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OPERATIONS RESEARCH
1.0 Definition, Origin and Applications
Literally, the word ‘Operation” is an action applied to some problems while “Research” refers to
an organized process of seeking out facts about the same. Operations Research is a powerful
scientific decision making technique which deals on how best to design and operate systems under
conditions requiring allocation of scarce resources. Also, it can be defined as the application of
scientific methods, techniques and tools to problems involving the operations of systems so as to
provide those in control of the system with optimum solutions.
Operations Research was first coined in 1940 by McClosky and Trefthen in a small town, Bowdsey
of United Kingdom. The technique came into existence in military context. During World War II,
military management called on scientists from various disciplines and organized them into teams
to assist in solving strategic and technical problems to improve the execution of various military
projects. By their joint efforts and experience they suggested systematic and scientific study of the
operations of the systems and showed remarkable progress. Following the end of the war, the
success of military team attracted the attention of industrial managers who were seeking solutions
to complex problems that seemed to be similar to the problems of the armed forces.
All areas of Operations Research applications cannot be listed specifically because new areas of
its application emerges every day in today’s world, however, irrespective of the area the technique
is applied, all Operations Research practitioners face similar typical problems of the following
nature;
 In what sequence should parts be produced on a machine in order to minimize the changeover time?
 How can a dress manufacturer lay out its patterns on rolls of cloth to minimize wasted
material?
 How many elevators should be installed in a new office building to achieve an acceptable
expected waiting time?
 What's the most efficient route for a long-distance telephone call?
 What is the lowest-cost formula for chicken feed which will provide required quantities of
necessary minerals and other nutrients?
Hence, the Operations Research techniques offers solution to problems involving resources
allocation, control, inventory, monitoring, queuing, replacement, random process, searching,
scheduling and sequencing etc. Operations Research is a vital complex problem solving tool of
Applied Mathematics, Economics, Computer Science and Engineering, Industrial Engineering,
Systems Engineering and other related fields.
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1.1 Phases of Operations Research






Formulation of the problem—study the situation to identify objectives (for example,
minimize cost or maximize profit), alternative actions, constraints on the solutions and
data requirements.
Construction of Model - translate the problem from verbal and qualitative description into
a mathematical, quantitative model. The model will be an abstraction or simplification of
the real situation. Some elements (unimportant, we hope) may be ignored in order to
simplify the model.
Deriving a solution to the model - select and use an appropriate computational method to
solve the model.
Evaluate the validity of the solution — access whether the solution is it practically
reasonable and also whether some important requirements are ignored?
Establishment of control –some considerable monitoring procedures should be placed on
the solution to detect any significant changes in the conditions upon which the model is
based and to get the model revised to reflect the change.
Implementation – educate the management of the system studied on the recommended
solution of the problem for easy implementation.
1.2 Operations Research Tools/Techniques
 Linear Programming
 Integer Programming
 Goal Programming
 Dynamic Programming
 Non linear programming
 Network Analysis
 Inventory Theory
 Queuing (waiting line) Theory
 Game theory
 Decision Theory
 Replacement Theory
 Markov Analysis
 Simulation Technique
 Work Study
2.0 LINEAR PROGRAMMING (LP)
Linear programming techniques are used for solving operations problems which involve
determination of the best allocation of scarce resources among competing alternative uses such as
allocation of limited production line to competing products that can be produced and sold;
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allocation of component materials so as to minimize the cost of the product when the proportion
of its components may vary; scheduling the allocation of products at one location to users at some
other locations etc.
Most real life applications of linear programming techniques fall into the four major variant
categories;
a. Product-mix problem/Production planning problem
b. Transportation problem
c. Assignment problem.
The LP models for product-mix and production planning determines the best allocation of limited
production capacities to competing products that can be produced and sold as well as component
materials mix of products so as to minimize production cost or maximize profit. Transportation
model deals with determination of minimum cost plan for transporting a single commodity from a
number of sources to destinations while assignment model determines minimum cost plan for
assigning a number resources to tasks (such as machines to jobs or workers to machines).
Application of Linear programming requires that the following conditions must be met:
i.
ii.
iii.
iv.
v.
vi.
vii.
viii.
There must be a well-defined objective of the organization such as maximization of
profit/quantities manufactured or cost minimization to be achieved.
There must be constraints on the availability of resources such as availability of raw
materials, plant capacity, power storage capacity, product demand etc. required in
achieving the objective function.
The objective function and constraints must be expressed as linear functions of the
decision variables involved - linearity.
The contribution to the objective function from each decision variable must be
proportional to the value of the decision variable - Proportionality.
The contribution to the objective function for any variable is independent of the values
of the other decision variables.
It is assumed that the decision variables can take fractional values - Divisibility.
The values of decision variables must be greater than or equal to zero - Non-Negativity.
It is assumed that all the parameters of the program are known with certainty. Unlike
proportionality, additivity, and divisibility, the certainty assumption is not particularly
“linear”
Mathematical Representation of Linear Programming Problems
The three basic steps in formulating a linear programming model are as follows:
1. Identify all decision variables to be determined and assign algebraic symbol to each.
2. Identify the objective function (criterion) to be optimized in the given problem and express
them as linear function in terms of the decision variables.
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3. Identify all limitations or constraints in the given problem and express them as linear
inequalities or equations in terms of the decision variables.
If there are n decision variables and m constraints in the problem, the mathematical model of the
LP problem is generally given as follows;
Optimize (Max. or Min.): 𝑍 = ∑𝑘𝑖=1 𝑐𝑖 𝑥𝑖
(Objective Function)
𝑘
𝑘
Subject to the constraints: ∑𝑖=1 𝑎𝑖𝑗 𝑥𝑖 ≤ 𝑏𝑖 or ∑𝑖=1 𝑎𝑖𝑗 𝑥𝑖 ≥ 𝑏𝑖
(Constraints)
𝑥𝑖 ≥ 0
(Non-negativity restriction)
𝑡ℎ
Where; 𝑥𝑖 = Quantity of 𝑖 decision variable of interest.
𝑐𝑖 = Constant representing per unit contribution of 𝑖 𝑡ℎ decision variable to the objective
function𝑎𝑖𝑗 = Constant representing exchange coefficients of 𝑖 𝑡ℎ decision variable in the
𝑗 𝑡ℎ constraint.
𝑏𝑖 = Constant representing 𝑗 𝑡ℎ constraint requirement of availability.

Formulation and Solution Procedures for Product-Mix/Allocation Models
Specific algorithms involved in formulating and solving LP models of these categories of problems
are presented in examples 1 to 11. The models are solved using two major variant methods,
graphical and algebraic (simplex) methods.
