Linear Programming

3.1 INTRODUCTION

Linear programming (LP) is a quantitative method for solving problems concerning

resource allocation. This study is about managing the resources – how the management

should allocate the resources efficiently so that maximum benefits can be obtained from

the limited resources available. Examples of resources are raw materials, employees,

labor hours, machine hours, financial budgets, etc. Since the resources are limited, the

management must utilize and allocate them efficiently so that the firm could obtain

maximum profit.

3.2 REQUIREMENTS OF A LINEAR PROGRAMMING PROBLEM

The linear programming problems have four important properties in common.

1. All Linear Programming problems in practice seek to maximize or to minimize

certain quantity, usually profit or cost. We refer to this property as the objective

function of a Linear Programming problem.

2. All problems have certain restrictions, or constraints that limit the degree to

which we can achieve our objective. Therefore, we want to maximize or minimize

the objective function, subject to some limited resources available (constraints).

3. There must be some alternative courses of action to choose from in the

solution. If there is no alternative solution to choose from, then we would not

need LP.

4. The objective and constraints in Linear Programming problems must be

expressed in terms of linear equations or inequalities. The linear mathematical

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relationship requires all terms used in the objective functions and the constraints

involved to be of the first degree.

3.3 FORMULATING LINEAR PROGRAMMING PROBLEMS

Three components of Linear Programming are the variables, objective and constraints.

The variables are continuous, controllable, and non-negative. X1 represents the first

variable, X2 represents the second variable, and so on; X1, X 2, …X n… > 0.

The objective of the Linear Programming problem is a mathematical representation of

the goals in terms of a measurable objective such as the amount of profit, cost, revenue,

or quantity. It must be in a linear function and is represented by one of these two forms:

i) Maximize C1X1 + C 2X 2 +….. + C nX n,

or ii) Minimize C 1X 1 + C 2X 2 +….. + C nX n.

The constraints are a set of mathematical expressions representing the restrictions

imposed on the controllable variables and thereby limiting the possible values of those

variables. The mathematical expressions of the constraints can be given either as linear

equations (=) or as linear inequalities (<, >, >, <).

A linear programming model will be structured to find the values of controllable

variables that maximize or minimize a defined linear objective function, subject to a set

of defined linear constraints and subject to the non-negativity restrictions.

Example 3.1 The owner of a small company that manufactures clocks must decide how many clocks

of each type to produce daily in order to maximize profit. The company manufactures

two types of clocks, that is regular clocks and alarm clocks. The resources required to

produce the clocks are labor and processing machine. The limited resources available

every day are as follows:

Labor hours = 1,600 hours

Machine hours = 1,800 hours

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The requirement for resources for each type of clocks is as follows:

To produce one unit of regular clock requires 2 hours of labor, and 6 hours of machine

time. While to produce one unit of alarm clock requires 4 hours of labor, and 2 hours of

machine time. The profit for each unit of regular clock sold is RM 3.00, while the profit

for each unit of alarm clock sold is RM 8.00. Determine how many units of each type of

clock should be manufactured every day so that the profit is maximum?

Solution:

First of all we need to define the decision variables involved:

Let X1 = number of regular clocks to manufacture per day,

Let X2 = number of alarm clocks to manufacture per day.

Now we should construct a table to present the total resources available, the required

resources for each clock, and the profit obtained if each unit of the respective clocks is

sold.

Refer to Table 3.1. The manager must now determine the values of X1 and X2.

Table 3.1: The resources, the requirements, and the profitsResourcesrequired

X1

(regular clock)X2

(alarm clock)Total resources

available

Labor hours 2 4 1600

Machine hours 6 2 1800

Profit per unit 3 8

The objective function is to maximize the profit: Maximize profit 3X1 + 8X2

Subject to the constraints: Labor hours constraint : 2X1 + 4X2 < 1600

Machine hours constraint : 6X1 + 2X2 < 1800

non-negativity conditions : X1 > 0, X2 > 0.

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This is a two-variable Linear Programming problem. The manager’s job is to determine

how many regular clocks (X1) and how many alarm clocks (X2) to produce in order to

maximize the profit. We can solve this problem using a graphical method.

Example 3.2

A manufacturer produces two types of cotton cloth: Denim and Corduroy. Corduroy is a

heavier grade cotton cloth and, as such, requires 3.5kg of raw cotton per meter whereas

Denim requires 2.5kg of raw cotton per meter. A meter of Corduroy requires 3.2 hours of

processing time, and a meter of Denim requires 3 hours. Although the demand for Denim

is practically unlimited, the maximum demand for Corduroy is 510 meters per month.

The manufacturer has 3250kg of cotton and 3000 hours of processing time available each

month. The manufacturer makes a profit of RM4.50 per meter from Denim and RM6 per

meter from Corduroy. The manufacturer wants to know how many meters of each type of

cloth to produce to maximize profit.

(a) Formulate a linear programming model for the above problem.

Solution:

Let X1 be the quantity of denim cloth type to be produced

Let X2 be the quantity of corduroy cloth type to be produced

Resources required

X1

(Denim)X2

(Corduroy)Total resources

available

Cotton cloth 2.5 3.5 3250

Processing time 3.0 3.2 3000

Profit 4.5 6.0

Maximize Profit = 4.50X1 + 6.00X2

Subject to constraints:

2.5X1 + 3.5X2 < 3250 (cotton cloth constraint)

3.0X1 + 3.2X2 < 3000 (processing time constraint)

X1 > 0

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0 < X2 < 510

(the production of courdroy should not exceed the maximum demand of 510 meters)

Example 3.3

A non-profit organization is raising fund for AIDS victims. They plan to sell two types of

soft drinks namely cherry fizz and lemon pop. They decided to set up a soft drink stall in

Kuala Lumpur. The resources required to produce the drinks are ingredient A and B.

To produce one batch of cherry fizz requires 5 liters of ingredient A and 5 liters of

ingredient B, while to make one batch of lemon pop requires 3 liters of ingredient A and

11 liters of ingredient B. There are 30 liters of ingredient A and 55 liters of ingredient B

available for use. Due to some marketing strategy, the organization has decided not to

produce more than 4 batches of lemon pop at any one time.

Profit is made at RM 3 for each batch of cherry fizz and RM 4 for each batch of lemon

pop. How many batches of each type should be produced every time so that the profit is

maximum?

Solution:

Define the decision variables:

Let X1 = number of batches of cherry fizz to produce,

and X2 = number of batches of lemon pop to produce.

Resources required Cherry fizz(X1)

Lemon pop(X2)

ResourcesAvailable

Ingredient A 5 3 30

Ingredient B 5 11 55

Profit 3 4

From the table we can write the mathematical inequalities as follows:

Constraints :

5X1 + 3X2 < 30;

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5X1 + 11X1 < 55;

X1 > 0,

X2 < 4 (not more than 4 batches of lemon pop to be produced)

Objective function :

The total profit made is 3X1 + 4X2, and this is to be maximized

Thus, the linear programming model may be written as:

Maximize 3X1 + 4X2

subject to:

5X1 + 3X2 < 30;

5X1+ 11X2 < 55;

X2 < 4;

X1 > 0.

