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Operations Research 1

1. Introduction

1.1. Definition of Operations Research

There have been various definitions for Operations Research (O.R.) like applieddecision making, quantitative common sense, making of economic decision, etc.

Here we pick one of the most appropriate;

Definition: Operations research

Operations research is the application of up to date scientific methods,

techniques by inter-disciplinary teams to problems involving control of or-

ganized systems so as to provide solutions which best serve the purposes of

the organization as a whole.

Operations Research provides executive departments with a quantitative basis

for decision making regarding the operations under their control.

1.2. Characteristics of Operations Research

Interdisciplinary team approach:- it is developed by a team of scientists drawn

from various disciplines. e.g. mathematicians, statisticians, economists, engi-

neers, etc.

Systems approach:-emphasis is on the overall approach to a system in order to

get the optimum decisions.

Helpful in improving the quality of solutions-Doesnt give perfectanswers but

merely givesbadanswers to the problems which otherwise have worst answers.

Scientific method:- Operations Research uses techniques of scientific research

Goal Oriented Optimum Solution:- Tries to optimize a well-defined function

subject to given constraints (optimization).

Uses models:- Operations Research uses models built by quantitative measure-

ment of the variables concerning a given problem. -It also derives solutions

from the models.

Requires willing executives:- For experiments of alternative solutions.

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Reduces complexity:- Simplifies the work of executives especially due to prior

experimentations.

Operations Research is therefore both a science and an art.

1.3. Methodology of Operations Research

(a) Problem identification: Recognition that a problem exists is very

important in any management decision-making process, but in

practice its timing may be critical (resolve an existing problem

or to forestall a predicted problem).

(b) Formulating the problem: Once it becomes apparent that a prob-

lem exists-and a solution is required-the problem must be explicitlyformulated in terms of;

The perceived boundaries or limits to the problem.

The objectives of the investigation.

The defined roles of those involved in the investigation.

The decision variables making up the problem are within the

control of the decision-maker and which are not.

(c) Constructing the model: This step involves constructing the model

or the mathematical expressions describing inter-relations of all

variables and parameters in the study. The model must include an

objective function which defines the measure of effectiveness of the

system and the constraints or the restrictions.

(d) Deriving the solution: This involves finding the optimal values

of the controlled (independent) variables that produce the best

performance of the system for specified values of the uncontrolled(dependent) variables. An optimum solution is determined on the

basis of the various equations of the model satisfying the given

constraints and optimizing the objective function.


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(e) Testing the model validity: The solution values of the model, ob-

tained at solutions stage are then tested against actual observa-

tions. In other words, effort is made to test the validity of the

model used. A model is supposed to be valid if it can give reliable

prediction of the performance of the system represented through

the model. In effect, performance of the model must be compared

with the policy or procedure that it is meant to replace.

(f) Controlling the solution: This step of an O.R establishes control

over the solution by proper feed-back of information on variables

which might have deviated significantly.

(g) Implementing the results: Implementing the results constitutes the

last step of an OR study. Because the objective of O.R. is not

merely to produce reports but to improve the performance of sys-

tems, the results of the research must be implemented, if they are

acceptable.

The nature of a problem dictates the Operations Research method to be used

among the available ones.

1.4. Some useful definitions

(a) Model: This is a representation or abstraction of the real/actual

object.

(b) Objective function: This is a mathematical function decision to be

optimized.

(c) Constraints: Equations or inequalities representing the restrictions

imposed on the decision variables.

(d) Model formulation: The process of determining the objective func-

tion and constraints each expressed in terms of decision variables.

(e) Feasible solution: Set of the decision variables which satisfy all the

constraints.


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2. Classification of Problems in O.R.

Broadly speaking, problems in O.R. can be categorised; Some of the categories

are;

1. Allocation: Allocation problems involve the allocation of resources to

activities in such a manner that some measure of effectiveness is opti-

mized. E.g. Jobs to applicants, Money to investment projects

2. Replacement: Replacement problems are concerned with situations

that arise when some items (such as machines, electric light bulbs, etc.)

need replacement because the same may deteriorate with time or may

break down completely or may become out-of-date due to new devel-

opments. Replacement problems thus occur when one must decide the

optimal time to replace equipment for one reasons or the other.

3. Sequencing: Sequencing problems are the problems concerned with

placing items in a certain sequence or order for service. For instance,

N-jobs requiring different amounts of time on different machines must

each be processed on M-machines in the same order with no passing

between machines, then the question: How should the jobs be ordered

for processing to minimize the total time to process all of the jobs on allof the machines constitutes an example of a sequencing problem.

4. Routing: Routing problems are problems related to finding the optimal

route from an origin to a destination when a number of alternative routes

are available. For example, a salesman may wish to visit each of N-

cities once and only once before returning to his headquarter, then his

problem is: In what order should he visit the cities so that the overall

distance travelled is minimized? Such a problem is referred to as a

routing problem.