Graphical Method: The steps involved are;
1. Formulate the appropriate LP model.
2. Plot the constraint inequalities by treating each inequality as equality and determining the
value of each variable in the equation at zero value of the other.
3. Identify the feasible region which satisfies all constraints simultaneously. For “less than or
equal to” constraints this region is found below the lines while for “greater than or equal
to” constraints breeds feasible region above the lines.
4. Locate the solution points on the feasible region which always occur at the corner points
of the feasible region.
5. Evaluate the objective function at each of the corner or solution points to identify the
optimum value.
Examples 12 and 13 illustrate how to solve LP models by graphical methods.
Simplex Method: The optimum solution obtained by graphical method is always associated with
corner points of the feasible solution region. Fundamentally simplex method is based on this
principle but lack the visual advantage of graphical solution; it employs iterative process that starts
at a feasible corner point, normally origin and systematically moves from one feasible extreme to
another until the optimum is eventually reached. There several algorithms for solving linear
programming problems by simplex method, but the easiest is the one which applies Gauss-Jordan
reduction process presented below. The steps involved are as follows;
1. Formulate the appropriate LP model.
2. Put the formulated model in a standard form by converting all inequality constraints to
equations. Converting inequality constraints to equations involves introduction of slack or
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surplus variables to each constraint as appropriate. An inequality of the “less than or equal
to” is transformed to an equality by the addition of non-negative slack variable while
“greater than or equal to” inequality goes with negative surplus variable.
3. Draw up the starting simplex matrix-initial feasible solution.
4. Identify the basic variables and non basic variables in the matrix. Non-basic variables are
variables that can be made equal to zero such as slack or surplus variables. The remaining
are basic variables.
5. Use Gauss-Jordan’s reduction method to obtain the model solution (optimum simplex
matrix). Optimum matrix is reached when the coefficients of all the basic variables are
unity in their respective equation and also eliminated or made zero other equations.
6. Interpretation of optimum simplex matrix and sensitivity analysis. Unlike the graphical
method, the optimum simplex matrix contains many important information about the
system studied, the least of which are the optimum values of basic variables (optimum
solution). The information that can be obtained from the optimum simplex matrix either
directly or with simple additional calculations include;
(a) The optimum solution
(b) Status of the resources
(c) Unit-worth of each resources
(d) Sensitivity of the optimum solution to changes in availability of resources
(e) Sensitivity of the optimum solution to changes in the usage of resources by activities.
(f) Sensitivity of the optimum solution to changes in the coefficient of the objective function.
The first item can easily be gotten from the optimum simplex matrix while the remaining five
require additional calculations that are based on the information in the optimum solutions.
Resources status:
Constraints may be classified as either scarce or abundant, depending respectively on whether or
not the optimum solution consumes the entire available amount of the associated resources. Only
constraints of the type ≤ represent resource restriction, since they imply that there is a maximum
limit on its availability. Constraints of the type ≥ cannot physically represent resource restriction
since they only imply that the solution must meet certain requirements such as satisfying minimum
demand or specification. In this case under study, there four constraints of the type ≤. The first
two representing ingredient usage as actual resources restriction while the last two dealing with
demand limitations imposed by the market condition are equivalent resources restriction. The last
two constraints in this problem are resources restrictions since increasing demand limits is
equivalent to expanding the company’s share in market which in terms of naira has the same effect
as increasing availability of physical resources (such as ingredients) through allocation of
additional funds. A zero slack shows that the entire amount of the resource is consumed by the
activities of the model; thus, it is scarce while a positive slack means that the resource is not used
completely and that it is abundant. Any increase in an abundant resource will simply make them
more abundant without affecting the optimum solution whereas a scarce resource can be increased
for the purpose of improving the solution.
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Unit-Worth of Resources:
The rate of improvement in the optimum value of the objective function as a result of increasing
the available amount of a resource is known as the unit worth of the resource. That monetary values
were associated with the unit worth of a resource does not mean that one should think of it in the
same term as the real price that can be paid to buy the resource. Instead, it is an economic measure
that quantifies the unit worth of a resource from view point of the optimal value and this varies as
the constraints changes even with the same physical resources. The unit worth of a resource is
sometimes referred to as the shadow price or more technically dual price.
Duality and Sensitivity Analysis:
Every linear programming problem has another related unique LP problem involving the same
data which also describes original problem. The original problem is called primal while the other
is known as the dual. The primal can be solved by transposing or reversing the rows and columns
of the statement of the problem. Reversing the rows and columns in this way gives us dual problem,
all maximization LP primal problems transform minimization LP dual problems and vice versa.
Solution to dual program can be found in the same way as the primal. Duality is an extremely
important instrument of linear programming technique due to the following;
1. Optimal solution of the dual gives complete information of the primal and vice versa.
2. Converting the original problem to dual simplifies mathematical computation if the primal
problem contains large number of constraints in form of rows and comparatively lesser
number of variables in the form of column.
3. Duality provides information on how the optimal solution changes with variations in the
coefficient and formation of the problem- Sensitivity Analysis
4. Duality revealed linear programming problem as a two-person zero-sum game.
The following conditions apply as primal LP problem is converted dual;
1. There is a dual constraint for every primal variable.
2. There is a dual variable for every primal constraint.
3. The objective of the dual is minimization with “greater than or equal to” constraints with
variables unrestricted in sign when the primal problem is maximization.
4. The objective of the dual is maximization with “less than or equal to” constraints with
variables unrestricted in sign when the primal problem is minimization.
5. Constraint coefficient of a primal variable becomes the left side of coefficients of the
corresponding dual constraint while objective coefficient of the same variable becomes the
right side of the dual constraint.
These relationships between the primal and dual LP problem are algebraically tabulated as follows;
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S/No.
1
Primal Problem
Maximize:
∑𝑘𝑖=1 𝑐𝑖 𝑥𝑖 Subject
∑𝑘𝑖=1 𝑎𝑖𝑗 𝑥𝑖 ≤ 𝑏𝑖
𝑥𝑖 ≥ 0
2
Minimize:
∑𝑘𝑖=1 𝑐𝑖 𝑥𝑖 Subject
∑𝑘𝑖=1 𝑎𝑖𝑗 𝑥𝑖 ≥ 𝑏𝑖
𝑥𝑖 ≥ 0
𝑍=
to
𝑍=
to
Dual Problem
Minimize: 𝑊 = ∑𝑘𝑖=1 𝑏𝑖 𝑦𝑖
Subject to ∑𝑘𝑖=1 𝑎𝑖𝑗 𝑦𝑖 ≥ 𝑐𝑖
𝑦𝑖 ≥ 0
Maximize: 𝑊 = ∑𝑘𝑖=1 𝑏𝑖 𝑦𝑖
Subject to ∑𝑘𝑖=1 𝑎𝑖𝑗 𝑦𝑖 ≤ 𝑐𝑖
𝑦𝑖 ≥ 0
The relationships between the solutions of primal and dual models are as follows:
1. The dual solution, W = Optimal solution, Z.
2. Difference between the left and right sides of the dual constraints associated with starting
primal variables are equal to the respective coefficient of each variable in the optimum
primal equation.