3.4 GRAPHICAL SOLUTION TO A LINEAR PROGRAMMING PROBLEM

If a linear programming problem involves only two decision variables (X1 and X2), the

solution may be found using the graphical technique. In practice, this technique is not

commonly employed since most LP problems involve more than two variables. However,

the technique is illustrated using the problem in Examples 3.1.

In the graphical solution method, what you need to do is:

a) Obtain the linear equation for each constraint in the problem

b) Plot the linear equations of the constraints on the graph

c) Determine the feasible region based on the inequalities in (a)

d) Obtain the feasible solution points based on the feasible region in (c)

e) Obtain an optimal solution for the problem by applying each feasible solution point

into the objective function.

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3.4.1 THE FEASIBLE REGION AND CORNER POINTS METHOD

Example 3.4

Consider the problem in Example 3.1.

Find the optimal solution to the problem by using the graphical method.

The objective is to maximize the profit:

Maximize 3x1 + 8x2.

Subject to:

The Constraints:

labor hours : 2x1 + 4x2 < 1600 -----(1)

machine hours : 6x1 + 2x2 < 1800 -----(2)

non-negativity conditions : x1, x2 > 0.

Solution:

Firstly, plot all linear equations of the constraints on the graph. Secondly, shade the area

covered by the respective inequality (, etc) of the constraints. Thirdly, determine the

feasible region, that is the common region which belongs to the inequalities. The feasible

region is the common area or the area that overlapps all constraints.

Find all coordinates (X1 and X2) that fall in the feasible region. The coordinates that fall

in the feasible region are the feasible solutions for the problem at hand. Solve all

intersections between constraints and also determine their coordinates.

Finally, input each coordinate of the feasible solution into the objective function and

determine its value.

For the maximization objective function, the optimal solution occurs at the feasible

solution that yields the maximum value.

And for the minimization objective function, the optimal solution occurs at the feasible

solution that yields the minimum value.

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Applying to Example 3.4.

The first thing to do is to draw both equations (Equation 1 and Equation 2) on the same

graph. In order to draw any equation, we need to obtain at least two coordinates from the

respective equation.

Constraint Equation (1)

2X1 + 4X2 = 1600

Let X2 = 0, then we have 2X1 + 4(0) = 1600 ↔ 2X1 = 1600 or X1 = 800

Then (800, 0) will be the first coordinate on the graph.

Let X1 = 0, then we have 2(0) + 4X2 = 1600 ↔ 4X2 = 1600 or X2 = 400

Then (0, 400) will be the second coordinate on the graph. Now, you can draw

the linear equation 2X1 + 4X2 = 1600 by connecting the two coordinates.

Constraint Equation (2)

6X1 + 2X2 = 1800

Let X2 = 0, then we have 6X1 + 2(0) = 1800 ↔ 6X1 = 1800 or X1 = 300

Then (300, 0) will be the first coordinate on the graph.

Let X1 = 0, then we have 6(0) + 2X2 = 1800 ↔ hence 2X2 = 1800 or X2 = 900

Then (0, 900) will be the second coordinate on the graph. Now, you can draw

the linear equation 6X1 + 2X2 = 1800 by connecting the two coordinates.

X2

(0,900) 900

(0,400) 400

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(300,0) (800,0)0 X1

Once we have plotted the two constraints, we will find the two equations intercept with

one another. In this case, we have to solve the two equations simultaneously to determine

the coordinate where the two equations intercept.

How to estimate the coordinate of interceptions between the two equations:

The equation of constraints:

Machine Hours : 6X1 + 2X2 = 1800

Labor Hours : 2X1 + 4X2 = 1600

Non negativity conditions : X1 > 0, X2 > 0

Divide the second equation (labor hours) by 2, we have:

1/2 [2X1 + 4X2 = 1600] » X1 + 2X2 = 800

Machine Hours : 6X1 + 2X2 = 1800 ----(1)

Labor Hours : X1 + 2X2 = 800 ----(2)

Subtract the two equations (1) – (2), we obtain:

6X1 + 2X2 = 1800 ----(1)

(-) X1 + 2X2 = 800 ----(2)

5X1 = 1000

Therefore X1 = 200

Then, substitute X1 = 200 into equation (2):

200 + 2X2 = 800

2X2 = 800 – 200

2X2 = 600

Therefore X2 = 300

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So the interception point beween the two equations occurs at (200, 300)

Figure 3.1: Feasible Region as defined by all constraints for Example 3.4

X2

(0,900)

900

(0,400) 400

(200,300)

(0,0) (300,0) (800,0)0 X1

300 800

From Figure 3.1, the feasible solutions are (0, 400), (200, 300), and (300, 0).

For your information, the feasible solution only occurs at the coordinates in the feasible

region. The feasible region is the region covered by all constraints requirement in the

problem (in this case there are two constraints).

Applying the objective function Maximize 3X1 + 8X2 on the feasible solution,

we have:

At (0, 400) the profit is 3(0) + 8(400) = RM3,200 *

At (200, 300) the profit is 3(200) + 8(300) = RM3,000

At (300, 0) the profit is 3(300) + 8(0) = RM900

The objective of the above linear programming problem is to maximize the profit. The

maximum profit RM3,200 occurs at (0, 400). Hence, the company should decide to

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produce X1 = 0 (regular clock) and X2 = 400 (alarm clock) in order to maximize the profit

based on the current constraints.

3.4.2 SOLVING THE MINIMIZATION PROBLEMS

Sometimes the problem is stated in terms of the cost of materials used in production

instead of the profit contribution for each product. In this case the objective function

should be changed to minimizing the cost.

Example 3.5

A firm produces 2 types of chemicals namely AA and BB. Chemical AA costs RM2 per

liter and chemical BB costs RM3 per liter to manufacture. Over the week, at least 6

liter of AA and 2 liter of BB must be produced. One of the raw materials needed to make

each chemical is in short supply and only 30gm are available. Each liter of AA requires

3gm of this material while each liter of BB requires 5gm. How many liters of AA and BB

should the firm produce in order to minimize the total cost?

Solution:

Define decision variables: Let x1 = number of liters of chemical AA to be produced Let x2 = number of liters of chemical BB to be produced

Define the objective function; Minimize total cost = 2x1+ 3x2

Define the constraints; x1 > 6 (at least 6 litres of AA should be produced) x2 > 2 (at least 2 litres of BB should be produced) 3x1+ 5x2 < 30;

Thus, the problem becomes: Minimize total cost = 2x1 + 3x2

Subject to constraints: x1 > 6 (at least 6 liters of x1 to be produced)

x2 > 2 (at least 2 liters of x2 to be produced)

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3x1+ 5x2 < 30 (each x1 requires 3gm while each x2 requires 5gm and the

supply is limited to 30gm in total)

x1, x2 > 0 (the non-negativity constraints)

Again, we should plot the equation of constraints, obtain the feasible region, the feasible

solution points, and finally apply the objective function on the feasible solution points to

determine the optimal solution.

