The constraints for 37.5 mm sieve:

0.95 至 0.90x) 十 0.95x2 + 1.00x3 + 1.00x4 三1.00X 1 + X2 + x3 + X4

This can be written as:

0.05x) - 0.05X3 - 0.05X4 三 O

and 0.10x) + 0.05X2 注。


0.35 ~三 0.50x) + 0.65x2 + 0.80X3 + 1.00x4 三 0.70x) + X2 + X3 + X4


0.15xl + 0.30X2 + 0.45X3 + 0.65x4 ~ 0

(1)

(2)

(3)


and - 0.20x) - 0.05x2 + 0.10x3 + 0.30x4 ~三 O


0.10 ~三 0.00x) + 0.25x2 + 0.40x3 + 0.90x4 三 0.40x) + x2 + x3 + x4


(4)

- 0.10x) + 0.15x2 + 0.30x3 + 0.80x4 ~ 0 …一一………一一.. (5)

and - 0.40x) - 0.15x2 + 0.50x4 ~ 0 個…一一…一…一 (6)


o 三 0.00x) + 0.00x2 + 0.03x3 + 0.30x4 三 0.05x) + x2 + x3 + x4


0.03x3 + 0.30x4 ~ 0 “………… (7)

and - 0.05x) - 0.05x2

- 0.02x3

+ 0.25x4 ~三。 …呵呵呵呵呵 …叮叮叮叮叮叮… (8)

The constraints for demand:

x) + x2

+ x3

+ x4 = 150 .“白白白呵呵，“… 一叮叮叮叮叮叮個 …………一一一 (9)

Also, X) , x2，莉， x4 ~ 0

The solution for this model is given below:

Total minimum costs 1,690.16

x) = 65.26 tonnes

x2 = 10.04 tonnes

x3 = 55.22 tonnes

x4 19.48 tonnes

3.7 Liquid Blending Problem

In Example 3.6, we have seen a problem involving aggregate blending. For

blending liquids, the linear programming model can be formulated, in most

cases, in the very same way. However, we will now see an example of


optimizing liquid blending problern in which the requirernents are specified in a

slightly different way.

Example3.7

A paint manufacturing company produces three enamels types 1, 11 and 111 by

mixing acrylic polyrners in three different forrnulations , types A, B and C.

Type A polyrner contains 55% solids and 45% solvent, type B polymer 45%

solids and 55% solvent, type C polyrner 35% solids and 65% solvent. Polymers

A, B and C cost $6 per litre, $7.5 per litre and $9 per litre respectively.

Type 1 enamel must contain at least 30% solids and at least 50% solvent, type 11

enamel at least 40% solids but no more than 60% solvent, type 111 enarnel not

more than 60% solids and not more than 70% solvent.

The p也nt manufacturing company has got a rush order for 800 litres of type 1

enamel, 950 litres of type 11 enamel and 650 litres of type 111 enamel. How

many litres of each polyrner should the company purchase for producing the

required enarnels?

Solution 3.7 The problem can be presented diagramatica11y as shown in Fig. 3.7.

800 950 650

Fig.3.7 A liquid blending problem

36 L的ear Optimization in Applications

The objective function:

Minimize P = 6(xA1 + XA2 + XA3) + 7.5(xB1 + XB2 + XB3)

+ 9(xc1 + XC2 + XC3)

Constraints:

The constraint for enamel 1 solids:

0.55xA1 + 0.45xB1 + 0.35xc1 ~ 0.30 XA1 + XB1 + XC1

i.e. 0.25xA1 + 0.15xB1 + 0.05xc1 ~。一切一一一一一一…一一一一一…--- (1)

The constraint for enamel 1 solvent:

0.45xA1 + 0.55xB1 + 0.65xcl 主 0.50XA1 + XB1 + XC1

i.e. -0.05xA1 + 0.05xB1 + 0.1 5xc1 去。


0.55xA2 + 0.45XB2 + 0.35xc~ ~ 0.40 XA2 + XB2 + XC2

(2)

i.e. 0.15xA2 + 0.05XB2 - 0.05xC2 ~ 0 …………………一一…… (3)

The constraint for enamel 11 solvent:

0.45XA2 + 0.55xB2 + 0.65xcL 三 0.60XA2 + XB2 + XC2

i.e. -0.15xA2 - 0.05XB2 + 0.05xC2 至 O


0.55xA3 + 0.45xB3 + 0.35xc1 三 0.60XA3 + XB3 + XC3

(4)

i.e. -0.05XA3 - 0.15xB3 - 0.25xC3 三。一一“一一一一一 (5)