Example 1: The management of a coal-fired electric power generating plant is studying the plant
operational set-up in order to comply with the latest emission standard under the air pollution
control laws which fixed maximum sulphur oxide and particulate emissions of 3000ppm and
12kg/h respectively. Coal is brought to the plant by rail road and dumped into stockpiles near the
plant where it is carried by a conveyor belt to the pulverizing unit which pulverizes and feeds it to
the combustion chamber at a desired rate. The conveyor loading capacity is 20ton/h regardless of
the type of coal loaded. The heat produced in the combustion chamber is used to make steam to
drive the turbine. Two types of coal are used in the plant, Grade A which is hard, clean burning,
low sulphur content and expensive, and Grade B which is relatively cheap, soft, smoky coal with
high sulphur content. The specific pulverization rates, thermal values and emission of each coal
are as tabulated.
Coal
A
B
Pulverizing rate
(Ton/h)
16
24
Steam Production
(lb/Ton)
24000
20000
Sulphur
oxide
emission (ppm)
1800
3800
Particulate emission
per ton
0.5
1.0
Knowledge of the maximum possible steam/electricity output of the plant will enable the
management to determine the margin of safety available to meet peak demand for power.
Formulate this problem as a linear programming model.
Solution
Decision variables: 𝑥1 and 𝑥2 are respective quantities of coal A and B used per hour
Objective function: what is the combination of A and B that maximizes steam output, Z?
The total quantity of steam produced per hour, Z = 24000𝑥1 + 20000𝑥2
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Constraints:
Particulate emission; 0.5𝑥1 + 𝑥2 ≤ 12
.
Conveyor capacity;
𝑥1 + 𝑥2 ≤ 20
.
Pulverizer capacity; 0.063𝑥1 + 0.042𝑥2 ≤ 1
Sulphur oxide emissions; 1800 𝑥
𝑥1
1 +𝑥2
.
+ 3800 𝑥
𝑥2
1 +𝑥2
≤ 3000 or 1200𝑥1 + 800𝑥2 ≥ 0
.
Non-Negative Condition: x1  0; 𝑥2 ≥ 0
The complete mathematical model (LP model) of this problem is summarized as follows;
Maximize:
Subject to:
Z = 24000𝑥1 + 20000𝑥2
0.5𝑥1 + 𝑥2 ≤ 12
𝑥1 + 𝑥2 ≤ 20
0.063𝑥1 + 0.042𝑥2 ≤ 1
1200𝑥1 + 800𝑥2 ≥ 0
x1  0; 𝑥2 ≥ 0
Example 2: A company produces a product that has a fixed total weight say 10 pounds bag of
grass seed. The particular product could be a mixture of Rye, R and Fescue, F seed. Assume that
specifications state that the product must contain at least 3 pounds of Fescue and at least 4 pounds
of Rye. If the objective is to minimize total cost of production when Rye seed costs N60 per pound
and fescue N40 per pound, formulate the LP model of this problem.
Solution
Minimize:
C = 60𝑅+ 40𝐹
Subject to:
𝑅 + 𝐹 = 10
𝑅≥4
𝐹≥3
𝑅 ≥ 0; 𝐹 ≥ 0
Example 3: A small electrical manufacturing company manufactures two types of electrical
switches in one of its Department, a two pole switch and a three pole switch. The net profit on a
two pole switch is 45kobo per switch and 30kobo per switch for a three pole switch. Each of the
two pole switches requires 2minutes of machine time while the three pole switches requires
1minute per switch. The Department has five machines each of which provides 200 usable minutes
per day. Each of them requires 1minute of direct labour and two labourers are employed for this
work each providing 400 usable minutes of labour per day. The company can sell up to 400 two
pole and 700three pole switches per day. Formulate the LP model of this problem.
Solution
Decision variables: 𝑥1 and 𝑥2 are respective numbers of 2-pole and 3-pole switches produced
Maximize:
Z = 0.45𝑥1 + 0.30𝑥2
Subject to:
2𝑥1 + 𝑥2 ≤ 1000
𝑥1 + 𝑥2 ≤ 800
𝑥1 ≤ 400
𝑥2 ≤ 700 ; x1  0; 𝑥2 ≥ 0
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Example 4: Four different models of a certain products, P, Q, R and S. The most critical operations
are in machine centres, A, B and C. Each of the products must go through each of the machine
centre with varying amount of time required as tabulated. The total available time in hour for
centers A, B and C are 2000 8000 and 5000 respectively. If the profits per unit generated by the
models are N5.24, N7.30, N8.44 and N4.18 for P, Q, R and S respectively, formulate the LP model
of this problem.
Machine Centre
A
B
C
P
1.5
1
1.5
Required Time (Hour)
Q
R
1
2.4
5
1
3
3.5
S
1
3.5
1
Solution
Decision variables: 𝑥1 ,𝑥2 ,𝑥3 and 𝑥4 are respective numbers of P, Q, R and S produced
Maximize:
Z = 5.24𝑥1 + 7.30𝑥2 + 8.34𝑥3 + 4.18𝑥4
Subject to:
1.5𝑥1 + 𝑥2 + 2.4𝑥3 + 𝑥4 ≤ 2000
𝑥1 + 5𝑥2 + 𝑥3 + 3.5𝑥4 ≤ 8000
1.5𝑥1 + 3𝑥2 + 3.5𝑥3 + 𝑥4 ≤ 5000
x1  0; 𝑥2 ≥ 0; x3  0; 𝑥4 ≥ 0
Example 5: KAN international company owns a small wine factory that produces both alcoholic
and non-alcoholic wines on wholesale distribution. Two basic ingredients, A and B are used to
produce the wine. The maximum availability of A is 8 tons per day while that of B is 10 tons per
day. The daily requirements of the ingredients per ton of alcoholic and non-alcoholic wine are
summarized in the following table.
Ingredient
A
B
Tons of Ingredient per ton of wine
Non-alcoholic
Alcoholic
2
3
3
2
Maximum
Availability
8
10
A survey has established that the daily demand for alcoholic wine cannot exceed that non-alcoholic
wine than2 tons. The survey also shows that the demand for alcoholic wine is limited to 2 tons
daily. Wholesale prices in naira per ton of alcoholic and non-alcoholic wine are thirty and forty
thousand respectively. It is required to determine how much alcoholic and non-alcoholic wine the
company will produce to maximize gross income. Formulate the LP model of this problem.