Figure 3.2: Feasible Region for Example 3.5

From the graph (Figure 3.3), the feasible solution points occur at

(6, 12/5), (6, 2), and (20/3, 2)

Applying the objective function 2X1 + 3X2, we have:

At (6, 12/5) : 2(6) + 3(12/5) = RM19.2

At (6, 2) : 2(6) + 3(2) = RM18.0 (minimum cost)

At (20/3, 2) : 2(20/3) + 3(2) = RM19.3

Hence the optimal solution occurs at X1 = 6, X2 = 2

x1 = number of liters of chemical X to be produced

x2 = number of liters of chemical Y to be produced

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Example 3.6

AGM Super-Bike Company (ASC) has the latest product on the upscale toy market;

boys’, girls’ and toddlers’ bike in bright fashion colors, a strong padded frame, chrome-

plated chairs, brackets and valves, and a non-slip handle bar. Due to the best sellers

market for high quality toys, ASC is confident that it can sell all the bicycles produced, at

the following prices; boys’ bike at RM 420, girls’ bike at RM 350 and toddlers’ bike at

RM 150.

The company’s accountant has determined that direct labor costs will be 45% of the price

ASC receives from the boys’ model, 40% of the price ASC receives from the girls’ model

and 30% from the toddlers’ model. Production costs other than labor, but excluding

painting and packaging are RM 45 per boys’ bike, RM 35 per girls’ bike and RM 20 per

toddlers’ bike. Painting and packaging cost is RM 25 per bike regardless of the model.

ASC plant’s overall production capacity is 400 bicycles per day. However realizing that

the demand for boys’ bikes are more than girls’ bikes, the marketing manager insists that

ASC must produce not more than 200 units of girls’ bikes, and not more than 100 units of

toddlers’. Each boys’ bikes requires 3 labor hours, each girls’ bike requires 2.4 hours and

each toddlers’ bike requires 2 hours to complete. ASC currently employs 120 workers,

who work 8-hours per day. Assume the company does not retrench workers since the

company believes its stable work force is one of its biggest assets.

(a) Formulate the above Linear Programming problem.

Solution: First of all, we need to define the decision variables as follows:

Let:

X1 be the number of boys’ bikes produced per day.

X2 be the number of girls’ bikes produced per day.

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X3 be the number of toddlers’ bikes produced per day.

Variables X1 X2 X3

Selling price (RM) 420 350 150

LaborCost

-189 -140 -45

Production Cost -45 -35 -20

PaintingCost

-25 -25 -25

Net Profit 161 150 60

Requirements Variables Resources

X1 X2 X3 X1 X2 X3 400X2 - X2 - 200

X3 - - X3 100

Labor 3.0 2.4 2.0 960 hours120 workers x 8 hours

Objective function is:

Maximize profit: 161X1 + 150X2 + 60X3

Subject to (constraints) :

X1 + X2 + X3 < 400 (capacity constraints)

X2 < 200 (production constraint for X2 girl’s bike)

X3 < 100 (production constraint for X3 toddler’s)

3 X1 + 2.4 X2 + 2 X3 < 960 (labor hours constraint)

X1, X2, X3 > 0 (non-negativity constraints)

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Example 3.7

Toy Race Car Company manufactures three different race cars called Racer_1, Racer_2,

and Racer_3. The resources required to make the products are technical services, labor

and administration. Table below provides the requirements for the production. The profits

for each unit sold for Racer_1, Racer_2, and Racer_3 are RM10, RM6, and RM4

respectively.

Product TechnicalServices (hour)

Labor(hour)

Administration(hour)

Unit profit(RM)

Racer 1 1 10 2 10

Racer 2 2 4 2 6

Racer 3 3 5 6 4

The resources available for each week are 100 hours of technical services, 600 hours of

labor, and 300 hours of administration.

(a) Formulate the problem as a linear programming problem.

Solution: LetX1 be the number of Racer_1 to be produced.

X2 be the number of Racer_2 to be produced.

X3 be the number of Racer_3 to be produced.

The limited resources available are technical services (100 hours), labor (600 hours), and

administration (300 hours).

The above informations are summarized in the following table:

Resources X1 X2 X3 Total Resources

Technical 1 2 3 100

Labor 10 4 5 600

Administration 2 2 6 300

Profit 10 6 4

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The objective function: Maximize profit 10X1 + 6X2 + 4X3

Subject to (constraints):

1X1 + 2X2 + 3X3 < 100 (Technical services constraints)

10X1 + 4X2 + 5X3 < 600 (Labor constraints)

2X1 + 2X2 + 6X3 <300 (Administration constraints)

X1, X2, X3 > 0 (Non-negativity constraints)

3.4.3 SPECIAL CASES IN LINEAR PROGRAMMING (WHEN USING THE GRAPHICAL APPROACH)

Case 1: Infeasibility

A condition that arises when there is no solutions to a linear programming that

satisfies all of the constraints; graphically, no feasible region exists.

Case 2: Unboundedness

A linear programming problem will not have a finite solution; graphically, the

feasible region is open-ended.

Case 3: Redundancy

The presence of redundant constraints; simply one that does not affect the

feasible solution region.

Case 4: Alternate Optimal Solutions

Two or more feasible solutions that give the same optimal value; graphically,

when the slope of the objective function is parallel to one of the constraints

(infinitely many solutions).

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3.5 THE SIMPLEX METHOD

In the previous section, we saw how a linear programming problem with only two

variables (X1 and X2) was solved by using the graphical method. However, if the linear

programming problem has more than two variables (X1, X2, ….,Xn), the graphical

method is impossible to apply since normal graphs only possess two dimensions.

In this section, an algebraic method is developed to solve Linear Programming (LP)

problems with more than two variables.

The technique, called the simplex method, is an iterative process that begins with

the initial solution and, by means of computational routines, moves to an improved

solution until the optimal or final solution is reached.

In particular, for a two-variable problem, we have seen that it is only necessary to

examine the feasible solution points obtained in the feasible region, and this is precisely

what the simplex method does. Essentially the method repeatedly searches for a better

solution point until the objective function cannot be improved further even if we move to

another feasible solution point.

In this section, we will not discuss the process in details since its complexity requires a

longer medium of instruction. In addition only Mathematicians and Statisticians have the

capacity and patience to discuss such process. For us non-Mathematicians and non-

Statisticians, it is relevant for us to be more interested in how to set-up the actual problem

into the initial simplex table, and the interpretation of the results in a final table in term of

management problem; assuming the process would be carried out by a computer package

called “QM for Windows” (manual computation is very tedious). The computer package

is available in the market and is being used by managers to analyze their data. The

manager only needs to know how to intepret the results in the final table for their decision

making.

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3.5.1 HOW TO SET UP THE INITIAL SIMPLEX SOLUTION Before setting up the initial simplex table, it is essential to write the problems in their

standard form:

a) All decision variables (X1, X2, ….,Xn) must be non-negative (> 0).

b) The objective function is either to maximize (for profit) or to minimize (for cost)

c) The Right-Hand-Side of each constraint is a non-negative (quantity of resources).

d) All constraints must be in the form of mathematical equations.

e) We need to introduce the slack variables to represent the “unused resources”.

* Slack variables will become an Identity Matrix in the initial solution.