5. Inventory: Inventory problems are problems with regard to holding

or storing resources. The decisions required generally entail the deter-

mination of how much of a resource to acquire or when to acquire it.

The problem of deciding how much of a certain commodity to hold in


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inventory is one of real concern to business and industrial houses. In-

ventory problem is the problem to determine the level of inventory that

will optimize the measure of effectiveness.

6. Queuing: Queuing problems or what are known as waiting-line prob-

lems are problems that involve waiting for service, Queuing problems

encircle us from the time we rise in the morning until we retire at night

In business world several types of interruptions occur: facilities break

down and require repair, power failures occur, workers or the needed

material do not show up where and when expected. Allocation of facili-

ties Considering such interruptions be done and to do so means solving

a queuing problem.

7. Competitive: arise when two or more people are competing for a

certain resource which may range from an opponents king in a game of

chess to a larger share of the market in business world

8. Search: Search problems are problems concerned with searching for

information that is required to make a certain decision, The problem

concerning exploring for valuable natural resources like oil or some other

mineral is an example of a search problem.

2.1. Operations Research Techniques

Mathematical models have been constructed for the above categorized O.R.

problems and methods for solving the models are available in many cases.

Such methods are usually termed as O.R. techniques. Some of the important

O.R. techniques often used by decision-makers in modern times are;

(a) Programming (Linear, Non - linear Programming, Dynamic, Heuris-

tic, Integer, Algorithmic etc.)

(b) Queuing theory/waiting line

(c) Inventory Analysis

(d) Network analysis


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(e) Simulation

(f) Game theory

(g) Decision theory

Exercise 1.Read and make some notes on

1.Significance of Operations Research

2.Limitations of Operations research

Begin Quiz

Answer each of the following questions before you proceed.1.(2mks) OR is because it is developed by math-

ematicians, statisticians, economists, engineers, etc.

2.(2mks) is the process of determining the objec-

tive function and constraints each expressed in terms of decision variables.

3.(2mks) is the first of the five steps of an OR

inquiry

4.(2mks) Under which category of OR problems is Decision theory?

End Quiz

Marks: Percent:

Comments:


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3. Linear Programming

Linear Programming deals with problems in which linear functions are to be

optimized(maximized or minimized) subject to constraintswhich are usually

specified by linear inequalities, linear equations and to the condition that all

the variables must assume negative values.

The general formulation of linear programming problems is as follows.

Optimize the objective function

Z=c1x1+c2x2+...+cnxn

Subject to the linear constraints

a11x1+a12x2+...a1nxn{, =,}b1

a21x1+a22x2+...a2nxn{, =,}b2

.

.

.

am1x1+am2x2+...amnxn{, =,}bm

and to the non-negative condition

x1 0, x2 0, x3 0,...xn 0

There are many problems in engineering management and social sciences that

can be formulated this way.

In matrix form we can write as

OptimizeZ=c X subject to A X{, =,}B, X0.

where the underline denotes a vector and bold denotes a matrix.

Optimal feasible solution

In a linear programming problem a set of the values of the decision variables

x1, x2,...xnthat satisfy the linear constraints and the nonnegative condition is

called a feasible solution


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Definition: Optimal feasible solution

Optimal feasible solution is the set of decision variables which is feasible

and optimizes the objective function.

In order to obtain the Optimal feasible solution, we need to have some concepts

from linear algebra.

3.1. Convex analysis

Definition: A line in Rn or n-dimensional space

The set

L(x1, x2) ={X|X=x1+ (1 )x2, (0, 1)

That is any point X on the line joining x1 andx2 can be written as

X=x1+ (1 )x2 whereis a parameter such that 0 < } and s2 = {X

|aX < }


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IfHis included in the spaces s1and s2, then we have the closed half spaces.

sc1 = {X|aX} and sc2 = {X|a

X}

Example. Show that a closed half space sc1 ={X|aX} is a convex set.

LetX1 and X2 be two points in sc1 thena

X1 and aX2

then any point X on the segment joining x1 and x2 is given by

X=X1+ (1 )X2 (0, 1)

multiplying both sides bya gives

aX = aX1+ (1 )aX2

+ (1 )

aX +

aX

which is a member ofsc1. SinceX1 andX2 were arbitrarily chosen, the set

sc1 is a convex set.

Theorem 3.1 The intersection (null set) of convex sets is also convex. Let

x1 andx2 be two points in C=c1 c2... cn (Ci is a convex set i). SinceCi

is convex for all i, then the line segment joiningx1 andx2 is entirely in setCi

for all i. This line is entirely inC C.

