PRODUCTION AND OPERATIONS MANAGEMENT
BUS 5040
TOPIC: LINEAR PROGRAMMING
1.0 Introduction
2.0 Formulation of LP
3.0 Solution of LP using Graphical method
4.0 Solution of LP using Simplex method
5.0 Conclusion
1.0 LINEAR PROGRAMMING
Linear generally describes a relationship among two or more variables which are directly
and precisely proportional. ‘LINEAR’ therefore, refers to a situation where
EQUATIONS and INEQUALITIES in the problem must be of the first degree and the
relationship among the variables must be directly and proportionally precise.
‘PROGRAMMING’ refers to ‘computing’ or ‘calculating’ some unknowns in a set of
equations and/ or inequalities MATHEMATICALLY EXPRESSED.
Consequently, ‘LINEAR PROGRAMMING’ can be said to be a mathematical technique
used in determining the best possible allocation of a firm’s scarce resources to the several
alternatives using linear relationship among variables.
Initially, the use of Linear Programming was limited to problems involving military
operations. Currently, the technique is extensively applied in business and industry.
1.1 Application of LP in Operations Management.
a) Scheduling school buses to minimize the total distance traveled when carrying
students.
b) Allocating Police Patrol units to high crime areas in order to minimize response time
to 911 calls.
c) Scheduling tellers at the bank so that needs are met during each hour of the day while
minimizing the total cost of labor.
d) Selecting the product mix in a factory to make best use of machine- and labor hours
available while minimizing the firm’s profit.
e) Picking blends of raw materials in feed mills to produce finished feed combinations at
minimum cost
f) Determining the distribution system that will minimize total shipping cost from several
warehouses to various market locations.
g) Developing a production schedule that will satisfy future demands for a firm’s product
and at the same time minimize total production and inventory costs
1
h) Allocating space for a tenant mix in a new shopping mall so as to maximize revenue to
the leasing company.
1.2 Basic Requirements of Linear Programming
a) Well defined objective function: The objective function must be well defined to
clearly show the relationship between variables and the firm’s goals. The
objective is either a MAXIMIZATION of benefits/ profits or MINIMIZATION of
losses / costs.
b) Alternative courses of action: Since a firm uses many/ alternative/or
combination of resources – MAN, MATERIALS, MACHINE, MONEY,
METHODS, MARKETS, MEASUREMENTS, MANAGEMENT – to achieve its
chosen goal there must be alternative courses of action to achieve these goals.
c) Linearity: All equations and inequalities must describe the problems in a linear
form, where only equations are acceptable. This assumes constant returns to
scales without the possibility of economies or diseconomies of scale.
d) Mathematical formulation: In order to have accurate means of measurement and
also to accurately translate the relationship among the variables used, the firm’s
objectives, equations and inequalities must be mathematically expressed.
e) Limited resources: There must be restrictions on the amount of resources used to
attain the objective. Hence, the resources must be finite, economically
quantifiable and limited in supply.
1.3 Advantages of Linear Programming
a) Optimum use of factors of production: Linear programming allows users to
effectively use the firm’s factors of production and also to effectively select and
distribute the factors.
b) Improves managerial decision making: Linear programming being factual and
PRACTICAL allows managers to collect only pertinent data which enables them
to study and consider problems in their realistic and true nature. This improves the
quality of decision making and helps in amplifying the analytic abilities of
managers.
c) Prepares future leaders: Since LP can deal with complex problems in a
systematic manner, it offers an opportunity to young managers to improve their
skill in examing, evaluating and solving such problems.
1.4 Limitations of LP
a) Practical application of models: Because of internal and external factors in an
organization, there may be a well formulated theoretical LP model which may not
have practical applications. For example, the behavior in the market in the market
being dynamic may not render itself applicable to a well formulated theoretical
model.
b) Cost: It might have been a complete waste of time and money to use original data
collected as the current conditions might have rendered them out –of-date.
2
c) Difficulties in combining objective functions: All linear programming objective
functions are either MAXIMIZATION or a MINIMIZATION one. However, in
practical situations, an organization may find itself faced with a situation where it
has to maximize some benefits while minimizing some costs. This may be
difficult to combine in a single objective function.