Solution
Decision variables:𝑥1 and𝑥2 are respective tons of non-alcoholic and alcoholic wines to be
produced.
Maximize:
Z = 4𝑥1 + 3𝑥2
Subject to constraints of:
2𝑥1 + 3𝑥2 ≤ 8
3𝑥1 + 3𝑥2 ≤ 10
−𝑥1 + 𝑥2 ≤ 2
𝑥2 ≤ 2
x1  0; 𝑥2 ≥ 0
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Example 6: Nelson Company produces two types of locomotive parts. The company buys castings
that are drilled, bored and finished. Castings for part A costs N3 and sell for N7 each while part B
costs N4 and sell for N9 each. The running cost per hour for drilling, boring and finishing machines
are N30 and N18 and N20. If the capacity limits of each machine with respect to each part is as
tabulated below and the management desired to know the product mix for maximum profit,
formulate the problem as LP model.
Machine
Drilling
Boring
Finishing
Capacity Limit per hour
PART A
PART B
30
60
36
36
40
20
Solution
Decision variables:𝑥1 and𝑥2 are respective quantities of Part A and B to be produced per hour.
Determination of profit per part produced:
Cost of drilling
Cost of boring
Cost of finishing
Cost of cast
Total Cost
Selling Price
Profit
PART A
30/30 =1.0
18/36 =0.5
20/40 = 0.5
3
5
7
2
PART B
30/60 = 0.5
18/36 = 0.5
20/20 = 1.0
3
6
9
3
Thus, total profit per hour, Z is given as; 2𝑥1 + 3𝑥2
𝑥
𝑥
𝑥
𝑥
Capacity limits: Drilling:- 301 + 302 ≤ 1; Boring:- 361 + 362 ≤ 1; Finishing:-
𝑥1
40
𝑥
+ 202 ≤ 1.
Multiplying these constraints through to clear fraction gives the LP model for this problem as;
Z = 2𝑥1 + 3𝑥2
2𝑥1 + 𝑥2 ≤ 60
𝑥1 + 𝑥2 ≤ 36
𝑥1 + 2𝑥2 ≤ 40
x1  0; 𝑥2 ≥ 0
Example 7: Sampson Company has the option of using one of four different types of production
processes. First and second processes yield items of product A while the third and fourth yield
items of product B. The inputs of each process are cartons of material M, kilogram of material N
and labour measured in man-weeks. The profitabilities of the process differ even for processes
producing the same item, since each process varies in its input requirements. The company
deciding on a weeks production schedule is limited in the range of possibilities by the available
amount of manpower and of both kinds of raw materials. The full technology and input restrictions
are given in the table below.
Maximize:
Subject to:
Item
M in cartons
N in kg
Man-Week
Unit Profit
Production Level
One item of product, A
4
5
8
6
1
1
5
6
𝑥1
𝑥2
One item of product, B
10
18
4
2
1
1
10
12
𝑥3
𝑥4
Total Availability
120
150
20
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It is required to find the unknown production levels,𝑥1 ,𝑥2 ,𝑥3 and 𝑥4 which maximizes profit.
Formulate the LP model of this problem.
Solution
Complete mathematical model for the problem is summarized as follows;
Maximize:
Z = 5𝑥1 + 6𝑥2 + 10𝑥3 + 12𝑥4
Subject to:
4𝑥1 + 5𝑥2 + 10𝑥3 + 18𝑥4 ≤ 120
8𝑥1 + 6𝑥2 + 4𝑥3 + 2𝑥4 ≤ 150
𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 ≤ 20
x1  0; 𝑥2 ≥ 0; x3  0; 𝑥4 ≥ 0
Example 8: Richman Company may purchase and mix of one or more of three types of grains,
each containing different amount of four nutritional elements; the data are given in table below.
The management specifies that any feed mix for the live stock must meet at least minimum
nutritional requirement and they seek the least costly among all such mixes. Suppose their planning
horizon is two weeks period. Formulate the LP model of this problem.
Nutrient
Ingredient A
Ingredient B
Ingredient C
Ingredient D
Cost per unit weight
Weight level
Grain I
3
1
4
0.5
42
𝑥1
One unit weight
Grain II
Grain III
4
8
2
0
2
0
0.3
1
36
100
𝑥2
𝑥3
Minimum Total Requirement
in two weeks
1200
300
800
210
Solution
The problem of this company can be the stated in other words as determine the unknown weight
level of𝑥1 ,𝑥2 and 𝑥3 which minimizes cost. The complete mathematical model for the problem is
summarized as follows;
Minimize:
C = 42𝑥1 _ + 36𝑥2 + 100𝑥3
Subject to:
3𝑥1 _ + 4𝑥2 + 8𝑥3 ≥ 1200
𝑥1 _ + 2𝑥2 ≥ 300
4𝑥1 _ + 2𝑥2 ≥ 800
0.5𝑥1 _ + 0.3𝑥2 + 8𝑥3 ≥ 210
𝑥1 ≥ 0; x2  0; 𝑥3 ≥ 0
Example 9: A company runs by Jim and Jane produces tables and chairs. Each chair requires
4hours to assemble and 2 hours to finish while each table requires 2 hours to assemble and 4hours
to finish. Jim works only on the assembly operation while Jane works only on the finishing
operation and each puts in not more than 8 hours per day. There is a variable cost per day of N30
per chair and N20 per table. The chair and table sales cannot exceed 3 and 4 units respectively.
While the total sales for the chair and table cannot exceed 5 units. The chairs and tables require
drying after gluing. A chair requires one hour while a table 5 hours in the drying room. Because
start-up is costly, operating the room is expensive unless it is used for at least 5 hours each day.
There are also requirements with regard to packing, 5 machine runs are required for a chair and 2
for a table. It is impractical to set up this machine unless at least 10 runs are to be made. If these
young entrepreneurs wish to minimize cost, formulate the LP model of this problem.
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Solution
Decision variables: 𝑥1 and 𝑥2 are respective numbers of chairs and tables to be produced
Minimize:
C = 3𝑥1 + 2𝑥2
Subject to:
4𝑥1 + 2𝑥2 ≤ 8
2𝑥1 + 4𝑥2 ≤ 8
x1  3; 𝑥2 ≤ 4
𝑥1 + 𝑥2 ≤ 5
𝑥1 + 5𝑥2 ≥ 5
5𝑥1 + 2𝑥2 ≥ 10
x1  0; 𝑥2 ≥ 0
Example 10: A person has N12,000,000 which he want to invest in different types of investment
opportunities. He has consulted investment consultant to advise him on this who gave him advice
to invest up to 40% on government sector, N2, 000,000 on companies share A and B put together
and at least 25% on mutual fund. If the anticipated rates of return from these investments are 0.1,
0.2, 0.25, 0.15 and 0.12 for government sector, share A, share B and mutual fund respectively,
formulate the LP model of this investment dilemma.