Convex polytype- This is a set which is an intersection of a finite number of

closed half-spaces (e.g. For a polygon in R2. X Y

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Theorem 3.2 A set of convex combinations of the points x1, x2,...,xn C

where C in convex is also convex. LetCbe the set of all convex combinations

of x1, x2,...,xn, let Y1 and Y2 be two points in C, then Y1 =n

i=1

ixi and

Y2 =n

i=1

ixi

i=

i =1 i, i 0. Letx0 be any point on the

line segment joining Y1 and Y2, then x0 can be written as x0 = Y1 + (1

)Y2 (0, 1)

x0 = Y1+ (1 )Y2

=n

i=1

ixi+ (1 )n

i=1

ixi

=n

i=1

(i+ (1 )i)xi =n

i=1

ixi

i= i+ (1 )i and (0, 1)

Sincei 0, i 0 theni 0 and again

n

i=1

i=n

i=1

i+ (1 )i

=

i+(1 )

i

=+ 1 = 1

Thusx0 =n

i=1

ixi wherei 0andn

i=1

i= 1. the line segment joiningY1 and

Y2 inCis therefore entirely contained inC andC is convex.

Definition: convex combination

A convex combination of n+1 points is known as n-simplex. i.e. if it has 3

points (non-linear) then it is a 2-simplex

Example. Prove that any point inside the triangle is a convex combination

of x1, x2andx3 Let x0 be any point inside the triangle. By definition x0 =

x1+ (1 )Y (0, 1)

But

Y =x2+ (1 )x3(on linex2, x3) (0, 1)


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Therefore

x0= x1+ (1 )[x1x2+ (1 1)x3]

=x1+1x2+ (1 1)x3 1x2 (1 1)x3

=x1+ ( 1)x2+ (1 1 +1)x3.

x0=3

k=1

kxk

where1 =

2 = 1

3 = 1 +1

x0 =3

k=1

kxk = 1x1+ 2x2+ 3x3. Since 0 < < 1and0 < 1 < 1 then

is 0 for all i (i = 1, 2, 3).

Now1+2+2= +1+11+1 = 1 therefore3

k=1

k = 1 x0

is a convex combination ofx1, x2 andx3.

3.3. Extreme points of a convex set

Let C be a convex set. A point Z is an extreme point in C if and only if its not

possible to find points X and Y in C such thatZ=X+(1)Y

(0, 1).We note that the extreme points are actually the vertices or corner points.

Example

Consider the figure

figure

It is not possible to locate two distinct points in or on the above figures(Convex

sets) with the property that the line joining these points will include the ex-

treme points (E) and be included in the convex set.

Theorem 3.3 The convex set of feasible solution of linear programming prob-

lem has at least one extreme point and at most a finite number of extreme points.

The objective function is optimized at one of the extreme points.


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Note that the convex set in a linear programming problem is the constraint

set.

3.4. Optimal solutions to linear programming (LP) problems

In Linear Programming problem we are required to optimize the objective

function

p= CX

under a constants set. Suppose that R is the constraint set of the given

problem, then the optional solution is an extreme point ofR

Theorem 3.1 A basic feasible solution of a linear programming problems is

an extreme point of the constraint set

3.5. Formulation and solutions to LP problems

Formulation of Linear Programming models is the process of deter-

mining the objective function and the set of constraints into a mathematical

model each stated in terms of the decision variables. Solution is then the de-

termination of the optimal feasible solution. This can be done graphically (for

a maximum of 2 decision variables or using simple algorithm.

Graphical method

This involves representing all the constraints on a graph. The region of inter-

section represents the feasible region and by theorem (3.1), one of the extreme

points gives the optimal solution.

Example. The manager of a theatre which has a capacity of 300 seats sells

tickets to children and adults at Shs.10 and Shs.50 respectively. To cover his

rental expenses, he has to take at least Shs. 2500 for each show.

It is the companys policy to have at least 100 seats spared for children.Formulate this problem for profit maximization and solve it graphically.

Solution: Letx and y be the number of 10/= and 50/= seats occupied respec-

tively. Then we need to;

Maximize 10x+ 50y subject to;


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x+y 300 (restriction on the hall capacity)

10x+ 50y 2500 (restriction on total collection)

x 100 (restriction on seats for children)

y 0

Graph

Figure 1: One point at the vertex (corner) maximizes the

profit. At point;, Testing for optimal solution. At point

A(100, 30), z= 10(100) + 50(30) = 2500

B(100, 200), z= 10(100) + 50(200) = 11000

C(300, 0), z= 10(300) + 50(0) = 3000

D(250, 0), z= 10(250) + 50(0) = 2500

The maximum occurs at x = 100,y = 200, the firm should therefore sell 100seats/tickets for children and 200 for adults in order to make a maximum

profit of Ksh 11,000.