2.0 FORMULATION OF LP
2.1 Steps followed when formulating a LP mathematical model
i) Understand the problem clearly and also be sure whether the objective is
MAXIMIZATION or a MINIMIZATION.
ii) Decide on what the decision variables are to be and also give a precise and correct
definition
iii) Identify and represent all constraints for all the limited resources and be sure to
include all the restrictions identified.
iv) Collect all the relevant data and define all the parameters of the model as numerical
constraints.
v) Solve the problem using the appropriate technique.
2.2 To properly formulate a LP mathematical model one needs to ask and answer
the following THREE QUESTIONS.
a) What goal is to be achieved if an optimum solution is to be eventually determined
i.e. what is the objective being sought? Is it MAXIMIZATION or a
MINIMIZATION one?
b) What are going to be the decision variables? i.e. what don’t we know in this
problem that if we knew then the problem would be solved? The decision
variables are the unknowns in the problem.
c) What are the CONSTRAINTS or RESTRICTIONS or LIMITATIONS being
imposed on the problem?
Note
Once these questions are successfully answered, the mathematical model can then be
clearly set.
2.3 Formulating LP problems
One of the most common LP applications is the product – mix problem. Two or more
products are usually produced using limited resources. The company would like to
determine how many units of each product it should produce in order to maximize overall
profit given its limited resources
3
2.3.1 Class activity one..
An electronic company produces two products: (i) Apex Walkman, a portable AM / FM
CD player, and (ii) Apex Watch TV, wristwatch - sized television. The production
process of each product is similar in that both requires a certain number of hours of
electronic work and a certain number of labor hours in the assembly shop. Each
Walkman takes 4 hours of electronic work and 2hours in the assembly shop. Each watchTV requires 3 hours in electronics and 1 hour in assembly. During the current production,
240 hours of electronic time are available and 100 hours of assembly department time are
available. Each Walkman sold yields a profit of $7; each watch TV produced may be sold
for a $ 5 profit.
The company’s problem is to determine is to determine the best possible combination of
Walkmans and Watch-TVs to manufacture in order to reach the maximum profit. This
product – mix situation can be formulated as LP problem
2.3.2 Class activity two
A clothing manufacturer that produces men’s shirts and pajamas has two primary
resources available: sewing-machine time ( in the sewing department ) and cutting –
machine time ( in the cutting department ). Over the next month, the company can
schedule up to 280 hours of work on sewing machines and up to 450 hours of work on
cutting machines. Each shirt produced requires 1.00 hour of sewing time and 1.50 hours
of cutting time. Producing each pair of pajamas requires 0.75 hours of sewing time and
2.00 hours of cutting time. Each shirt sold yields a profit of $3 and each pajama sold
yields a profit of $5. Formulate a LP for this problem.
2.3.3 Class activity three
A Turkey Ranch is considering buying two different types of turkey feed. Each feed
contains, in varying proportions, some or all of the three nutritional ingredients essential
for fattening Turkeys. Brand Y feed costs the ranch $0.02 per pound. Brand Z cost $ 0.03
per pound. The Rancher would like to determine the lowest – cost diet that meets the
minimum monthly intake requirements for each nutritional ingredient.
Composition of each pound of feed:
Ingredient
Brand Y feed
Brand Z feed
A
B
C
Cost/pound
5oz
4oz
0.5oz
Ksh. 170
10oz
3oz
0.3oz
Ksh. 250
Minimum monthly
requirement.
90oz
48oz
1.5oz
4
3.0 SOLUTION OF LP USING GRAPHICAL METHOD
If a Linear programming model involves only two variables, it is possible to solve it
using graphical method. However, when there are three or more variables, the graphical
method becomes either impractical or impossible to solve.
A graph normally has four regions but the graphical solution is confined to the first
quadrant. This is because of the non – negativity requirement. In a graphical method,
constraints which are inequalities must be graphed by temporarily relaxing the
inequalities. Let all the lines on the graph be straight lines, by finding their co-ordinates.