The constraint for enamel III solvent:

0 .45XA3 + 0.55xB3 + 0.65xc ::; 0.70 XA3 + XB3 + XC3

i.e. -0.25xA3 - 0.15xB3 - 0.05xC3 ::; 0

The constraints for quantities of enamels:

(6)

XA1 + XB1 + XC1 = 800 ………..._..“一……"… (7)

XA2 + XB2 + xC2 = 950 。……………。"……_..-... (8) XA3 + XB3 + XC3 = 650 叮叮叮叮一一…何呵呵… …….... (9)

and all xij 主 o where i = A, B, C

and j = 1, 2, 3


Total minimum purchasing costs 15,000

XA1 = 600

XA2 = 950 XA3 = 650

XB1 0 XCI 200

XB2 = 0 XC2 = 0

XB3 = 0 XC3 = 0

3.8 Wastewater Treatment Optimization

Example3.8

37

An industrial company has three factories located along three streams as shown

in Fig. 3.8.

Q1

Fig. 3.8 An industrial wastewater treatment problem

Factory A generates a daily average of 1000 m3 industrial wastewater of 800

mgllitre of BOD5 (biological oxygen demand 剖 5 days, a measure of degree of


pollution), factory B 1500 m3 of wastewater of 600 mgllitre of BO叭， and

factory C 1800 m3 of wastewater of 1000 mgnitre of BOD5• Before the

wastewater is discharged into the stre缸肘， the industria1 company has to build

in-house treatment plants at each of the factories so as to remove the pollutants

to a level which is acceptable at the downstream waters.

The costs for treating 1 kg of BOD5 at factory A is $100, factory B $110 and

factory C $120. The rates of flow per day in the streams are QI' Q2 and Q3

which are 0.2 million m3, 0.22 million m3 and 0.25 million m3 respectively. The

flows in the streams are assumed to have no pollutants unless they 訂e

contaminated by the discharges from the factories.

The river requirement is that no water in any p訂t of the streams will exceed an

average standard of 2 mgnitre of BOD5• It can a1so be assumed that 20% of the

pollutants discharged at factory A will be removed by natural processes (e.g.

sunshine, oxidation etc.) before they reach factory B, and that 15% of the

pOllutants be removed by natural processes between factory B and factory C.

The company wishes to know the optima1 treatment capacities of the in-house

treatment plants at the three factories.

Solution 3.8 Let x1 = kg of BOD5 to be removed daily at factory A

x2 = kg of BOD5 to be removed daily at factory B x3 = kg of BOD5 to be removed daily at factory C


Minimize P = 100x1 + 110x2 + 120月

subject to the following constraints:

The constraint for factory A:

Daily BOD5 generated by factory A = 800 mglL x 1000 m3

= 800 kg


.'. Daily BODs released into stream = 800 - X 1

River standard = 2 mglL and Q2 = 0.22 mi l1ion m3

:.Allowable pollutants in stream 2 after passing factory A = 440 kg

Hence, 800 - X 1 三 440

and Xl 三 800

The constraint for factory B:

Daily BODs generated by factory B = 600 mg/L X 1500 m3

= 900 kg

'. Daily BODs released into stream = 900 - X2

River standard = 2 mg/L and Ql + Q2 = 0.42 million m3

(1)

(2)

:.Allowable pollutants in stream 1 after passing factory B 840 kg

20% of pollutants are removed by natural processes

Hence, 0.8(800 - Xl) + (900 - X2) 主 840

and X2 至 900

The constraint for factory C:

Daily BODs generated by factory C = 1000 mg/L X 1800 m3

= 1800 kg

:. Daily BODs released into stream = 1800 - X3

River standard = 2 mg/L and Ql + Q2 + Q2 = 0.67 million m3

(3)

(4)

.'. Allowable pollutants in stream 1 after passing factory C 1340 kg

15% of pollutants are removed by natural processes

Hence,

0.85[0.8(800 - Xl) + (900 - X2)] + (1 800 - X3) 至 1340 一…… (5)

and X3 至 1800 一一一 (6)

Other constraints:

Xl 三 0 ， X2 2三 0， x3 ~ 0


Total minimum treatment costs = 222,200

39

40

x) = 360 x2 = 412

x3 = 1174

Linear Optimization in Applications

3.9 Critical Path of a Precedence Network

A precedence network is usually refe叮ed to as an activity-on-node network, that

is, an activity is denoted by a node in a network diagram which shows the

sequence of the activities in a logica1 manner. There are of course methods

other than linear programming for tracing a critica1 path in a network. The

following example, however, shows how linear programming can do such a job.