Example 3.8 (APR 2001)

Obtain the initial simplex table for Example 3.7

Objective function: Maximize profit 10X1 + 6X2 + 4X3

Subject to (constraints):

1X1 + 2X2 + 3X3 < 100 (Technical services constraints)

10X1 + 4X2 + 5X3 < 600 (Labor constraints)

2X1 + 2X2 + 6X3 < 300 (Administration constraints)

X1, X2, X3 > 0 (Non-negativity constraints)

Solution:

Since we have 3 resources in each of the equations, we need to introduce three slack

variables namely S1, S2, and S3 to represent the unused resources 1 (technical services

hours), resources 2 (labor hours), and resources 3 (administration hours) respectively.

Now, we write the above equation using the standard form of simplex method.

Maximize profit 10X1 + 6X2 + 4X3 + 0S1 + 0S2 + 0S3

1X1 + 2X2 + 3X3 + 1S1 + 0S2 + 0S3 = 100

10X1 + 4X2 + 5X3 + 0S1 + 1S2 + 0S3 = 600

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2X1 + 2X2 + 6X3 + 0S1 + 0S2 + 1S3 = 300

Profit contribution for each variable (X1, X2, X3, S1, S2, S3)

10 6 4 0 0 0

X1 X2 X3 S1 S2 S3 Quantity

1 2 3 1 0 0 100

10 4 5 0 1 0 600

2 2 6 0 0 1 300

The initial simplex table for Example 3.8

Cj

profit

SolutionMix

10 6 4 0 0 0 ResourcesQuantity

x1 x2 x3 S1 S2 S3

0 S1 1 2 3 1 0 0 100

0 S2 10 4 5 0 1 0 600

0 S3 2 2 6 0 0 1 300

Zj 0 0 0 0 0 0

Cj - Zj 10 6 4 0 0 0

Example 3.9

Solve the following problem using the simplex method.

Maximize (profit): 3x1 + 8x2

Subject to 2x1 + 4x2 < 1600 (labor hours constraint)

6x1 + 2x2 < 1800 (machine hours constraint)

x2 < 350 (materials constraint)

x1, x2 > 0;

To prepare the problem for the initial simplex table, the model should be written as:Maximize (profit): 3x1 + 8x2 + 0S1 + 0S2 + 0S3

Subject to 2x1 + 4x2 + 1S1 + 0S2 + 0S3 = 1600 6x1 + 2x2 + 0S1 + 1S2 + 0S3 = 1800 0x1 + 1x2 + 0S1 + 0S2 + 1S3 = 350

x1, x2, S1, S2, S3 > 0.

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Where S1, S2, S3 are called Slack Variables (representing unused resources)

The above equations are now expressed in Table 3.2.

Table 3.2: Initial Simplex Table for Example 3.9

Cj

profit

SolutionMix

3 8 0 0 0 ResourcesQuantity

x1 x2 S1 S2 S3

0 S1 2 4 1 0 0 1600

0 S2 6 2 0 1 0 1800

0 S3 0 1 0 0 1 350

Zj 0 0 0 0 0 0

Cj - Zj 3 8 0 0 0

Always remember that the optimal solution occurs at the identity matrix.

From the above table, we can intepret the initial solution as follows:

S1 = 1600, S2 = 1800, S3 = 350

All resources remain in original quantity (unused) since the production has not started

The solution for decision variables x1 = 0, x2 = 0

The total profit = 0.

The Value of Zj and Cj - Zj row:

The Zj value in each column in the table is the sum of each number in the C j column

multiplied by the corresponding number in the decision variable column.

For example:

The Zj value under variable x1 is the summation of 0(2) + 0(6) + 0(0) = 0

The Zj value under variable x2 is the summation of 0(4) + 0(2) + 0(1) = 0

The Zj value for the quantity column provides the profit obtained in the current solution.

In this case the profit is 0(1600) + 0(1800) + 0(350) = 0

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Meanwhile, the Cj - Zj value in each column represents the net profit for each variable.

To compute these Cj - Zj values, simply deduct the Zj value obtained for each decision

variable from the Cj value at the top of every variable.

For example, the Cj - Zj under x1 is 3 – 0 = 3. The the Cj - Zj under x2 is 8 – 0 = 8. The

same steps are used to compute the values for S1, S2, and S3 respectively.

3.5.2 SIMPLEX SOLUTION PROCEDURES

The Simplex Algorithm

1. For the maximization problem, if there are variables that have positive signs in row

Cj- Zj, select the highest number – and the variable which corresponds to this is called

the entering variable. (If all numbers in the row Cj - Zj are either 0 or negative values,

an optimal solution has been reached).

2. Find the ratios of the resources quantity to the numbers in the column of the entering

variable. Select the smallest ratio. The variable which corresponds to this smallest

ratio, will be the variable to be omitted from the solution mix. This variable is called

the leaving variable.

3. The pivot number is the number at the intersection of the column of the entering

variable and the row of the leaving variable. Divide every number in the pivot row by

the pivot number and this row (R) is then written in the next table in the same row

position as it was in the previous table.

4. To obtain the remaining rows in the new table, it should be firstly noted that the

column of a variable in the current solution mix are all zeros except for a 1 in the row

corresponding to this variable. This condition must now be obtained for the variable

that is entered in (1). This is obtained by adding and subtracting suitable multiples of

row R from the rows of the previous table. When this is done, return to (1).

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For the problem at hand, this algorithm should be applied twice to obtain a final

(third) simplex table. The final simplex table (using simplex algorithm) is shown in

Table 3.3

Table 3.3 Final Simplex Table for Example 3.9

Cj SolutionMix

3 8 0 0 0 ResourcesQuantity

x1 x2 S1 S2 S3

3 x1 1 0 0.5 0 -2 100

0 S2 0 0 -3 1 10 500

8 x2 0 1 0 0 1 350

Zj 3 8 1.5 0 2 3100

Cj - Zj 0 0 -1.5 0 -2

3.5.3 INTERPRETATION OF SIMPLEX SOLUTION

From the final table (Table 3.3), the optimal solution for this production is as follows:

X1 = 100, X2 = 350, S1 = 0, S2 = 500, S3 = 0, with total profit = $ 3100.

i.e. To maximize the total profit, the company should produce 100 units of X1 and 350

units of X2.

S1 = 0 and S3 = 0, mean the labor hours and assemblies hours are all used (balance = 0).

S2 = 500 means 500 hours of processing time left unused at the end of production

process. (i.e. only 1800-500 = 1300 hours of processing are used).

For the next production, the management should allocate only 1300 hours of processing

time. This optimal solution (as expected) agrees with those found by graphical method. It

is clear that with the number of arithmetic calculations involved in the simplex method,

an arithmetic error would occur when solving an LP problem manually. For this reason,

we will be solving the LP problems with the aid of a computer package. Nevertheless, it

is important to understand how the simplex algorithm works, since this knowledge is very

useful when performing a sensitivity analysis on the solution.

75

3.5.4 SPECIAL CASES IN LINEAR PROGRAMMING

FIVE special cases may arise when solving an LP problem using the simplex method.