First plot all the feasible solutions in the feasible space or area. A feasible area contains
all the possible solutions to the problem. This solution satisfies all the constraints. Hence,
A graph of a linear inequality does include:
 A straight line
 All points in one side of that line that satisfies the constraint.
Decision variables refer to choices available to a decision maker. Feasible region refers
to the set of all feasible combinations of decision variables.
3.1.1 Class activity 4.0
In order to find optimal solution to a LP problem, we must first identify a set, or region,
of feasible solutions. The first step in doing so is to plot the problems constraints on a
graph.
In the example in activity 1.0, the variables X1 (Walkmans in the example) is usually
plotted as the horizontal axis of the graph, and the variable X2 (Watch TV’s) is plotted as
the vertical axis.
The complete problem may be restated:
Maximize profit = $ 7 X1 + $ 5 X2
Subject to the constraints:
4X1 + 3X2 ≤ 240 (Electronics constraints)
2X1 + 1X2 ≤ 100 (Assembly constraints)
These two constraints are also called the non-negativity constraints
X1 ≥ 0 (Number of walkmans produced is greater than or equal to 0)
X2 ≥ 0 (Number of watch TV’s produced is greater than or equal to 0 )
The first step in graphing of the problem is to convert the constraint inequalities into
equalities ( or equations )
Constraint A:
Constraint B:
4X1 + 3X2 = 240
2X1 + 1X2 = 100
5
The conclusion made after plotting the graphs is that it would be feasible to manufacture
30 Walkmans and 20 Watch TV’s ( X1 = 30 , X2 = 20 ), but it would violate the
constraints to produce 70 Walkmans and 40 Watch TV’s.
3.1.2 Class activity 5.0
This is a class activity based on determining a graphical solution to activity 3.0. Note that
this is a MINIMIZATION problem
3.2 OTHER METHODS OF SOLVING LP PROBLEMS
3.2.1 Corner – point solution method
This technique is simpler in concept than the iso – profit line approach, but it involves
looking at every point of the feasible region. The mathematical theory behind linear
programming states that an optimal solution to any problem ( that is the values of X1, X2
that yield the maximum profit ) will lie at the corner point, or extreme point, of the
feasible region.
3.2.2 Class activity 6.0
Solution of class activity 3.0 using the corner point method.
3.2.3 Iso – profit line solution method
Iso-profit line method is an approach for solving a linear programming maximization
problem graphically. It is used once the feasible region has been established.
3.2.4 Class activity 7.0
Solution of class activity 3.0 using the Iso-profit line method
4.0 SOLUTION OF LP USING SIMPLEX METHOD
Is an alternative method used in solving problems with more than two variables? It is an
interactive or step by step method that begins from the origin then systematically moves
from the one feasible solution to another until the optimum solution is found.
The origin from which the method starts is normally referred to as the STARTING or
INITIAL solution. It is important to note that each move is expected to improve the
objective function.
4.1 Terminologies used in Simplex method.
Basic variable: These are the variables which in the current step do have positive values.
6
Non – Basic Variables: These are those variables which currently are not in the basic
and have zero values. To determine the number of these variables use the following
formula: n – m = Non –basic variables
Where: n = number of variables after having converted the original inequalities to
equalities.
m = number of constraint equations
Entering variables: This is one of the current non – basic variables which can improve
the objective function better than the other non – basic variables, if its value is increased
above zero. If there is no other non –basic variable to be increased, stop the operation
because the current basic solution is optimal.
In case there is a tie in the variables to be selected in order to basic, the selection can be
done arbitrarily. However, this may result into several computations before an optimum
solution is found. In order to reduce the number of computations, use the following rules:
a) Where there is a tie between variables, choose arbitrarily.
b) If there is a tie between slack variables, choose arbitrarily.
c) If there is a tie between a decision variable and a slack variable, choose the decision
variable as the entering variable.
Note:
i) For a maximization problem, select that non-basic variable to enter the basis, whose coefficient in the objective function is the most NEGATIVE. If none exists, the current
solution is optimal, and the operation terminates.
ii) For a minimization problem, select that non-basic variable to enter the basis, whose
co-efficient in the current objective function is the most POSITIVE. If none exists, the
current solution is optimal and the operation terminates.