Example3.9

Fig. 3.9 An activity-on-node network

The durations of the activities of a project are shown in the network diagram in

Fig. 3.9 and are as follows:

Activity Acti vity duratio且~豆豆ks}

10 5

20 4

30 3

40 6

50 2

60 9

70 3


Find the shortest project duration (critical path) by the use of linear

programmmg.

Solution 3.9 Let x lO = start time of activity 10

X20 = start time of activity 20

X 70 start time of activity 70

X END = project duration

The objective is to:

Minimize P = X END

subject to

X 30 - X)。這 5

X40 - X20 ~ 4

X SO - X2。這 4

Xω - X 30 ~ 3

Xω- X40 ::三 6

X70 - Xs。這 2

X END - Xω~ 9

X END - X 70 ~ 3

一“…(1)

一.. (2)

(3)

........... (4)

m …一一一………........- (5)

(6)

(7)

一“ (8)

All Xj 這 o for i = 10, 20, 30, 40, 50, 60, 70, END


Shortest project duration = minimum X END = 19 weeks X lO = 2

X20 = 0

X 30 = 7

X40 4

XSO 14

Xω= 10

X70 = 16

X END = 19


Readers can a1ways check whether the project duration is 19 weeks or not using

the critical path method.

3.10 Time-Cost Optimization of a project Network

In Example 3.9, we have seen how the critical path or the shortest duration in

a project network can be found using the linear programming method. Now,

we shall see how the time-cost optimization can be done if the normal duration

& cost and the crash duration & cost of each activity in the network are given.

Example 3.10

The normat duration and cost and those of the crash for the activities of the

network in the previous example (Fig. 3.9) are as follows:

Normal Crash Activity Duration (R) Cost (U) Duration (Q) Cost (V)

10 5 weeks $100 4 weeks $120

20 4 150 3 170

30 3 150 3 150

40 6 300 4 400

50 2 200 2 200

60 9 550 5 990

70 3 100 2 150

The indirect cost (overheads and so on) of the project is $60 per week. How

should the activities be compressed so that an optimal project duration can be

achieved with the minimum total direct and indirect costs of the project?

Solution 3. 10 Before formulating the linear programming model, some preliminary theories of

the normal point and the crash point of an activity must be discussed.

Formulating Unear Optimization Problems

Activity cost

Crash v l-----------------

M

U 一

Q

Normal

Activity duration

43

When additional resources (hence additional cost) is used to carry out an

activity, it wi1l usually be completed faster than its normal duration. The crash

point is reached when the duration of the activity cannot be further shortened

even if extra resources are input. We denote the normal duration and crash

duration by R and Q respectively, and normal cost and crash cost by U and V

respectively.

The cost slope of an activity between normal and crash = C = 芷主1lR-Q

Now, we can calculate C of each activity:

Activity C= 主主Jl

R-Q

10 20

20 20

30

40 50

50

60 110

70 50

An activity can be compressed to any duration between R and Q. We assume

that D is the activity duration after compressing (R 三 D 三 Q) such that the time

and cost of the overall project is optimized. M is the corresponding cost

associated with D.


By similar triangles, we can easily derive that

D = R -已 (M - U)

Therefore, for each activity:

DIo=5-L(Mlo-lO0) lU 20 ,_._/U

D20=4.L (M20.150) 山 2。“

D30 = 3

D的 =6--L(M的- 300) 制 50 叫

D50 = 2

D印 =9. 」-(Mω- 550) 110 山

D7OU-4(M7。 "100)50 '~-'V

Using the formulation discussed in Example 3.9, the activity duration

constramts are:

X30 - X\O ;;::: D \O (Remember that Xj is the

X40 - X2。這 D20 start time of activity i)

X50 - X50 ;;::: D20

Xω - X3。這 D30

Xω - X4。這 D40

X70 - X50 三 D50

XEND - Xω;;:::Dω

XEND - X70 ;;::: D70

These can be written as:

30 - ^l。這 5 -土 (M\O - 100) ……一一…一…一一一 (1) 20

的- ^2。這 4-4(Mo-150) 一一一一一一一一一…一一 (2) 20 ' ~


。- ^2。三 4- 一 (M20 - 150) 一 (3)20

Xω- x3。三 3 一個 (4)