Solution
Decision variables: 𝑥1 ,𝑥2 ,𝑥3 , 𝑥4 and 𝑥5 are respective investment on government sector, share A,
share B and mutual fund.
Complete mathematical model for the problem is summarized as follows;
Maximize:
Z = 0.1𝑥1 _ + 0.2𝑥2 + 0.25𝑥3 + 0.15𝑥4 + 0.12𝑥5
Subject to:
𝑥1 _ + 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 ≤ 12,000,000
𝑥1 ≤ 4,800,000
𝑥2 + 𝑥3 ≤ 2,000,000
𝑥4 + 𝑥5 ≥ 3,000,000
𝑥1 ≥ 0; x2  0; 𝑥3 ≥ 0; x4  0; 𝑥5 ≥ 0
Example 11: Production of a certain chemical mixture should contain 80mg of chloride, 28mg of
nitrate and 36mg of sulphate per kilogram. The company can use two substances and a costless
base. Substance X contains 8mg of chloride, 4mg of nitrate and 6mg of sulphate per gram.
Substance Y contains 10mg of chloride, 2mg of nitrate and 2mg of sulphate per gram. Both
substances cost N20 per gram. It is required to produce the mixture using substances X and Y so
that cost is minimized. Formulate the LP model of production process.
Solution
Decision variables: 𝑥1 and 𝑥2 are respective units of substances X and Y to be used
Minimize:
C = 20𝑥1 + 20𝑥2
Subject to:
8𝑥1 + 10𝑥2 ≥ 80
4𝑥1 + 2𝑥2 ≥ 28
6𝑥1 + 2𝑥2 ≥ 36
x1  0; 𝑥2 ≥ 0
13
Example 12: Solve the following linear programming problem using graphical method
Maximize:
Z = 13𝑥1 + 15𝑥2
Subject to:
6𝑥1 + 5𝑥2 ≤ 54
3𝑥1 + 4𝑥2 ≤ 36
x1  0; 𝑥2 ≥ 0
Solution
Step 1: Formulate the appropriate LP model:
The LP is as given.
Step 2: Plotting of the equations:
For first constraint: - When 𝑥1 = 0; 𝑥2 = 10.8while 𝑥1 = 9 when 𝑥2 = 0;- (0, 10.8) (9,0)
For second constraint: - When 𝑥1 = 0; 𝑥2 = 9while 𝑥1 = 12 when 𝑥2 = 0;- (0, 9) (12, 0)
The plotting of the two lines on 𝑥1 and 𝑥2 axis is shown below;
Step 3: Identification of feasible region:
The feasible region is the area enclosed be the two lines, thus, ODEA is the feasible region.
Step 4: Locate the solution points:
The solution points constitutes the coordinates of the feasible region which are points
O(0,
0); D(0, 9); A(9, 0) and E(4, 6)
Step 5: Evaluate the objective function at each of the corner or solution points:
The objective function; Maximize: Z = 13𝑥1 + 15𝑥2
Solution point O (0, 0); 𝑥1 = 0; 𝑥2 = 0; 𝑍 = 0
Solution point D (0, 9); 𝑥1 = 0; 𝑥2 = 9; 𝑍 = 135
Solution point A (9, 0); 𝑥1 = 9; 𝑥2 = 0; 𝑍 = 117
Solution point E (4, 6); 𝑥1 = 4; 𝑥2 = 6; 𝑍 = 142
The solution point E yields maximum profit of N142, thus the manufacture should produce or
allocate 4 units of 𝑥1 and 6 units of 𝑥2 .
Example 13: Solve the following linear programming problem using graphical method
Minimize:
Z = 25𝑥1 + 50𝑥2
Subject to:
30𝑥1 + 5𝑥2 ≥ 120
2𝑥1 + 10𝑥2 ≥ 30
10𝑥1 + 9𝑥2 ≥ 90
x1  0; 𝑥2 ≥ 0
Solution
Step 1: Formulate the appropriate LP model:
The LP is as given.
Step 2: Plotting of the equations:
For first constraint:- When 𝑥1 = 0; 𝑥2 = 24while 𝑥1 = 4 when 𝑥2 = 0;- (0, 24) and (4,0)
For second constraint:- When 𝑥1 = 0; 𝑥2 = 3while 𝑥1 = 15 when 𝑥2 = 0;- (0, 3) and (15, 0)
For third constraint:- When 𝑥1 = 0; 𝑥2 = 10while 𝑥1 = 9 when 𝑥2 = 0;- (0, 10) and (9, 0)
The plotting of the two lines on 𝑥1 and 𝑥2 axis is shown below;
14
Step 3: Identification of feasible region:
The feasible region is the area upward direction of the lines, thus, BGHC is the feasible region.
Step 4: Locate the solution points:
The solution points constitutes the coordinates of the feasible region which are points B(0, 24);
G(2.87, 6.82); H(7.7, 1.46) and C(15, 0).
Step 5: Evaluate the objective function at each of the corner or solution points:
The objective function; Maximize:
Z = 25𝑥1 + 50𝑥2
Solution point B (0, 4); 𝑥1 = 0; 𝑥2 = 24;
𝑍 = 1200
Solution point G (2.87, 6.82); 𝑥1 = 2.87; 𝑥2 = 6.82; 𝑍 = 412.75
Solution point H (7,7, 1.46); 𝑥1 = 7.7; 𝑥2 = 1.46; 𝑍 = 265.5
Solution point C (15, 0); 𝑥1 = 15; 𝑥2 = 0;
𝑍 = 375
The solution point H yields minimum cost of N265.50, thus the manufacture should produce or
allocate 7.7 units of 𝑥1 and 1.46 units of𝑥2 .
The graphical method can be extended to models with three or four decision variables using three
and four axis plots respectively, however simplex method is most appropriate for solving models
with more than two decision variables.
Example 14: Solve the following linear programming problem by simplex method
Maximize:
Z = 8𝑥1 + 10𝑥2
Subject to:
3𝑥1 + 4𝑥2 ≤ 15
4𝑥1 + 𝑥2 ≤ 8
x1  0; 𝑥2 ≥ 0
Solution
Step 1: Formulate the appropriate LP model
The LP is as given.
Step 2: Put the formulated model in a standard form.
Maximize:
Z = 8𝑥1 + 10𝑥2 or Z -8𝑥1 - 10𝑥2 = 0
Subject to:
3𝑥1 + 4𝑥2 + 𝑆1 = 15
4𝑥1 + 𝑥2 + 𝑆2 = 8
x1  0; 𝑥2 ≥ 0; 𝑍 ≥ 0; S1  0; 𝑆2 ≥ 0
Step 3: Draw up the starting simplex matrix-initial feasible solution.