Example. A farmer has 50 of land to on which to plant maize and beans. He

has a workforce of 150 laborers and it takes 4 laborers to work on 1 ha of maize

and 2 laborers to work on 1 ha of beans. He has a capital of $4500 and 1 ha

of maize requires $50 to cultivate (inputs) while 1 ha of beans requires $100.

Suppose that the farmer wishes to maximize the profits and the profit per ha

is $30 for maize and $40 for beans, set up a linear programming problem and

solve it.

Solution


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Let x andy be the number of ha on maize an beans respectively then, we

have

Maximizep = 30x+ 40y subject to

x+y 50 (Constraint on land)

4x+ 2y 150 (Constraint laborers)

50x+ 100y 4500 (Constraint on capital)

x, y 0 (Non-negativity condition on decision variables)

We can represent the constraints in a graph as follows

Figure 2: Testing for optimal solution. At point

A(0, 0), p= 30(0) + 40(0) = 0, B(0, 45), p= 30(0) + 40(45) = 1800

C(10, 40), p= 30(10) + 40(40) = 1900 B(25, 25), p= 30(25) + 40(25) = 1750

(37.5, 0), p= 30(37.5) + 40(0) = 1125The maximum occurs at x = 10, y= 40, The farmer should therefore

cultivate 10 ha of maize and 40 ha of beans order to make a maximum profit

of $1900.

3.6. The simplex method

This is a step by step method which is used to solve linear-programming prob-

lems with any given number of decision variables

Method

Set objective function p = 0 and decision variables X to zero (0). (No pro-

duction).

Move progressively till no more combination can be made. i.e. Each step


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should give a better Solution as you increase or decrease the decision vari-

ables.

Definition: Slack variable

Slack variable a non-negative variable which when added to a less that or

equal to inequality makes it an equation.

Example.

x+y 7

but

x+y+s1= 7

Thens1 is a slack variable

Definition: Surplus variable

Surplus variable a non-negative variable which when added to a greater

than or equal to inequality makes it an equation.

Example.

x+y 7

but

x+y = s2+ 7

Thens2 is a surplus variable

Let x and y be the number of ha on maize an beans

respectively then, we have

Maximize p = 30x + 40y subject to

x + y

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which simplifies to;

Maximize p = 30x + 40y subject to;

x + y < = 5 0

2 x + y < = 7 5

x + 2y = 0

Adding slack variables s1, s2 and s3 and constructing the

initial tableau leads to

Tableau #1

Basis x y s1 s2 s3 p Solution

s1 1 1 1 0 0 0 50

s2 2 1 0 1 0 0 75

s3 1 2 0 0 1 0 90

p -30 -40 0 0 0 1 0

This solution Not optimal because there exists -ve values in

p row. Entering variable corresponds to the most negative value

p-row =>y Maximum y=min(50/1, 75/1, 90/2)=45

from 90/2, which implies that pivot value = 2 and departing

variable is s3

New row z=s3/2

Reducing all the elements of y column to zero using Jordan gausscomputations gives Tableau #2

Tableau #2


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s1 1/2 0 1 0 -1/2 0 5

s2 3/2 0 0 1 -1/2 0 30

y 1/2 1 0 0 1/2 0 45

p -10 0 0 0 20 1 1800

This solution Not optimal because there exists -ve values in

p row. Entering variable corresponds to the most negative value

in p-row =>x

Maximum x=min(5/(1/2),30/(3/2), 45/(1/2))=10 from 5/(1/2),

which implies that pivot value = 1/2 and departing variable is s4

New row z=s1/(1/2)

Reducing all the elements of x column to zero using Jordan gauss

computations gives Tableau #3

Tableau #3


x 1 0 2 0 -1 0 10

s2 0 0 -3 1 1 0 15

y 0 1 -1 0 1 0 40

p 0 0 20 0 10 1 1900

the corresponding solution is s1=0, s2=15, s3=0, x=10, y=40 and

Max p=1900.

The farmer should cultivate 10 ha of maize 40 ha of beans

in order to make a maximum profit of $1900

Example. A company manufactures two products A and B. The profit per

ton of the two products are $50 and $60 respectively. Both products require

processing in three machines M1, M2 andM3. With the following table giving

the details of (Hrs per 1 tonne)


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A B Total available (Hrs) pwk

M1 4 2 600

M2 3 4 500

M3 4 6 800

Assuming that no other constraints, find A and B that maximizes the

weekly profit

Solution: The solution is as follows

Maximize p = 50x + 60y subject to

4x +2y

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2. A company produces two types of leather belts; say A and B. Belt A is

of superior quality and belt B is inferior. Profits on the two are sh40

and shs 30 per belt respectively. Each belt of type A requires twice as

much time as required by a belt of type B. If all belts were of type B, the

company could produce 1,000 belts per day. But the supply of leather

is sufficient only for 800 belts per day. Belt A requires a fancy buckle

and only 400 of them are available per day. For belt B only 700 buckles

are available per day. How should the company manufacture two types

of belts in order to have a maximum overall profit? (solve it graphically

and using simplex method).