ω- 的這 6-4(Mω- 300) 一………一…………--一…… (5)50

X70 - x50 ~ 2 …… …一………叮叮叮叮一 ~....-“一一… (6)

xm-MM--L(Mω- 550) 110ω

(7)

7OM-L (M70-lO0) “ ……呵呵… …_._._.-.-......-… (8) 50

The second set of constraints is the minimum cost (or normal cost) constraints:

MJO ~ 100 “ (9)

M2。這 150 (10)

M30 ~ 150 (1 1)

M4Q ~ 300 ……白白白曰“……一… (12)

扎150 ~ 200 . (13)

Mω~550 • (14)

扎170 ~ 100 _.....一………………………(15)

The third set of constraints is the maximum cost (or crash cost) constraints:

MJO 三 120 .. (1 6)

M20 ~ 170 (17)

弘130 三三 150 (1 8)

M4Q至 400 “一…(19)

M50 至 200 (20)

Mω 三三 990 - (21)

M70 三三 150 (22)


Other constraints are:

M j ~三 o for i = 1, 2, .. ..., 7

and Xj ~三 o fori=I , 2,....., 7, END


Minimize P = M \O + M20 + M30 + M40 + Mso + Mω+ M70 + 60XEND


Total minimum project cost = 2,640

X\O = 0 M \O = 100

X20 0 M20 = 170

X30 = 5 M30 = 150

X40 = 3 M40 = 350

XSO = 3 Mso = 200

Xω= 8 Mω= 550

X70 = 14 弘170 = 100

X END = 17

Project duration = 17 weeks

We can observe that only M20 and M40 in the result are different from their

norma1 cost U20 and U40 respectively. By comparing M20 and U20, we can know

that activity 20 is shortened by 1 week. Similarly, we can know that activity 40

is also shortened by 1 week. The normal (origina1) project duration is 19 weeks

(see Example 3.9) and the optima1 project duration is found to be 17 weeks,

because the two activities (20 and 40) are each shortened by 1 week.


Exercise

1. Formulate the following trans-shipment problem as a linear programming model.

+12 Q) þ 0 Þ ø -4

x xJ; @-6 +10 G)' ;@' ;0) -8

The unit cost from route to route are as follows:

hZL 3 4 5 6 7 33 45

2 39 28 . .

3 . 13 16 20 4 . 19 18 11

2. A manufacturing process involves 4 stages 1, 11, 111 and IV by inputting two

different raw materials A and B. There are four final products 1, 2, 3 and 4

coming out from the processes as shown in the following diagram:


Raw Material A Raw Material B

250 m3 and 200 m3 of raw materials A and B respectively are available per

week. Raw material A costs $2.6 x 103 per m3 and raw material B $3.2 x 103

perm3

Plant 1 has a capacity of 400 m3 per week. In plant 1, processing costs for

process 1 and process II respectively are $2.5 per m3 and $5.0 per m3.

Plant 2 has a capacity of 200 m3 per week. In plant 2, processing costs for each

final product and the yields associated with the processes are:

Finalorodl Process III Process IV FP 2 0.6 ' FP 3 0.3 0.5 FP4 . 0.4 Loss 0.1 0.1

Processing cost/m3 $2.5/m3 $10.5/m3

The company has to supply in the next week a minimum of 50 m3 of final

product 1, a minimum of 25 m3 of final product 2, a minimum of 60 m3 of final

product 3, and a minimum of 40 m3 of final product 4 to its customers. If the

Formulatíng Unear Optímízatíon Problems 49

production of final product 3 is in excess, it will cost the company to store it at

$250 per m3• The selling price of the final products 訂e:

Final Product FP 1 FP2 FP3 FP4

Formulate a linear programming model for the problem so that the financial

return of the company is maximized.

In Example 3.1 of Chapter 3, we have seen what a typical transpo付ation

problem is. Examples 3.3, 3.4 and 3.5 are also transportation problems

although they are less obvious than example 3.1. Transportation problems can

be solved, of course, by the simplex method. However, there is an algorithm

which can also solve transp。吋ation problems without using the techniques of

the simplex method. In this chapter, we will discuss the techniques of this new

algorithm and will see how to use it to solve Examples 3.1 , 3.3, 3.4 and 3.5.