𝑥1
3
4
-8
𝑥2
4
1
-10
𝑆1
1
0
0
𝑆2
0
1
0
Z
0
0
1
15
8
0
Step 4: Identify the basic variables and non basic variables in the matrix.
The basic variables are𝑥1 ,𝑥2 and Z while non-basic variables are 𝑆1and 𝑆2
Step 5: Using Gauss-Jordan’s reduction method to obtain the optimum simplex matrix.
15
Operation 1:
𝑅1
:3
𝑥1
R1
R2
R3
Operation 2: 𝑅2 − 4𝑅1 and 𝑅3 + 8𝑅1 :-
R1
R2
R3
𝒙𝟐
𝑆1
1
4
-8
4/3
1
-10
1/3
0
0
1
0
0
4/3
-13/3
2/3
1/3
-4/3
8/3
𝑥1
𝑥1
3
Operation 3:𝑅2 (− 13):-
R1
R2
R3
1
0
0
𝑥1
4
𝒙𝟐
2
Operation 4: 𝑅1 − 3 𝑅2 and𝑅3 − 3 𝑅2 :-
R1
R2
R3
1
0
0
𝒙𝟐
4/3
1
2/3
𝑆1
𝑆2
0
1
0
Z
0
0
1
𝑆2
5
8
0
Z
0
1
0
0
0
1
15
-12
40
𝑆1
𝑆2
Z
1/3
4/13
8/3
0
-3/13
0
0
0
1
𝑆2
Z
4/13
-3/13
2/13
0
0
1
𝒙𝟐
𝑆1
0
1
0
-1/13
4/13
96/39
5
36/13
40
51/39
36/13
496/13
Thus, the manufacture should produce or allocate 51/39 units of 𝑥1 and 36/13 units of𝑥2 to make
a maximum profit of N496/13.
Example 15: Lincon Company produces two types of feed for pigs and birds. The basic
ingredients, A and B are used to produce the feed. The maximum availability of A is 6 tons per
day while that of B is 8 tons per day. The daily requirement of the ingredients per ton of pig feed
and bird feed are as tabulated.
Ingredient
Tons of ingredient per ton of feed Maximum
Availability (ton)
Pig feed
Bird feed
1
2
6
A
2
1
8
B
Market survey has revealed that maximum demand for bird feed is limited to 2 ton daily and that
the daily demand for pig feed cannot exceed that of bird feed by more than one ton. The price per
ton is N3 thousand for bird feed and N2 thousand for pig feed.
Solution
The LP model for this problem is stated as follows;
Maximize:
Z = 2𝑥1 + 3𝑥2
Subject to:
𝑥1 + 2𝑥2 ≤ 6
2𝑥1 + 𝑥2 ≤ 8
−𝑥1 + 𝑥2 ≤ 1
𝑥2 ≤ 2
x1  0; 𝑥2 ≥ 0
Where 𝑥1 and 𝑥2 are respective tons of pig feed and bird feed while Z is the profit. The standard
form of this model is as follows;
16
Maximize:
Subject to:
Z - 3𝑥1 - 2𝑥2
𝑥1 + 2𝑥2 + 𝑆1 = 6
2𝑥1 + 𝑥2 + 𝑆2 = 8
−𝑥1 + 𝑥2 + 𝑆3 = 1
𝑥2 + 𝑆4 = 2
𝑥1 ≥ 0; x2  0 𝑥1 ≥ 0; S1  0; 𝑆2 ≥ 0; S 3  0; 𝑆4 ≥ 0
The initial
feasible matrix;
𝑥1
1
2
-1
0
-3
𝑥2
2
1
1
1
-2
𝑆1
𝑆2
𝑆3
1
0
0
0
0
0
1
0
0
0
0
0
1
0
0
𝑆1
-1/3
2/3
-1
-2/3
1/3
𝑆2
2/3
-1/3
1
1/3
4/3
0
0
1
0
0
𝑆4
0
0
0
1
0
Z
0
0
0
0
1
6
8
1
2
0
The optimum simplex matrix
𝑥1
1
0
0
0
0
𝑥2
0
1
0
0
0
𝑆3
𝑆4
0
0
0
1
0
Z
0
0
0
0
1
10/3
4/3
3
2/3
38/3
Optimum solution: Produce 10 and 4 tons of pig and bird feeds respectively to get maximum profit
of N38,000.
Resources status: The status of the resources in this problem can be obtained from the optimum
simplex matrix as follows;
Resources
Equation
Ingredient A
𝑥1 + 2𝑥2 + 𝑆1 = 6
Ingredient B
2𝑥1 + 𝑥2 + 𝑆2 = 8
Limit on excess of pig feed
over bird feed
Limit on demand for pig
feed
−𝑥1 + 𝑥2 + 𝑆3 = 1
𝑥2 + 𝑆4 = 2
Computation
10
4
+ 2 ( ) + 𝑆1 = 6; 𝑆1 = 0
3
3
10
4
2 ( ) + ( ) + 𝑆2 = 8; 𝑆2 = 0
3
3
10 4
−
+ + 𝑆3 = 1; 𝑆3 = 3
3 3
4
2
+ 𝑆4 = 2; 𝑆4 =
3
3
Resources
Status
Scarce
Scarce
Abundant
Abundant
Unit-Worth of Resources: The unit worth of each resource in the problem is tabulated as follows;
Resources
Unit Worth
Ingredient A
Ingredient B
Limit on excess of pig feed over bird feed
Limit on demand for pig feed
1/3 thousand naira per ton
4/3 thousand naira per ton
0
0
It is obvious that the last two resources are abundant from their zero unit worth and this confirmed
the results of the analysis on resources status.
17
Example 16: Write the dual of the following primal problem.
Maximize:
Z = 4𝑥1 + 10𝑥2 + 3𝑥3
Subject to:
𝑥1 + 3𝑥2 + 𝑥3 ≤ 15
3𝑥1 − 𝑥2 + 2𝑥3 = 12
𝑥1 ≥ 0; x2  0; 𝑥3 ≥ 0
Solution: The standard form of this primal LP is as follows;
Maximize:
Z = 4𝑥1 + 10𝑥2 + 3𝑥3 + 0𝑆1
Subject to:
𝑥1 + 3𝑥2 + 𝑥3 + 𝑆1 = 15
3𝑥1 − 𝑥2 + 2𝑥3 + 0𝑆1 = 12
x1  0; 𝑥2 ≥ 0;
x3  0; 𝑆1 ≥ 0
The dual equivalent is given as;
Minimize:
𝑊 = 15𝑦1 + 12𝑦2
Subject to:
𝑥1 : 𝑦1 + 3𝑦2 ≥ 4
𝑥2 : 3𝑦1 − 𝑦2 ≥ 10
𝑥3 : 𝑦1 + 2𝑦2 ≥ 3
𝑆1 : 𝑦1 + 0𝑦2 ≥ 0 (implying that𝑦1 ≥ 0)
𝑦1 , 𝑦2 unrestricted
Removing redundancy of 𝑦1 unrestricted which is dominated by 𝑦1 ≥ 0, the dual problem
summarizes as follows;
Minimize:
𝑊 = 15𝑦1 + 12𝑦2
Subject to:
𝑦1 + 3𝑦2 ≥ 4
3𝑦1 − 𝑦2 ≥ 10
𝑦1 + 2𝑦2 ≥ 3
𝑦1 ≥ 0
𝑦2 unrestricted
Example 17: Write the dual of the following primal problem.