Leaving Variables: This is one of the current basic variables which must be made nonbasic when the entering variables become basic, hence it leaves the base for the entering
variable. Such a variable reaches its maximum value.
Entering or Pivot Column: This is the column which contains the entering variables.
Leaving row: This is the row in the current simplex tableau containing the leaving
variables.
Pivot element: This is the positive co-efficient at the intersection of the entering column
and the leaving row.
New Pivot Equation: This is the equation to be taken by the entering variables. It is
found by dividing the co-efficients in the leaving row by the pivot element.
Old pivot equation: This is the equation of the leaving variable.
7
Optimality condition: This is the condition used for determining the entering variable. It
states that: “In a maximization (minimization) case the entering variable is the current
non – basic variable with the most NEGATIVE (POSITIVE) coefficients in the objective
equation. When all the non-basic coefficients in the objective equation are non-negative (
non- positive ), the optimum is reached”.
Feasibility condition: This is the condition used for determining the leaving variable. It
states that for both maximization and a minimization problem, the leaving variable is the
current basic variable with the smallest ratio in the right – hand side or the simplex
tableau. The coefficients used in computing ratios must be positive ones in the entering
column. Ignore zero or negative co-efficients. If there is a tie, break it arbitrarily.
4.2 Summary of steps in the simplex procedure for a maximization problem.
STEP I: Formulate the problem as on LP mathematical model. This gives the Primal
model.
STEP II: Introduce the supplemental variables to convert the primal model into
STARNDARD FORM. In this case the inequalities will be converted into
equations.
STEP III: Construct the SIMPLEX TABLEAU and enter the equalities into the tableau.
STEP IV: Calculate the objective value for this initial solution.
STEP V: Determine the entering variables by choosing the highest negative value
column. This is the variable with the highest negative coefficient in the
objective function.
STEP VI: Determine the row to be replaced from the minimum ratio column. Only
Compute the ratio for those rows whose elements are positive.
STEP VII: Compute the NEW PIVOT EQUATION for the entering variable by
dividing:
OLD PIVOT EQUATION
-----------------------------PIVOT ELEMENT
STEP VIII: Compute the value for the remaining rows including the objective values.
STEP IX: Calculate the objective value for this solution.
STEP X: If there is still a Negative value in the objective function, return to STEP V to
Determine the entering variables.
8
STEP XI: If there is no negative value in the objective function, the optimal solution has
been obtained. Hence stop the operation.
4.2.1 Class activity 7.0
Solution of a LP problem using the simplex method.
4.2.1 Class activity 8.0
Students to demonstrate independent ability to solve LP problems using the simplex
method.
6.0 CONCLUSION
The development of LP has been ranked among the most important scientific advances of
the mid – 20th Century. Its impact since just 1950 has been extraordinary. Today, it is the
standard tool that has saved many thousands or millions of dollars for most companies or
businesses of even moderate size in various countries of the world; and its use in other
sections of society has been spreading rapidly. A major proportion of all scientific
computation on computers is devoted to the use of LP.
7.0 ADDITIONAL CLASS ACTIVITIES
Activity 1:
Chad’s Pottery Barn has enough clay to make 24 small vases or 6 large vases. He has
only enough of a special glazing compound to glaze 16 of the small vases or 8 of the
large vases. Let X1 = the number of small vases and X2 = the number of large vases.
9
The smaller vases sell for $3 each, and the larger vases would bring $9 each.
(a) Formulate the problem
(b) Solve the problem graphically
Activity 2:
A fabric firm has received an order for cloth specified to contain at least 45 pounds of
cotton and 25 pounds of silk. The cloth can be woven out of any suitable mix of two
yarns A and B. They contain the proportions of cotton and silk (by weight) as shown in
the following table:
Cotton
Silk
A
30%
50%
B
60%
10%
Material A costs $3 per pound, and B costs $2 per pound. What quantities (pounds) of A
and B yarns should be used to minimize the cost of this order?
10