4.1 The General Form of a Transp。前ation Problem

Before discussing the general form of a transportation problem, we shall look at

the particular form again which has been illustrated in Example 3. 1. This

ex缸nple is now reproduced below for easy reference. Its transportation

tableau is as follows:

Customer 2 Supply

之j !QJ X l1 X12 12

~ 斗W缸e- 2 X21 X22 20

house 主j 叫

3 X31 X32 30

Demand 29 33

The warehouses 1, 2 and 3 supply 12, 20 and 30 tonnes of stock per week

respectively to customers 1 and 2 who demand 29 and 33 tonnes of stock per

week respectively. The cost of transporting one tonne of stock from warehouse


i to customer j is given in the small box at the left top corner of the decision

variable Xij. As already explained in Example 1.3, the linear programming

model for this typical transportation problem is:

Minimize P 9Xll + 10x12 + 6X21 + 7X22 + 8X31 + 11x32

subject to

Xll + X12 三 12

X21 + X22 三 20

X31 + X32 三 30);(1;ts

Xll + X21 + X31 = 29

X12 + X22 + X32 = 33

all xij 三 o for i = 1, 2, 3 and j = 1, 2

The above is a particular case of transportation problems. We can generalize it

and a general transportation tableau is shown in Fig. 4.1 below.

Demand stations

2 n Supply

叫心| 心lXll X12 Xln SI

臼 83 區|Supply 2 X21 X22 X2n S2

statlOns

巨型| 凶凶m Xml Xm2 Xmn Sm

Demand D1 D2 Dn

Fig. 4.1 A general transportation tableau

We can see that there are m supply stations and n demand stations. Cij is the

unit cost of transportation from supply station i to demand station j. Si is the

quantity available at supply station i and Dj is the quantity required at demand

Transportation Problem and Algorithm 53

station j. C ij , Si and Dj are given quantities in the problem. The decision

variables Xij are the ones which we want to find out by solving the problem. The

linear programming model can be formulated as:

Minimize P= C))X)) + C 12X12 + . .... ... + CijXij + . ........ + CmnXmn

口

z ECA (in a more compact form)

subject to

XII + X I2 + ……一一 + Xln 三 SI …一一……一………(1)

X21 + X22 + _...._._--… 一 + X2n 三 S2 一白白白白_..... “……… (2) I m supply

I constraints Xml + Xm2 +…一…叮叮叮 + Xmn ~三 Sm

Xll + X21 + 叮叮叮一…… + Xml Dl …(m+l)

X I2 + X22 +…一一… + Xm2 = D 2 (m~2) I叫nd

I constraints Xl n + X2n +一一一一一 + Xmn = Dn (m+n)

and all Xij ~三 O for i = 1, 2, ....., m and j = 1, 2, ....., n

4.2 The Algorithm

As mentioned before, there is an algorithm other than the simplex method to

solve transportation problems. The algorithm is usually referred to as

transportation algorithm. It will be described in this section using examples.

Example4.1

Use the transportation algorithm to solve the transportation problem in Example

3. 1.

54 Línear Optimization in Applications

Solution 4. 1

There are four steps in the a1gorithm.

品且i

Draw the transportation tableau as shown in Fig. 4.2.

Customer 2 Supply

之| 叫12

~ l工JW訂e- 2 20

house

3 主J i山

30

Demand 29 33 62 正ω~alt

Fig. 4.2 Transportation tableau for Example 4.1

It should be observed that in this example, the tota1 supply (12 + 20 + 30 = 62) is equal to the total demand (29 + 33 = 62). This will make our problem simpler.

Cases where total supply is not equa1 to total demand will be discussed later.

We must firstly find an initia1 feasible a11ocation. It is reasonable to start by

allocating as many tonnes of stock as possible to the route i-j with the lowest C吵

and in our ex剖nple， this is route 2-1 (C21 = 6). The maximum quantity that can

be allocated to this route is 20, because warehouse 2 can supply only 20 tonnes.

This is followed by the a110cation of the remaining demand to customer 1, who

will demand a further 9 tonnes, making a tot叫 of 29 tonnes. The most

inexpensive route to do so is through route 3-1 (C31 = 8).

After allocating 9 tonnes to route 3-1 , the items of stock remained in warehouse

3 wi1l be reduced to 21 (i.e. 30-9). Then, a110cate 21 tonnes to route 3-2 so a11

30 tonnes of stock in warehouse 3 will be used up. Then, allocate 12 tonnes to

Transportation Problem and A旬orithm 55

route 1-2 so that the to叫 demand of customer 2 can be satisfied. Note that the

supply of warehouse 1 is also automatically satisfied because in this problem

total supply is equal to total demand.

Now, we have finished the initial allocation and this initial feasible solution is

shown in the transportation tableau in Fig. 4.3.