3. The manager of a hotel has sufficient money to buy a total of 100 cratesof soft drink of types A and B. He wants to buy at least twice as many

crates of type A as type B. He wants to buy maximum 80 crates of type

A and at least 10 crates of type B. Taking X to be the number of type

A crates and Y that of type B, write down all the inequalities based on

these facts. Show these inequalities on a graph and outline the region

in which (X, Y) must lie. The profit on a crate of type A is Shs. 60

and that of a crate of type B is Shs. 40. Find the number of crates of

each type that he should buy to make maximum profit and calculate this

maximum profit. (solve it graphically and using simplex method).

4. A chemical firm has 160 litres of solution A, 110 litres of solution B and

150 litres of solution C. To prepare a bottle of syrup X, 200ml of solution

A, 100ml of solution B and 100ml of solution C are needed. For a bottle

of syrup Y, 100ml of A, 200ml of B and 300ml of C are needed. Syrup

X sells at Shs. 60 per bottle and Syrup Y sells at Shs. 100 per bottle.

How many bottles of each type of Syrup should the firm make in order

to obtain the maximum amount of money? (solve it graphically).

5. A farmer requires 10, 12 and 12 units of chemicals A, B and C respec-

tively for his garden. A liquid product contains 5, 2 and 1 units of A, B

and C respectively per bottle. A dry product contains 1, 2 and 4 units

of A, B and C per carton while a paste product contains 1, 3 and 1 units

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of A, B and C per jar. If the liquid product sells at shs 30 per bottle,

the dry product sells at shs 20 per carton and the paste product at Ksh

25 per jar, How many of each should he purchase in order to minimize

the cost and meet the requirements? (Formulate and dont solve).

6. A farmer has 70 hectares of land available for growing maize and beans.

The cost of growing 1 ha of maize is $30 and the cost for growing 1 ha

of beans is $20 and the farmer has only $1800 available. The labor per

ha is 2 man days for maize and 4 man days for beans and a total of

240 man days of labour are available. If he makes a profit of $800 for 1

ha of beans and $700 for 1 ha of maize, formulate the underlying linear

program and solve it graphically and using simplex method.

7. OHagan Bookworm Booksellers buys books from two publishers. Duffin

House offers a package of 5 mysteries and 5 romance novels for $50, and

Gorman Press offers a package of 5 mysteries and 10 romance novels for

$150. OHagan wants to buy at least 2,500 mysteries and 3,500 romance

novels, and he has promised Gorman (who has influence on the Senate

Textbook Committee) that at least 25% of the total number of packages

he purchases will come from Gorman Press. Formulate the underlying

LP and solve it to determine the number of packages to be ordered from

each publisher in order to minimize the cost and satisfy Gorman? What

will the novels cost him? (Formulate and dont solve)

8. A company is test-marketing one of its new products in a particular

region and intends undertaking an extensive 1 week advertising campaign

to raise consumer awareness about the new product. The firm has an

advertising budget of500 000 and intends using regional TV, regional

radio and regional newspaper advertising. An advert on TV will cost

50 000 and is expected to reach 300 000 potential customers. For radio

the equivalent figures are 20 000 and 10 000 customer and for the

newspapers 2000 and 50 000 customers. The combination of adverts

on the three forms of media is flexible, although the company wants at

least 2 adverts on TV, 5 on radio and 10 in the newspapers. In addition


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the number of newspaper adverts must be no more than 2.5 times the

number of combined radio and TV adverts.Formulate the problem if the

company wishes to maximize the number of customers exposed to the

advertising campaign. (solve it graphically).

A company publishing textbooks is planning its production of the next book

scheduled to be printed. The book will be published in both paperback and

hardback format. The paperback sells for 10 per copy and costs 5 to pro-

duce and market. The hardback sells for 20 and costs 17. Market research

has indicated that total sales of the book are unlikely to exceed 10000 copies.

Of these at least 4000 are expected to be in paperback format with at least

2000 hardback. On the other hand, the company does not expect to sell morethan 4000 hardback copies. In addition, there are potential problems involved

in producing the paperback edition. The printing equipment is needed for

other paperback books and is available for printing this book for a period of

only 5000 hours. Each paperback takes 40 minutes to produce.

1. Formulate this problem and solve it assuming the company wishes to

maximize profit.(solve it graphically).

2. Assuming the company wishes to maximize revenue from sales refor-mulate this problem and solve it and computes the profit the company

would earn. (solve it graphically)

3. The production manager is strongly arguing that production should be

determined by costs. Reformulate the problem in terms of cost mini-

mization solve it . (solve it graphically).