Minimize:
Z = 7𝑥1 − 3𝑥2
Subject to:
−2𝑥1 + 2𝑥2 ≥ −5
3𝑥1 + 5𝑥2 ≥ 7
x1  0; 𝑥2 ≥ 0
Solution:
The standard form of this primal LP is as follows;
Minimize:
Z = 7𝑥1 − 3𝑥2 + 0𝑆1 + 0𝑆2
Subject to:
2𝑥1 − 2𝑥2 + 𝑆1 + 0𝑆2 = 5
3𝑥1 + 5𝑥2 + 0𝑆1 + 𝑆2 = 7
x1  0; 𝑥2 ≥ 0
x1  0; 𝑥2 ≥ 0; S1  0; 𝑆2 ≥ 0
18
The dual equivalent is given as;
Maximize:
Subject to:
𝑊 = 5𝑦1 + 7𝑦2
𝑥1 : 2𝑦1 + 3𝑦2 ≤ 7
𝑥2 : −2𝑦1 + 5𝑦2 ≤ −3
𝑆1 : 𝑦1 + 0𝑦2 ≤ 0 (implying that𝑦1 ≥ 0)
𝑆2 : 0𝑦1 + 𝑦2 ≤ 0 (implying that𝑦2 ≥ 0)
𝑦1 , 𝑦2 unrestricted
Removing redundancy of 𝑦1 and 𝑦2 unrestricted which is dominated by 𝑦1 ≥ 0 and 𝑦2 ≥ 0, the dual
problem summarizes as follows;
Maximize:
Subject to:
𝑊 = 5𝑦1 + 7𝑦2
2𝑦1 + 3𝑦2 ≤ 7
−2𝑦1 + 5𝑦2 ≤ −3
𝑦1 ≤ 0
𝑦2 ≤ 0
Example 18: Write the dual of the following primal problem.
Maximize:
Z = 6𝑥1 + 7𝑥2
Subject to:
𝑥1 + 3𝑥2 = 7
−2𝑥1 + 7𝑥2 ≥ 3
3𝑥1 + 7𝑥2 ≤ 9
𝑥1 unrestricted
𝑥2 ≥ 0
Solution: The standard form of this primal LP is as follows;
Maximize:
Z = 6𝑥1 / − 6𝑥1 // + 7𝑥2 + 0𝑆1 + 0𝑆2
Subject to:
𝑥1 / − 𝑥1 // + 3𝑥2 + 0𝑆1 + 0𝑆2 = 7
−2𝑥1 / + 2𝑥1 // + 5𝑥2 − 𝑆1 + 0𝑆2 = 3
3𝑥1 / + 3𝑥1 // + 7𝑥2 − 0𝑆1 + 𝑆2 = 9
𝑥1 / ≥ 0; x1  0; 𝑥2 ≥ 0; S1  0; 𝑆2 ≥ 0
The dual equivalent is given as;
Minimize:
𝑊 = 7𝑦1 + 3𝑦2 + 9𝑦3
Subject to:
𝑦1 − 2𝑦2 + 3𝑦3 ≥ 6
//
−𝑦1 + 2𝑦2 − 3𝑦3 ≥ −6 or (𝑦1 − 2𝑦2 − 3𝑦3 ≤ 6)
3𝑦1 + 5𝑦2 + 7𝑦3 ≥ 7
0𝑦1 − 𝑦2 + 0𝑦3 ≥ 0 (implying that𝑦2 ≥ 0)
0𝑦1 + 0𝑦2 + 𝑦3 ≥ 0 (implying that𝑦3 ≥ 0)
𝑦1 , 𝑦2 , 𝑦3 unrestricted
Removing redundancy of 𝑦2 and 𝑦3 unrestricted which is dominated by𝑦2 ≥ 0 and𝑦3 ≥ 0, gives
the dual problem as;
Minimize:
𝑊 = 7𝑦1 + 3𝑦2 + 9𝑦3
Subject to:
𝑦1 − 2𝑦2 + 3𝑦3 = 6
3𝑦1 + 5𝑦2 + 7𝑦3 ≥ 7
−𝑦2 ≥ 0;𝑦3 ≥ 0;𝑦1 unrestricted
19
Example 19: Find the dual solution to the following primal problem.
Maximize:
Z = 3𝑥1 + 2𝑥2
Subject to:
4𝑥1 + 3𝑥2 ≤ 12
4𝑥1 + 𝑥2 ≤ 8
4𝑥1 − 𝑥2 ≤ 8
𝑥1 ≥ 0; 𝑥2 ≥ 0
Solution: The optimum matrix of this primal obtained using simplex method is given as;
𝑥1
1
0
0
0
𝑥2
0
1
0
0
𝑆1
-1/8
½
2
5/8
𝑆2
3/8
-1/2
2
1/8
𝑆3
0
0
1
0
𝑍
0
0
0
1
3/2
2
4
17/2
The dual solution, W = Optimal solution of the primal, Z = 17/2. The optimum dual variables are
1
8
5
5
1
8
obtained from this table as follows; 𝑦1 − 0 = 8 ; y 2  0  ; 𝑦3 − 0 = 0, hence 𝑦1 = 8 ; y 2  ; 𝑦3 =
0
Example 20: Obtain the solution to the following primal problem by solving the dual.