Customer 2 Supply

主j |且l12 12

~ 工jW訂'e- 2 20 20

house

主J 叫3 9 21 30

Demand 29 33 62 (total)

Fig. 4.3 Initial feasible solution

The total transportation costs for this initial feasible solution is

20 x 6 + 9 x 8 + 12 x 10 + 21 x 11

= 543

這起且2

Next, we have to see if this total costs can be further reduced by using the

unused routes. So we examine the unused routes (1-1 and 2-2 in our example)

one by one.

For route 1-1: if one tonne is allocated to route 1-1 , then one tonne must be

subtracted from route 3-1 , one tonne added to route 3-2 and one tonne subtracted

from route 1-2 in order to maintain the supply and demand to be satisfied (see

Fig.4.4).


Customer 2 Supply

主j |凶12 - 1 +1 12 ~ |工J

W訂e- 2 20 20 house

主j 門3 9 - 1 30

Demand 29 33 62 (tota1)

Fig.4.4 Testing route 1-1

The change in cost when 1 tonne of stock is a110cated to route 1-1

= C11 - C31 + C32 - CI2

= 9 - 8 + 11 - 10

= +2 (a positive cost change)

We usua11y call such cost change (+2 in this case) the improvement index of

the unused route 1-1. A positive improvement index means that the use of route

1-1 will increase the transportation cost rather than reducing it. Therefore, we

reject using route 1-1.

For route 2-2 : if one tonne is a110cated to route 2-1 , then one tonne must be

subtracted from route 2-1 , one tonne added to route 3-1 and one tonne subtracted

from route 3-2. Fig. 4.5 shows this modification.

Transportation Problem and Algorithm

Customer 2

主j I!QJ 12

~ 斗W訂e- 2 20 - 1 +1

house 主j 叫 21 - 1 3 9+1

Demand 29 33

Fig. 4.5 T esting route 2-2

The improvement index of the unused route 2-2

= C22 - C21 + C31 - C32

= 7-6+8-11

= -2 (a negative cost change)

57

Supply

12

20

30

62 (total)

A negative improvement index indicates that a reduction in transportation cost is

possible by using the unused route 2-2.

S且主

Now, we try to utilize route 2-2 and use it as fully as possible. Let x be the

maximum number of tonnes of stock th剖 route 2-2 c組 be allocated (see Fig.

4.6).


Customer 2 Supply

之j 些l12 12

~ l斗W訂'e- 2 20 - x +x 20

house

主l i叫3 9+x 21 - x 30


Fig. 4.6 Allocating x to route 2-2

By inspection, the maximum x is 20 since route 2-1 cannot tak:e a negative va1ue

if x is large than 20. So we allocate 20 tonnes of stock to route 2-2. The new

(or second) feasible solution is shown in Fig. 4.7.

Customer 2 Supply

1.J 11旦l12 12

~ |斗W訂e- 2 20 20

house 主j |叫

3 29 30


Fig.4.7 Second feasible solution

The tota1 transportation costs for this second feasible solution is

29 x 8 + 12 x 10 + 20 x 7 + 1 x 11

503


We can see that 503 now is lower than 543 which is calculated from the initial

feasible solution.

品且.4

We now repeat step 2 to see if the unused routes 1-1 or 2-1 have negative

improvement index.

Improvement index of route 1-1

= C l1 - C31 + C32 - C12

= 9 - 8 + 11 - 10

= +2


= C21 - C31 + C32 - C22

= 6-8+11-7

= +2

There is no negative improvement index. Therefore, the second feasible

solution is the optima1 solution with total transportation costs equal to 503.

Readers should now comp缸'e this result with that of Example 3.1 in Chapter 3.

4.3 A Further Example

In Example 4.1 , the total supply of the three warehouses is equal to the total

demand of the two customers. We sha11 now see an example with different total

supply and total demand.

Example4.2

Redo Example 4.1 if the supply of stock in warehouses 1, 2 and 3 per week are

15 , 35 and 32 tonnes respectively while the weekly demands of the customers

remain unchanged.

60

Solution 4.2

品且i

Linear Optimization 的 Applications

Since the total supply is greater than the tota1 demand, we introduce a dummy

customer, customer 3, and so a new column (column 3) is added to the

transportation tableau. The C3 in the column is equa1 to zero. This is shown in

Fig.4.8.