4. Which of the three alternative objective functions do you think is most

appropriate?


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QuizSolve the LP below using dual simplex method

Maximize z= 150x+ 250y+ 450z

subject to2x+y+ 3z80

3x+y+ 4z60

x+ 2y+ 5z100

x 0, y 0 , z 0

Optimal Solution: p= ; x= , y= , z=

NOTE

This method of computation is known as Gauss - Jordan row operations.that is

New pivot row = Current pivot row Pivot element

All other rows:

Newrow = Current row - Pivots column coefficient New Pivot row

The rules for selecting the entering and Departing variables are referred to as

optimality and feasibility conditions.

Definition: Optimality condition

The entering variable (E.V) is the non-basic variable having the largest

negative coefficient in profits row. (ties are broken arbitrarily). Optimum

is reached where all profits row coefficients of non - basic variables are all

positive.

Definition: Feasibility conditionThe Departing variable (D.V) is the basic variable associated with the

smallest positive ratio of solution: pivot column elements (ties are arbi-

trarily broken)

Economic Interpretation


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1. If no slack variable do not appear in solution set, then all the resources

have been exhausted.

2. The solution quantity corresponding to si is the amount of resourceswhich were not utilized/abundant resource.

3. Values at profit row corresponding to Slack variables gives the marginal

value products of the corresponding resources (per unit contribution to

profit). They are also referred to as dual prices and in economics

shadow prices or imputed costs. They represent the worth per unit of a

resourcebi i = 1, 2...m.

In the farmers problem. s2 = 15 indicates that (15 2) = 30 laborers

will be un utilized.

4. The p row coefficients of s1, s2 and s3 are 20, 0 and 10 respectively

implying that;

Increasing land by 1 ha increases profit by $20

Increasing the number of employees has no effect on profit

Increasing the number of capital by $1 increases the profit by$10/50

The farmer should not pay more than $20 for every additional 1 ha hired.

It would not make sense to put more money in form of capital.

Simplex Method Algorithm

Step 1 - Determine a starting feasible solution.

Step 2 - Select the E.V using optimality condition. STOP if there is NONE.

Step 3 - Select Departing Variable using feasibility condition

Step 4 - Determine the new basic solution using the appropriate Gauss -

Jordan computation, Go to Step 2.


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Exercise 3. A carpenter makes boxes, tables and chairs. The profit contri-

butions of the three products are $20, $30 and $10 respectively. The carpenter

can afford to spend up to 40 hours per week working and takes two hours to

make a box, six hours to make a table and two hours to make a chair. Cus-

tomer demand requires that he makes at most a third as many boxes as the

total number of chairs and tables. The storage space available is 10 m2 and a

box requires 0.5 m2, a table takes up 1 m2 and a chair 0.4 m2. Formulate this

problem as a linear programming problem for profit maximization and solve

it using simplex method.

Exercise 4. A carpenter makes boxes, tables and chairs. The profit contri-

butions of the three products are $20, $30 and $10 respectively. The carpentercan afford to spend up to 40 hours per week working and takes 2 hours to

make a box, 6 hours to make a table and 2 hours to make a chair. Customer

demand requires that he makes at most a third as many boxes as the total

number of chairs and tables. The storage space available is 10 m2 and a box

requires 1/2 m2, a table takes up 1 m2 and a chair 2/5 m2. Formulate this

problem as a linear programming problem for profit maximization and solve

it using simplex method.

Example.A certain company produces 3 products A, B C which contributes

a profit of Shs 8, 5 and 10 respectively. The production machine has 400 hrs.

Capacity and each product uses 2, 3 and 1 machine hour respectively. There

are 150 units available of a special component with A using 1 unit and C using

1 Unit per unit. A special allow of 200kg is needed in this period and product

A and C uses 2kg and 4kg per unit. There is a limitation of production of unit

B to not more than 50. Advice the company in order to maximize the profit.

Solution

Let x, y and z be the number of units of products A, B and C to be

produced respectively.

The model is:


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Maxp= 8x+ 5y+ 10z

S/t 2x + 3y + z 400 (Hours)

x + z 150 (Special component)2x + 4z200 (Alloy)

y 50

x 0, y 0, z 0

Step 0: Introduce the Slack variables s1,s2 ands3 ands4 to get:

Maxp = 8x+ 5y+ 10z

S/t 2x+ 3y+z+s1= 400

x+z+s2= 150

2x+ 4z+s3= 200

y+s4 = 50

x 0, y 0, z 0

The initial solution is

Basis x y x3 s1 s2 s3 s4 Solution

S1 2 3 1 1 0 0 1 400

S2 1 0 1 0 1 0 0 150

S3 1 0 2 0 0 1 0 100

S4 0 1 0 0 0 0 1 50

p 8 5 10 0 0 0 0 0

Proceed and get the solution asx = 100,y = 50,x3= 0 and Maxp = 1050

3.7. Dealing with mixed on decision variables

If in a LP one or more constraint is given as xi k where k is a constant,

make the substitution yi= xi k 0 xi= yi+k.We then replacexiwithyi+kand adjust each constraint and the objective

function appropriately.