Case 1 : No feasible solution (infeasibility)

If there are no values of the decision variables that satisfy all of the constraints, then the

problem is said to have no feasible solution. This case can be recognized in the simplex

method when one or more artificial variables appear with a positive value in the final

table.

Case 2 : Alternative optimal solution

For some LP problems, there may be more than one set of values for the decision

variables that gives the optimal value of the objective function.

In the simplex method, this can be recognized when there is a zero value.

In row Cj - Zj of the final table under a variable that is not in the final solution mix, such

LP problems are said to have an alternative optimal solution. An alternative optimal

solution may be found by entering the said variable into a solution mix and performing a

further iteration on the table.

Case 3 : Unbounded solution

In some LP problems, the value of the objective function may be made as large

(maximization) or as small (minimization) as we please. Such problems are said to have

an unbounded solution. They can be recognized in the simplex method when, at any

iteration for a maximization (minimization) problem there is a positive (negative) number

in row Cj - Zj and zero and/or negative number in the column underneath.

Case 4 : Unbounded feasible region but bounded optimal solution

Although the feasible region is unbounded, the optimal solution is bounded; i.e. the

objective function has a finite optimal value. An unbounded feasible region can be

recognized in LP problem when in any column (excluding row Cj - Zj) of any table, there

are zero and/or negative numbers. The solution will be unbounded only if in addition to

76

this condition, there is a positive (negative) in row C j - Zj in that column for a

maximization (minimization) problem.

Case 5 : Degenerate solution

In Step (2) of the simplex algorithm, it was stated that if there was a tie for minimum

ratio to decide the variable leaving the solution mix then the tie should be broken

arbitrarily. At the next iteration, however, the values of those variables in the tie that did

not leave the solution mix will be zero. A solution in which one or more of the variables

in the solution mix has a value of zero is said to be degenerate.

This ‘degeneracy’ in LP problem may cause a situation, in which the simplex iteration

enter a loop and repeat the same sequence of iterations without reaching the optimal

solution, fortunately this does not happen very often in practice.

3.6 THE DUAL IN LINEAR PROGRAMMING

Every linear programming problem has a twin problem associated with it. One problem

is called “primal” while the other is called “dual”. In an easier term, the dual is simply

looking at the same problem but from the “opposite perspective”. For example, the

primal is looking at maximizing profit but the dual is looking at minimizing resources.

Looking at the “dual” perspective is useful for two reasons:

1) There is less computation required to solve the dual problem compared to solving

the primal problem

2) The solution of the dual problem has a meaningful economic interpretation. For

example, in the dual solution, we can have the actual prices that should be paid

for each unit of extra resources if the company wants to continue with the

production (shadow price).

77

3.6.1 THE RELATIONSHIP BETWEEN THE PRIMAL AND DUAL PROBLEMS

1. The primal problem is maximization, while the dual problem is minimization.

2. The maximization of primal problem has < (less or equal) constraints while the

minimization of dual problem has > (greater or equal constraints).

3. Each constraint of the primal would become a variable in the dual problem.

4. The quantity of resources of the constraints in primal problem would become the

coefficient of the objective function in the dual problem.

5. The coefficient of objective function in the primal would become the quantity of the

constraint in the dual

6. From a symmetry, the dual of the dual problem would be the primal problem.

Example 3.10

Primal problem Maximize 3X1 + 4X2 subject to

5X1 + 3X2 < 30 : Y1

5X1 + 11X2 < 55 : Y2

X2 < 4 : Y3

X1, X2 > 0.

Dual problem Minimize 30Y1 + 55Y2 + 4Y3

subject to

5Y1 + 5Y2 > 3

3Y1 + 11Y2 + Y3 > 4

Y1, Y2, Y3 > 0

Example 3.11

Maximize 3X1 + 2X2 + 4X3

subject to 2X1 + X2 + 5X3 < 40 : Y1

3X1 - 4X2 + X3 < 10 : Y2

X1, X2, X3 > 0.

78

The corresponding dual problem is

Minimize 40Y1 + 10Y2

subject to 2Y1 + 3Y2 > 3

Y1 – 4Y2 > 2

5Y1 + Y2 > 4

Y1, Y2 > 0.

* If the maximization of primal problem has ‘<’ constraints, then the

minimization of dual problem will have ‘>’ constraints.

* The optimal values of the primal and dual objective functions are equal.

3.6.2 THE OPTIMAL SOLUTION FOR DUAL VARIABLES

The optimal solution for the dual variables (Y1, Y2, …..Yn) can be obtained from the final

simplex table of primal problem under the resources columns (S1, S2, …..,Sn). Since the

objective function of dual is to minimize resources, the optimal solutions for Y1, Y2,

…..Yn can be found under S1, S2, …..,Sn respectively in the row Cj - Zj (ignore the

negative sign) of the final simplex table (primal).

Example 3.12

Using Example 3.9:

Maximize (profit): 3X1 + 8X 2 + 0S1 + 0S2 + 0S3

Subject to 2X1 + 4X2 + 1S1 + 0S2 + 0S3 = 1600 : Y1

6X1 + 2X2 + 0S1 + 1S2 + 0S3 = 1800 : Y2

0X1 + 1X2 + 0S1 + 0S2 + 1S3 = 350 : Y3

X1, X2, S1, S2, S3 > 0.

Table 3.4: Initial Simplex Table for Example 3.9 (problem in the primal form)Cj Solution 3 8 0 0 0 Resources

79

Mix QuantityX1 X2 S1 S2 S3

0 S1 2 4 1 0 0 1600

0 S2 6 2 0 1 0 1800

0 S3 0 1 0 0 1 350

Zj 0 0 0 0 0 0

Cj - Zj 3 8 0 0 0

The dual variable Y1 corresponds to the first primal constraint that contains the slack

variable S1 (unused resources 1).

The dual variable Y2 corresponds to the second primal constraint that contains the slack

variable S2 (unused resources 2).

The dual variable Y3 corresponds to the third primal constraint that contains the slack

variable S3 (unused resources 3).

Hence, the optimal value of Y1 can be obtained under column S1 in row Cj - Zj.

That is, Y1 = 0, Y2 = 0, Y3 = 0 in the initial simplex table (Table 3.4).

Table 3.5: Final Simplex Table for Example 3.9 (problem in the primal form)

Cj Solution

Mix

X1 X2 S1 S2 S3 Resources

Quantity(3) (8) (0) (0) (0)

3 X 1 1 0 0.5 0 -2 100

0 S2 0 0 -3 1 10 500

8 X 2 0 1 0 0 1 350

Zj 3 8 1.5 0 2 3100

Cj - Zj 0 0 -1.5

(Y1)

0

(Y2)

-2

(Y3)

* Y1, Y2, Y3 are the shadow prices for resources 1, 2, and 3 respectively

From the final simplex table of the primal, we can intepret the solution as follows:

X1 = 100 units, X2 = 350 units

80

S1 = 0 (all resource 1 is used)

S2 = 500 (500 units of resource 2 is unused)

S3 = 0 (all resource 3 is being used)

Maximum profit = 3 (100) + 8 (350) = RM 3,100.00 (Maximum)

3.6.3 INTERPRETATION OF THE DUAL VARIABLES

The corresponding dual problem for Example 3.12 is:

Minimize (resources): 1600Y1 + 1800Y2 + 350Y3

Subject to 2Y1 + 6Y2 > 3

4Y1 + 2Y2 + Y3 > 8

Y1, Y2, Y3 > 0

The optimal values for dual variables Y1, Y2, Y3 are located in row Cj - Zj under

S1, S2, S3 columns respectively in the final simplex table (Table 3.5). From Table 3.5, we

get Y1= 1.5, Y2 = 0 and Y3 = 2.