Customer 2 3 Supply

(dummv) 主j i立l 主j

15

豆」 l工j IJlJ W缸e- 2 35

house 主j |叫 IJLJ

3 32

Demand 29 33 20 82 (tot~n

Fig.4.8 Transportation tableau with dummy customer

We can observe that the demand of customer 3 (the dummy customer) is 20

tonnes, so that the total demand is made equa1 to the tota1 supply.

品且.2

Next, we use the same method as described in step 1 of the solution of Example

4.1 to obtain an initial feasible solution. In this case, the first a11ocation is to use

route 3-3 and put 20 tonnes in this route. Then the remaining procedures are the

same. The initial allocation is shown Fig. 4.9 below.


Customer 2 3 Supply

(dummv)

之j 凶 J?J 15 15

~ |工j 4 W缸'e- 2 17 18 35

house 主j 叫 IJLJ

3 12 20 32

Demand 29 33 20 82 (total)

Fig.4.9 Initial feasible solution

The unused routes are 1-1 , 1-3, 2-3 and 3-2. So we have to find the

improvement index of these routes.


= Cll - C21 + C22 - C12

= 9-6+7-10

= 0

For finding the improvement index of route 1-3, the method is a little more

complicated. The loop for it is not rectangular but irregul缸， as shown in Fig.

4.10.


Customer 2 3 Supply

(dummv) 之j |到15 - 1 IJ?J +1 15 ~ 叫“! |且|

W旺e- 2 35 house

主j UJ 主JI3 12 + 20 - 1 32


Fig. 4.10 The loop for calculating improvement index of route 1-3

The reason for using such a strange loop is that for finding the improvement

index of a particular unused route, no other unused routes should be involved.

Readers should have observed this point in the previous examples of calculating

improvement indices.

Now, the improvement index ofroute 1-3 can be calculated by:

C 13 - C 12 + C22 - C21 + C31 - C33

= 0 - 10 + 7 - 6 + 8 - 0

The calculations of improvement indices of routes 2-3 and 3-2 are rather simple,

and are shown as follows:


C23 - C21 + C31 - C33

= 0-6+8-0

+2

Transportation Problem and A旬orithm


C32 - C22 + C21 - C31

11-7+6-8

= +2

品且主

63

Now, we know that the improvement indices of unused routes 1-1 , 1-3, 2-3 and

3-2 征e 0, -1 , +2 and +2 respectively. We should use the unused route with the

most negative improvement index, and in our case it is route 1-3. So, we use

route 1-3 as fully as possible. Let x be the maximum number of tonnes of stock

that route 1-3 can be allocated (see Fig. 4.11).

Customer 2 3 Supply

(dummv) 主j l凶15 - x IJU +x 15 ~ 同時 +x IJU W缸e- 2 17 - x 35

house 主j 叫 JLJ

3 12 + x 32


Fig. 4.11 Allocating x to route 1-3

The maximum x is 15. So we allocate 15 tonnes of stock to route 1-3. The

second feasible solution is shown in Fig. 4.12.

64 L的ear Optimization 的 Applications

Customer 2 3 Supply

的ummv)

主j l到 IJU 15 15

~ |工j IJlJ W缸e- 2 2 33 35

house 主j 叫主l

3 27 5 32

Demand 29 33 20 82 (tota1)

Fig. 4.12 Second feasible solution

In the second feasible solution (Fig. 4.12), the unused routes are routes 1-1 , 1-2,

2-3 and 3-2. The improvement indices of these unused routes are ca1culated as

follows.

.

.且

•••

A hh3 1l mC VA EEA.

。

而‘J

ph

••

4‘J

mC AU n+ eEL--鬥U

句3

眩

C

副

Vl :l WAc h

= 9-8+0-0

= +1


= C I2 - C22 + C21 - C31 + C33 - C13 = 10 - 7 + 6 - 8 + 0 - 0

+1


C23 - C21 - + C31 - C33

= 0-6+8-0

= +2


句ru

司、d

e &E ••••

且

，圖，句、

d

mC TA "﹒?且，

OI F‘ ..

而F缸

白

C

M+8 Mn6 時

C+

臼

-7

V2. 。缸，

i

勻，

FCl+ h==

Since a11 the improvement indices are positive, there cannot be further

improvement. Therefore we stop here and conc1ude that the second feasible

solution is the optima1 solution. Curious readers may use the simplex method to

check the solution.

We should have observed that there are usua11y m+n-l used routes in a feasible

solution. If fewer routes are used and the supply and demand conditions are

satisfied, we ca11 such a case degeneracy. When a non-optima1 degenerate

feasible solution occurs, we have to add a zero a11ocation to a route and treat it

as a used route such that the total number of used routes is equa1 to m+n-l in

order that the algorithm can be continued.