Example


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Maximize Z= 5x1+ 3x2+ 4x3

Subject to 3x1+ 12x2+ 6x3 900

6x1+ 6x2+ 3x3 1350

2x1+ 3x2+ 3x3 390

x1 0, x2 20, x3 10

Sincex2 andx3 cannot be zero, we let

a= x1 a 0

b= x2 20 b 0

a= x3 10c 0

Adjusting the LP we get

Maximize Z= 5a+ 3b+ 4c+ 100

Subject to 3a+ 12b+ 6c 900 240 60 = 600

6a+ 6b+ 3c 1350 120 30 = 1200

2a+ 3b+ 3c 390 60 30 300

a,b,c 0

which is now solvable using the simplex method. Following the simplex algo-

rithm finally leads to the optimal table as

Basis a b c s1 s2 s3 Solutions1 0

52

12

1 0 12

50

s2 0 -1 -2 0 1 -1 100

a 1 32

32

0 0 12

150

Z 0 92

72

0 0 52

850

With the corresponding solution being a = 150, b = 0,c = 0 and maximum Z

= 750.

But under the initial conditions the optimal solution should be x1 = 150, x2

= 20,x3 = 10 with max Z = 850.

3.8. Minimization

For LPs which are not in standard form, the Simplex Method cannot be used

right away, because the initial point (the origin) is infeasible. It is advisable

to solve LPs when they are of the form


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MaximizeZ=CXsubject to

AXb, X0

If this is not, the case we need to transform the original problem to take

this standard form before we use ordinary simplex method

Example. Solve the LP below using simplex method

Minimize z= 20x+ 30y

subject to

x+y= 45

3x+ 4y 170

12x+ 6y 480

x 0, y 0

To write it in standard form,we

replace = with and inequalities

negate the objective function.

and solve it in the ordinary way as follows

Solution

Performing the adjustment we get

Maximize z = -20x -30y subject to

x + y < = 4 5

-x - y

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12x +6y + s4 = 480

x, y >=0

The first tableau is given by

Tableau #1

Basis x y s1 s2 s3 s4 z Solution

s1 1 1 1 0 0 0 0 45

s2 -1 -1 0 1 0 0 0 -45

s3 3 4 0 0 1 0 0 170

s4 12 6 0 0 0 1 0 480

z 20 30 0 0 0 0 1 0

Since all entries in the profit row are +ve, this solution is

optimal. But since s2=-45, the solution is infeasible

The pivot columns are those with -ve entries (column x and y)

to identify the entering variable take the ratios of the z-row

to the elements of the s2-row i.e (20/-1, 30/-1). The E.V is the

one corresponding to the smallest absolute value => x

Max x= min(+ve)(45/1, -45/-1, 170/3, 480/12)=40 from s4-row,

which implies that pivot value = 12

New row x=s4/12

Performing row reduction operations on x column gives

Tableau #2


s1 0 1/2 1 0 0 -1/12 0 5s2 0 -1/2 0 1 0 1/12 0 -5

s3 0 5/2 0 0 1 -1/4 0 50

x 1 1/2 0 0 0 1/12 0 40

z 0 20 0 0 0 -5/3 1 -800


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Since s2=-5, the solution is still infeasible

The pivot column is y and so y enters the solution

with pivot value -1/2

Max y=(-5/(-1/2)=10 (NOTE THAT THE OTHER RATIOS ARE NOT VALID)

New y-row=s2/(-1/2)

Reducing all the elements of y column to zero using Jordan

gauss computations gives

Tableau #3


S1 0 0 1 1 0 0 0 0

y 0 1 0 -2 0 -1/6 0 10

s3 0 0 0 5 1 1/6 0 25

x 1 0 0 1 0 1/6 0 35

z 0 0 0 40 0 5/3 1 -1000

The optimal and feasible solution is now found to be

x=35, y=10 and minimum z=-1000

Quiz: Solve the LP below using simplex method

Minimize z = 2x+y

subject to

x+y = 4

2x y 3

x 0, y 0

Optimal Solution: p= ; x= , y=


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Solutions to Examples

Exercise 3. The solution is as follows

Let x, y and z be the number of boxes, tables and chairs

to be produce respectively, then

Maximize profit p = 20x + 30y + 10z

subject to 2x + 6y + 2z

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Corresponding solution x=0, y=20/3, z=0, p=200 which is not

optimal. Entering variable corresponds to largest -ve value in

p-row =>x.