If we input the values of Y1= 1.5, Y2 = 0 and Y3 = 2 into the objective function:

Minimize (resources): 1600Y1 + 1800Y 2 + 350Y3 , we will get :

The total resources used = 1600 (1.5) + 1800 (0) + 350 (2) = RM 3100 (Minimum).

This value (RM3100) is equivalent to the maximum profit under the primal form.

The optimal value of a dual variable which is associated with a primal constraint gives

an indication of how much the objective function (profit) increases for every unit

increase in the resources provided that the current solution remains feasible and

optimal. That is, if the resources of a primal constraint is increased by one unit, the value

of the objective function is increased by Yi unit as long as the current solution mix is

feasible.

In other words, the optimal values of the dual variable (Y1, Y2, Y3) are called “shadow

prices” for resources 1, 2, and 3 respectively. These values indicate the actual worth of

each resources if the management wants to obtain additional resources to continue

production. The dual variable Yi may be interpreted as the per-unit contribution of the ith

resource (Si) to the value of the objective function.

81

In Example 3.12, the optimal values of the dual variables are Y1 = 1.5, Y2 = 0 and Y3 = 2.

Remember, the solution must always be greater than zero (Y1, Y2, Y3 > 0), hence all

negative values are changed to positive values.

The interpretation of Y1, Y2, Y3:

Each additional unit of the first resource (labor hours) would contribute RM1.50

to the profit in the primal objective function. Hence, if the management wants to obtain

additional labor (overtime hours), the cost should not be higher than RM1.50 per hour. In

other words, RM1.50 is the shadow price for labor hours.

Each additional unit of the second resource (processing hours) contributes RM0.

So, at the moment any increase in processing hour does not have any monetary value or

does not contribute anything to the profit. This is because of the current production which

has resulted in the unused resource 2 (S2 = 500). Any increase of this resources is

unnecessary.

Each additional unit of the third resource (alarm assemblies) would contribute

RM2.00. In other words, this additional resource should not cost more than RM2.00 per

unit since RM2.00 is the shadow price for resource 3.

EXERCISES

82

1 (APR 2001)

The Toy Race Car Company manufactures three different race-cars called Racer 1,

Racer 2, and Racer 3. The resources required to make the products are technical

services, labour and administration. The table below gives the requirements for the

production.

Product TechnicalServices (hour)

Labor(hour)

Administration(hour)

Unit profit(RM)

Racer 1 1 10 2 10

Racer 2 2 4 2 6

Racer 3 3 5 6 4

The resources available for each week are 100 hours of technical services, 600 hours

of labor, and 300 hours of administration.

(a) Formulate the problem as a linear programming problem.

(b) The optimal simplex table for the above problem is given below. The S1,

S2, S3 are the slack variables for technical services, labor, and administration

hours respectively.

Cj SolutionMix

10 6 4 0 0 0 ResourcesQuantity

X1 X2 X3 S1 S2 S3

6 X2 0 1 5/6 5/3 -1/6 0 66.67

10 X1 1 0 1/6 -2/3 1/6 0 33.33

0 S3 0 0 4 -2 0 1 100.00

Zj 10 6 20/3 10/3 2/3 0 733.32

Cj - Zj 0 0 -8/3 -10/3 -2/3 0

(i) Specify the optimal daily production levels for the company for the three

products. What is the total profit?

(ii) Which resource is not fully utilized? If so, how much is the spare capacity?

(iii) Write down the dual of the problem.

(iv)Based on the table, obtain the solution of the dual problem.

83

2 (SEPT 2002)

a) Summit Events Management plans to rent out sales booths during the Ramadan Food

and Fashion Fair. The booths are to be set up at a popular shopping mall. There should be

a maximum of 100 booths, comprising three different types: Type 1, Type 2, and Type 3.

The rental will be at a flat rate of RM500 each. However, the cost of setting up the booth

varies depending on the type of booth. The cost per unit and the number of hours to set

up each booth are as follows:

Summit has a RM32,000 budget to cover the cost. There are 4 workers available to set up

the booths. Each of them works 8 hours per day for 5 days. Formulate a linear

programming problem to determine the number of each type of booth to maximize profit:

Let:

X1 = Quantity of Type 1 booth S1=slack variable for the number of booths

X2 = Quantity of Type 2 booth S2=slack variable for the budget

X3 = Quantity of Type 3 booth S3=slack variable for man-hour

b) The following is the final simplex table for the above problem.

Cj Solution

Mix

100

X1

300

X2

200

X3

0

S1

0

S2

0

S3 Resources

200

0

300

X3

S2

X2

1

10

0

0

0

1

1

0

0

2

-40

-1

0

1

0

-1

10

1

40

8000

60

Zj

Cj-Zj

200

-200

300

0

200

0

100

-100

0

0

100

-100

i. State the optimal solution and interpret the values of all variables.

ii. Determine the maximum profit.

Type 1 Type 2 Type 3

Cost per unit RM 400 RM 200 RM 300

No. of man-hours 1 2 1

84

iii. How much of the budget will be used to reach the optimal solution?

iv. Formulate the dual and determine the optimal solution.

v. What would be the effect on the profit if 5 more man-hour is made available?

(c) Puncak Distribution packages distributes industrial supplies. A standard

shipment can be packaged in a class A container, a class B container, or a class C

container. One unit of class A container yields a profit of RM9, one unit of class B

container yields a profit of RM7, and one unit of class C container yields a profit of

RM8. Each shipment requires a certain amount of packing materials, material I and

material II ( in meters), and packing time (in hour).

The final simplex table of the problem is given below:

Cj SolutionMix

RM9X1

RM7X2

RM8X3

RM0S1

RM0S2

RM0S3

Resources

9

7

0

X1

X2

S3

1

0

0

0

1

0

0.8

0.8

-0.6

0.6

-0.4

-0.2

-0.4

-0.6

-0.2

0

0

1

32

12

6

Zj 9 7 12.8 2.6 0.6 0 372

Cj - Zj 0 0 -4.8 -2.6 -0.6 0

X1, X2 and X3 represent the number of class A, class B, and class C containers

respectively. S1, S2 and S3 are the slack variables representing the unused amount of

packing materials (I and II) and packing time respectively.

(i) What is the optimal number of each class of containers and the total profits?

(ii) Puncak is considering getting an extra packing material I at a cost of RM2.00

per meter. Should the firm do so?

(iii) What is the optimal solution of the dual problem? Interpret.