4.4 More Applications of Transp。吋ation Algorithm

In Sections 4.2 個d 4.3, we have seen how to use the transportation algorithm to

solve a typica1 transportation problem. However, some linear programming

problems which are less obvious compared with these two examples can also be

solved by the a1gorithm. They are problems in Examples 3.3, 3.4 and 3.5 of

Chapter 3.

4.4.1 Trans-shipment Problem

Firstly, let us see Example 3.3 of Chapter 3. It is a trans-shipment problem. The

problem is reproduced below for easy reference.

66

C25 (22)

C52 (1 8)

Linear Optimization 的 Applications

We can construct a transportation tableau for this problem. In the tableau, non-

feasible routes (such as 2-1 or 4-3) are excluded by assigning Cij as a big-M unit

cost. Routes from a node to the same node are assigned zero Cij where i = j. Fig.4.13 shows the required transportation tableau for this trans-shipment

problem (the problem without modification, see Solution 3.3).

s、u、4ur》cesetfn\atlOns 2 3 4 5 6 Supply

.!.Ql 凶凶凶坐i18

2 ~ 凶坐l 凶坐l

B

3 坐l U 盟凹凶

B - 14

但l 些l ~ w 凶4 B + 12

5 自i 主l 且l 主l 間

B

4B + 16 Demand B B B B 16 (total)

Fig. 4.13 Transpo吋ation tableau for trans-shipment problem (Example 3.3)

In this problem, only node 1 is strictly a source and node 6 strictly a destination.

A l1 other nodes have arrows both in and out so that they can be both sources and

destinations and are trans-shipment nodes. Node 1 has a supply of 18 and node

6 has a demand of 16 since they are strict1y source and destination. A buffer


stock (i.e. B) that is sufficiently large for feasible moves must be introduced at

the trans-shipment nodes. In this problem, the buffer stock B is equal to 30

because the total tonnes to be moved is 30 and is big enough for its purpose.

For demands, all the nodes except node 6 (a strict demand node) have been

given demands of B items (or 30 tonnes). But for supplies, they are a little

more complicated. Those nodes with no net increase or decrease in stock (nodes

2 and 5) have demand equal to supply, are assigned value B. However, node 3,

which has a net demand of 14, has been given a supply of B - 14 (or 30 -14 = 16) tonnes. Node 4, which has a net supply of 12 tonnes , has been given a

figure of B + 12 (or 30 + 12 = 42) tonnes.

Readers may use the transportation algorithm to solve this trans-shipment

problem. The optimal solution is shown in Fig. 4.14.

、s\ouEmheestSmiat、:ions 2 3 4 5 6 Supply

且凶坐l 且l 也l18 18

心j @Ql 且l 且l 坐l2 12 14 4 30

也l U 世l 叫凶3 16 16

且i 叫 U llJJ u1 4 30 12 42

區l 且l lu1 企」 l1Æ 5 26 4 30

136 Demand 30 30 30 30 16 (total)

Fig. 4.14 Optimal solution for Example 3.3

Readers may comp缸e this solution with the one given in Chapter 3 (Solution

3.3).


4.4.2 Earth Moving Problem

Now, let us look at Example 3.4 of Chapter 3. It is an earth moving

optimization problem. In fact, it is very similar to a typica1 transportation

problem. Its transportation tableau is shown in Fig. 4.15.

Fill Section

4 5 6 8 9 Supply

隍l 且l 起l ~ 隍l 自主l2 28

Cut μl 平l 起l 且l 凶i 且l3 32

sechon

平l 且l 但l 區l 每l 且l7 13

73 Demand 10 14 16 10 12 11 (total)

Fig. 4.15 Transportation tableau for earth moving problem (Example 3.4)

Readers may try solving this problem using the transportation a1gorithm and

comp征e the resu1t with that of Solution 3.4 in Chapter 3. The optimal solution

is shown in Fig. 4.16.

Fill Section

4 5 6 8 9 Supply

~ @1J 且i 凶 22l 10 主l2 28

Cut 起l 凶起l 起l ßllJ 且i3 6 16 10 32

sectIon 陣l 起l 起l 陸l 問2 問117 13

73 Demand 10 14 16 10 12 11 (total)

Fig. 4.16 Optimal solution for Example 3.4


4.4.3 Production Schedule Problem

Example 3.5 of Chapter 3 is a production schedule problem. Apparently it is

hard to do

linear optimization in applications -...

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