Maximum x=min((20/3)/(1/3),(20/3)/(10/3),(10/3)/(1/6))=2

corresponding to s2 row. Departing variable is s2,

pivot value =10/3

New row x=s2/(10/3)

Using the new row to reduce all other elements pivot column

to zero gives

Tableau #3

Basis x y z s1 s2 s3 p Soln

y 0 1 2/5 3/20 -1/10 0 0 6

x 1 0 -1/5 1/20 3/10 0 0 2

s3 0 0 1/10 -7/40 -1/20 1 0 3

p 0 0 -2 11/2 3 0 1 220

with corresponding solution x=2, y=6, z=0 and p=220 which is

not optimal. Entering variable corresponds to largest -ve

value in p-row =>z

Maximum z=min(6/(2/5), 2/(-1/5), 3/(1/10))=15,

corresponding to y row. Departing variable is y,

pivot value =2/5, New row z=y/(2/5)

Using the new row to reduce all other elements pivot column

to zero gives

Tableau #4

Basis x y z s1 s2 s3 p Solnz 0 5/2 1 3/8 -1/4 0 0 15

x 1 1/2 0 1/8 1/4 0 0 5

s3 0 -1/4 0 -17/80 -1/40 1 0 3/2

p 0 5 0 25/4 5/2 0 1 250


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which gives the optimal solution as x=5, y=0, z=15 and

optimal p=250.

The carpenter should therefore make 5 boxes, no tables

and 15 chairs in order to maximize his/her profit to $250

Exercise 3

Exercise 4. Let x, y and zbe the number of boxes, tables and chairs to be

produce respectively, then the formulated LP becomes

Maximizep = 20x+ 30y+ 10z

subject to

2x + 6y+ 2z 40, (restrictiononhours)

3x + 1y+ 1z 0, (customersrestrictions)

1/2x + 1y+ 2/5z 10, (restrictiononstoragespace)

x 0, y 0, z 0

We convert the problem to canonical form by introducing slack variables s1,

s2 and s3 to get

Maximizep = 20x+ 30y+ 10z

subject to

2x + 6y + 2z +s1 = 40,

6x + 1y + 1z +s2 = 0,

2x + 1y + 2/5z +s3 = 10,

x 0, y 0, z 0, s1 0, s2 0, s3 0

The initial tableau is

Basis x y z s1 s2 s3 Soln

s1 2 6 2 1 0 0 40

s2 3 1 1 0 1 0 0

s3 1/2 1 2/5 0 0 1 10

p 20 30 10 0 0 0 0


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This solution not optimal because there exists -ve values in the last row. En-

tering variable corresponds to the most negative value this row which in this

case is y

Maximum y = min(406, 01

, 101) = 0 which implies that pivot value = -1 and

departing variable is s2, New row y = s2/ 1.

Performing row operations using the pivot row to reduce all the elements

ofy column to zero leads to the following table.


s1 20 0 4 1 6 0 40

y 3 1 1 0 1 0 0

s3 7/2 0 (3)/5 0 1 1 10p 110 0 20 0 30 0 0

This solution not optimal because there exists -ve values in the last row.

Entering variable corresponds to the most negative value this row which in

this case is x.

Maximum x = min(4020 , 03

, 107/2) = 2 which implies that pivot value = 20

and departing variable is s1, New rowx= s1/20.


ofx column to zero leads to the following table.


x 1 0 (1)/5 1/20 3/10 0 2

y 0 1 2/5 3/20 (1)/10 0 6

s3 0 0 1/10 (7)/40 (1)/20 1 3

p 0 0 2 11/2 3 0 220

This solution not optimal because there exists -ve values in the last row. En-

tering variable corresponds to the most negative value this row which in thiscase is z

Maximum z= min( 2(1)/5

, 62/5

, 31/10

) = 15 which implies that pivot value =

2/5 and departing variable is y , New row z= y/2/5.


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ofzcolumn to zero leads to the following table.

Basis x y z s1 s2 s3 Solnx 1 1/2 0 1/8 1/4 0 5

z 0 5/2 1 3/8 (1)/4 0 15

s3 0 (1)/4 0 (17)/80 (1)/40 1 3/2

p 0 5 0 25/4 5/2 0 250

Since all entries in the last row are negative, this table gives the optimal

solution as;x = 5,z= 15,s3= 3/2 with a maximum objective function value

ofp = 250.

The carpenter should therefore make 5 boxes, no tables and 15 chairs in

order to maximize his/her profit to $250 Exercise 4
http://www.jkuat.ac.ke/elearning/http://www.jkuat.ac.ke/elearning/http://www.jkuat.ac.ke/elearning/

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Sma 2343 Introd