3 (MAR 2004)

85

Muarlite Company produces three types of desk lamps: Lamp A, Lamp B and Lamp

C. The lamps require three limited resources: fiberglass, plastic and wood. The

resource requirements for each lamp are as follows:

Lamp A requires 1 unit of fiberglass, 3 units of plastic and 2 units of wood. Lamp B

requires 2 units of fiberglass and 8 units of wood. Lamp C requires 1 unit of fiberglass

and 2 units of plastic. There are 500 units of fiberglass, 460 units of plastic and 840 units

of wood available each week. Profits from the sales of Lamp A, Lamp B and Lamp C are

RM30, RM10 and RM40, respectively.

(a) Formulate a linear programming model for this problem with the objective of

maximizing the weekly profit.

(b) The optimal solution to the above problem is given by the following simplex table.

Cj Solution

Mix30X1

10X2

40X3

0S1

0S2

0S3

Resources

0

40

10

S1

X3

X2

-1

1.5

0.25

0

0

1

0

1

0

1

0

0

-0.5

0.5

0

-0.25

0

0.125

60

230

105

Zj

Cj - Zj

62.5-32.5

100

400

00

20-20

1.25-1.25

X1, X2 and X3 are the respective productions of Lamp A, Lamp B and Lamp C.

S1, S2 and S3 are the respective amounts of slacks of fiberglass, plastic and wood.

(i) Obtain the quantity of each type of desk lamps to produce per week and determine

the total weekly profit.

(ii) State the resource(s) that is or are not fully utilized in the optimal production.

(iii) Formulate the dual of the problem.

(iv)Obtain the optimal solution of the dual and briefly explain the values.

4 (OCT 2004)

86

The objective of a linear programming problem is to maximize profit. The optimal

solution to the problem is given by the following simplex table.

Cj Solution

Mix

50

X1

40

X2

0

S1

0

S2

0

S3

Quantity

40

0

50

X2

S2

X1

0

0

1

1

0

0

0.32

-0.32

-0.2

0

1

0

-0.12

0.12

0.2

12

8

30

Zj 50 40 2.8 0 5.2

Cj - Zj 0 0 -2.8 0 -5.2

Where S1, S2 and S3 are the slacks for Resources 1, 2 and 3 respectively.

i) Write down the objective function for the above problem.

ii) Explain why this solution is already optimal.

iii) Obtain the optimal solution and determine the total profit.

iv) Are there any changes in the total profit when additional units of Resource 2 are

obtained? Explain your answer.

v) Is it worthwhile to obtain Resource 3 at RM5.00 per unit? Explain your answer.

5 (SEPT 2001)

A manufacturer produces two types of cotton cloth: Denim and Corduroy. Corduroy is a

heavier grade cotton cloth and, as such, requires 3.5kg of raw cotton per meter whereas

Denim requires 2.5kg of raw cotton per meter. A meter of Corduroy requires 3.2 hours of

processing time, and a meter of Denim requires 3 hours. Although the demand of Denim

is practically unlimited, the maximum demand for Corduroy is 510 meters per month.

The manufacturer has 3250kg of cotton and 3000 hours of processing time available each

month. The manufacturer makes a profit of RM4.50 per meter from Denim and RM6 per

meter from Corduroy. The manufacturer wants to know how many meters of each type of

cloth to produce to maximize profit.

(a) Formulate a linear programming model for the above problem.

(b) The final simplex table of the LP problem is shown below:

87

Cj SolutionMix

4.5 6 0 0 0 QuantityX1 X2 S1 S2 S3

0 S1 0 0 1 -0.83 -0.83 325

4.5 X1 1 0 0 0.33 -1.07 456

6 X2 0 1 0 0 1 510

Zj 4.5 6 0 1.5 1.2

Cj – Zj 0 0 0 -1.5 -1.2

X1 = number of meters of corduroy produced,

X2 = number of meters of denim produced,

S1, S2 and S3 are the slack variables for cotton usage, processing time and

demand for corduroy respectively.

(i) Determine the optimal production level and calculate the maximum profit obtained.

(ii) How many cotton (S1) and processing time (S2) are left over the optimal solution?

(iii) What is the maximum amount the manufacturer is willing to spend for an

additional processing time?

(iv) What will be the effect on the optimal solution if the manufacturer could only obtain

3000 kg of cotton per month? Justify your answer.

6 (MAR 2005)

a. A marketing company is doing research in order to maximize its profit of selling

products X, Y, and Z. These products will be sold from door to door. X is sold at RM7

per unit. It takes a sales person 10 minutes to sell one unit of X and it costs RM1 to

deliver the goods to the customer. Y is sold at RM5 each, and it takes a sales person 15

minutes to sell one unit. The goods are left with the customer at the time of sale. Z is sold

at RM12 each unit. It takes a sales person 12 minutes to sell one unit of Z, and it costs

RM0.80 to deliver it to the customer. During the week, a sales person is only allowed

delivery expenses of not more than RM75 and selling time is expected not to exceed 30

88

working hours. If the unit costs of product X, Y, and Z are respectively RM2.20,

RM1.80, and RM4.25, how many units of X, Y, and Z must be sold to maximize the total

sales? Formulate the problem as a linear programming model and develop an initial

simplex table for the problem (Do not solve the problem)

b. Indah Pottery Sdn Bhd, a manufacturer of products made from clay, is in the process

of introducing new design of vase namely Design 1, Design 2, and Design 3. The

products will be sold to the local market and to be exported to neighboring countries. The

production manager of the formulated the equation given below.

X1 = Quantity of Design 1 vase

X2 = Quantity of Design 2 vase

X3 = Quantity of Design 3 vase

Maximize Z = 12X1 + 18X2 + 10X3

Subject to: 2X1 + 3X2 + 4X3 < 50 X1 + X2 + X3 < 0

X2 + 1.5X3 < 0All X1, X2, and X3 > 0

i) State the dual for this problem

ii) The following is the final simplex table of the primal problem.

Complete the table by finding all values in the Zj and Cj – Zj rows.

Cj Solution

Mix

12

X1

18

X2

10

X3

0

S1

0

S2

0

S3

Quantity

12 X1 1 0 0.2 0.2 -0.6 0 10

18 X2 0 1 1.2 0.2 0.4 0 10

0 S3 0 0 2.7 0.2 0.4 1 10

Zj

Cj - Zj

iii) Give the optimal solution of the primal and briefly interpret it.

7 (NOV 2005)

89

A company supplies three components to automotive manufacturers. The components are

processed on 2 phases: Moulding and Polishing.

The duration in minutes required on each machine are as follows:

Component

Phase

Moulding Polishing

A 16 12

B 5 6

C 8 8

The moulding machine is available for 120 hours and the polishing machine is available

for 110 hours. No more than 200 units of component C can be sold, but up to 1000 units

of component B can be sold. In fact, the company already has orders for 500 units of

component A that must be delivered. The profit per unit for component A, B and C are

RM8.00, RM6.00 and RM9.00 respectively. Formulate the above Linear Programming

problem.

8 (APR 2006)

a) C-Bizz Mart is a retail catalog store specializing in cosmetics. Phone orders are taken

each day by a pool of computer operators, some of them are permanent while some are

temporary staff. A permanent operator can process an average of 75 orders per day, while

a temporary operator can process an average of 52 orders per day. The company receives

an average of at least 600 orders per day. A permanent operator will process about 1.3

orders with errors each day, while a temporary operator averages 4.2 orders per day with

errors. The store wants to limit errors to 24 per day. A permanent operator is paid RM60

per day including benefits and a temporary operator is paid RM40 per day. The company

wants to know the number of permanent and temporary operators to hire in order to

minimize costs.

Formulate a linear programming model for the above problem.

b) Consider the following linear programming problem:

90

Minimize Z = 2000X1 + 700X2 + 1600X3

Subject to:

5X1 + 2X2 + 2X3 > 20

4X1 + X2 + 2X3 > 30

X1, X2, X3 > 0

i) Write the dual for the above linear programming problem.

The following is an incomplete final simplex table for the dual.

Cj Solution Mix

20Y1

30Y2

0S1

0S2

0S3

Quantity

30 Y2 1.25 1 0.25 0 0 500

0 S2 0.75 0 -0.25 1 0 200

0 S3 -0.5 0 -0.5 0 1 600

Zj

Cj - Zj

ii) Complete the above simplex table

iii) State the optimal solution (including the minimum cost) for the dual.

iv) State the optimal solution (including the maximum profit) for the primal.

9 (OCT 2006)

a) Xlent Electronics manufactures several products including 25-inch GE25 and 29-inch

GE29 televisions. It makes a profit of RM50 on each GE25 and RM75 on each GE29.

During each shift, Xlent allocates up to 300 man-hours in its production area and 240

man-hours in its assembly area to manufacture the televisions. Each GE25 requires 2

man-hours in the production area and 1 man-hour in the assembly area while each

GE29 requires 2 man-hours in the production area and 3 man-hours in the assembly

area. Determine the production levels for GE25 and GE29 that would optimize the

profit per shift.

i) Formulate the above linear programming problem

ii) Solve the problem by using the graphical method

b) Complete the following initial simplex table:

91

Cj Solution

Mix

X1

(3)

X2

(5)

S1

(0)

S2

(0)

A1

(-M)

Quantity

Bi

S1 6 3 1 0 0 300

S2 5 7 0 1 0 200

A1 1 1 0 0 1 50

Zj

Cj-Zj

10. (APR 2007)

a) ACE Sdn Bhd manufactures three products X1, X2, and X3. The products must go

through two production processes: Machining and Assembly. The capacity in each

production process is limited by the number of labor hours available as shown in the

following table.

Process

Labor hour for each product Available

hoursX1 X2 X3

Machining 0.5 0.5 1.0 700

Assembly 0.5 0.5 2.0 600

Profits for the products X1, X2, and X3 are RM150, RM130, and RM250 respectively.

i) Formulate the above problem as a linear programming model

ii) Write the dual of the problem

iii) Set-up an initial simplex table for the problem

b) Given the following simplex table of a manufacturing firm:

92

Cj

Solutionmix

150 130 250 0 0 0 0 0 Quantity

X1 X2 X3 S1 S2 S3 S4 S5

0 S3 0 0 0 0.25 0.25 1 0 -0.5 56.25

150 X1 1 0 0 0 1 0 0 0 250.00

250 X3 0 0 1 -0.25 -0.25 0 0 0.5 143.75

0 S4 0 0 0 -0.25 -0.25 0 1 -0.5 243.75

130 X2 0 1 0 1 0 0 0 0 375.00

Zj 150 130 250 67.5 87.5 0 0 125

Cj - Zj 0 0 0 -67.5 -87.5 0 0 -125

i) What is the quantity of each product (X1, X2, and X3) that should be produced?

Determine the total profit.

ii) Intepret the shadow price for Resource 5

iii) State the resources that are not fully utilized.

iv) Obtain the solution for the dual problem.

v) Calculate the minimum resources used.

11. (OCT 2007)

Write down the standard form and construct the initial simplex table for the following

linear programming problem:

Maximize Z = 3X1 + 2X2 + X3

Subject to

X1 + 2X2 + 2X3 < 30

2X1 + 3X2 + X3 < 50

X1 – 2X2 > 0

X1, X2, X3 > 0

93

12. (OCT 2007)

a) Solve the following linear programming problem graphically, using the corner

point method.

Minimize Cost = X1 + 2X2

Subject to

1X1 + 3X2 > 900

8X1 + 2X2 > 1600

3X1 + 2X2 > 1200

X2 < 700

X1 X2 > 0

b) The following is the incomplete final simplex table for the dual, where S1, S2 and

S3 are the slack variables for the dual

Cj Solution

mix

10 24 50 0 0 0 Quantity

Y1 Y2 Y3 S1 S2 S3

10 Y1 1 0 0 4 -2 -1 4

24 Y2 0 1 0 0 2 -1 3

50 Y3 0 0 1 0 0 1 2

Zj

Cj - Zj

i. Complete the above simplex table

ii. Explain why this table gives the optimal solution

iii. What is the optimal solution for the primal? Interpret. What is the total cost?

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13. (APR 2008)

a) The optimal simplex table for the linear programming problem is given in the following table:

CjSolution

Mix12 15 14 0 0 0

QuantityX1 X2 X3 S1 S2 S3

S1 0 0 -1 1 2 -3 12X2 0 1 1 0 1 -0.5 16X1 1 0 0 0 -1 1 4ZJ

CJ-ZJ

Where,X1= number of units of Product 1 produced per weekX2= number of units of Product 2 produced per weekX3= number of units of Product 3 produced per weekS1= Slack variable for labor hours in Department AS2= Slack variable for labor hours in Department BS3= Slack variable for labor hours in Department C

i) Complete the optimal simplex table

ii) Determine the optimal production level and the maximum profit obtained.

iii) The firm is considering hiring an extra labor on a part time basis at a rate of RM3.50

per hour in Department C. What would be the daily profit obtained by the firm?

(Assume the firm operates 8 hours daily)

14. (OCT 2008)

a) A housing developer has purchased 20,000 m2 of land on which he is planning to

build two types of houses, detached and semi-detached within an overall budget of

RM3 million. A detached house costs RM60,000 to build and requires 650 m2 of land

and a semi-detached house costs RM40,000 to build and requires 300 m2 of land.

From past experience, the developer estimates the profits of a detached house and a

semi-detached house to be about RM9,000 and RM6,000 respectively.

i) Formulate a linear programming (LP) model for this problem.

ii) Write down the dual for this LP primal problem.

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b) The following is the optimal simplex table for the linear programming problem

of a company manufacturing two products X1 and X2. S1 , S2 and S3 are the slack

variables of the three resources A, B and C respectively involved in the manufacturing

process.

Cj

Solution

Mix

40 30 0 0 0

QuantityX1 X2 S1 S2 S3

X2 0 1 3.33 0 -2.22 20

S2 0 0 -0.67 1 0.44 1

X1 1 0 -1.67 0 2.78 25

Zj

Cj- Zj

i) Complete the above simplex table

ii) State the optimal solution, including the total profit.

iii) Is it worthwhile to purchase additional units of resource A at a cost of RM35.00 per

unit? Why?

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