Maximize:
Z = 3𝑥1 + 2𝑥2
Subject to:
4𝑥1 + 3𝑥2 ≤ 12
4𝑥1 + 𝑥2 ≤ 8
4𝑥1 − 𝑥2 ≤ 8
𝑥1 ≥ 0; 𝑥2 ≥ 0
The standard dual problem is as follows;
Minimize:
𝑊 − 12𝑦1 − 8𝑦2 − 8𝑦3 − 0𝑆1 − 0𝑆2 = 0
Subject to:
4𝑦1 + 4𝑦2 + 4𝑦3 − 𝑆1 − 0𝑆2 = 3
3𝑦1 + 𝑦2 − 𝑦3 − 0𝑆1 − 𝑆2 = 2
y1  0; 𝑦2 ≤ 0; 𝑦3 ≤ 0
The starting simplex matrix:
𝑦1
4
3
-12
R1
R2
R3
Operation 1:
𝑅1
; 𝑅2
4
R1
R2
R3
Operation 2: −
R1
R2
R3
𝑦2
4
1
-8
−𝑆1
1
0
0
−𝑆2
0
1
0
𝑊
0
0
1
3
2
0
𝑦3
1
-4
4
−𝑆1
1/4
-3/4
3
−𝑆2
0
1
0
𝑊
0
0
1
3/4
-1/4
9
𝑦3
-1
2
-4
−𝑆1
-1/8
3/8
3/2
−𝑆2
½
-1/2
2
𝑊
0
0
1
5/8
1/8
17/2
− 3𝑅1 ; 𝑅3 − 12𝑅1
𝑦1
1
0
0
𝑅2
; 𝑅1
2
𝑦1
1
0
0
𝑦3
4
-1
-8
𝑦2
1
-2
4
− 𝑅2 ; 𝑅3 − 4𝑅2
𝑦2
0
1
0
20
The primal solution, Z = Optimal solution of this dual, W = 17/2. The optimum primal variables
3
3
are obtained from this table as follows; 𝑥1 − 0 = 2 ; 𝑥2 − 0 = 2; hence 𝑥1 = 2 ; 𝑥2 = 2.
 Formulation and Solution Procedures for Transportation Problem
Formulation of a transportation model requires the following data;
1. Level of supply at each source.
2. Level of demand at each destination.
3. Unit transportation cost of the commodity from each source to each destination.
A necessary condition for feasible solution to any transportation problem is that the total sources
must be equal to total destinations. When the total sources equals total destinations the problem is
known as balanced transportation problem otherwise it is unbalanced. Introduction of dummy
source in an unbalanced transportation model in which the total capacities of the sources is less
than the total requirements of destinations and vice versa in order to establish equality between the
two capacities is required to make the model amenable for solution by transportation algorithm.
The three major variant methods of finding initial feasible solution of a transportation problem
include North-West Corner, Least-Cost and Vogel’s Approximation methods. Although, the
solution methods don’t normally yield optimal solution, Vogel’s method gives better initial
solution; hence, all results obtained by this method must be subjected to optimality test to confirm
that it is optimal.
North-West Corner Method: The steps involved are;
1. Prepare a balanced transportation showing the requirements of destinations and capacities
of sources
2. Make small squares at the right corners of the large squares and put respective cell
transportation costs in them.
3. Select the upper left hand (north-west) corner cell for shipment.
4. Make allocation as much as possible in the north-west cell to exhaust either the supply
available at one source or the demand at one destination without considering cost.
5. Adjust the supply and demand values. If all supply and demand values are exhausted then
stop, otherwise move one cell to the right or one cell down depending on the supply and
demand values to make allocation as in step four.
Least-Cost Method: The steps involved are;
1. Prepare a balanced transportation table showing the requirements of destinations and
capacities of sources
2. Make small squares at the right corners of the large squares and put respective cell
transportation costs in them.
3. Select the cell with least transportation cost for shipment.
4. Make allocation as much as possible in the least cost cell to exhaust either the supply
available at one source or the demand at one destination.
21
5. Adjust the supply and demand values. If all supply and demand values are exhausted then
stop, otherwise make allocation to the next least cost cell in succession until all supply and
demand values are exhausted.
Vogel’s Approximation Method: The steps involved are;
1. Prepare a balanced transportation showing the requirements of destinations and capacities
of sources
2. Make small squares at the right corners of the large squares and put respective cell
transportation costs in them.
3. Compute the penalty numbers for each row and column by taking the difference between
the lowest and second to the lowest unit transportation costs. The difference indicates the
penalty or extra cost which has to be paid for not assigning an allocation to the cell with
minimum transportation cost.
4. Select the row or column having the largest penalty number. If there is tie then it can be
broken by selecting the cell where maximum allocation can be made.
5. Allocate the maximum number of units possible to the cell with the lowest cost in the row
or column selected in step four.
6. Adjust the supply and demand values to eliminate any row or column for which availability
and requirement conditions are met.
7. Repeat step 3 to 6 until the entire available capacities at various sources and requirements
at destinations have been met before testing for optimality.
Optimality Test: The steps involved are;
1. Check whether the total number of allocations made in the basic feasible solution is equal
to m + n – 1, where m is the number of destinations and n is the number of sources.
2. Prepare the initial value table by putting the respective cost values in any cell that allocation
was made.
3. Calculate the column and row values. For this start by placing the first column value as
zero and obtain the first row value such that the sum of the column and row values gives
the cost value of first cell (1, 1). Obtain the remaining row and column values with respect
to the allocations in succession from the first.
4. Calculate the cost values of the cells without allocation in the basic feasible solution such
that the sum of the respective column and row values gives the value of cost value of each
cell.
5. Prepare test plan to see whether further improvement is required by subtracting the actual
real cost from the cost values obtained in step four. If the results are less than or equal to
zero, the solution is the optimum one while positive values in the test plan indicate that
improvement is possible.
6. When there is a positive value in the test plan, shift some allocation from cells where zero
occurred to the cell where positive value occur in the test plan and balance the matrix. The
22
shifting should only be done in that row in which positive value occurs. Repeat step 2 to 5
until the optimum condition is met.
Example 21: Farouk Company has plants in Akure, Aba and Kano. Its major distribution centres
are in Lagos, Enugu and Kano. The capacities of the three plants during the next quarter are 90,
20 and 70 trucks respectively. The quarterly demands at the three distribution centres are 75, 25
and 80 trucks. The transportation costs per truck in thousands naira from the plants to the
distribution centres are shown below;
From Plant
To Lagos
To Enugu
To Kaduna
3
8
1
Akure
2
6
4
Aba
5
7
3
Kano
Any plant can transport the trucks to any distribution centre up to its requirement.
(a) Determine the initial feasible solution using the North-West Corner method.
(b) Determine the initial feasible solution using the Least-Cost method.
(c) Determine the initial feasible solution using Vogel’s Approximation method.
(d) Test the optimality of the solution obtained by Vogel’s method.
Solution
(a) Determination of the initial feasible solution using the North-West Corner method.
Step 1:
To/From
Lagos
Akure
Aba
Kano
Requirement
Enugu
75
25
Kaduna
Capacity
80
90
20
70
180
Step 2:
To/From
Lagos
Enugu
Kaduna
Capacity
Akure
3
8
1
90
Aba
2
6
4
20
Kano
5
7
3
70
Requirement
75
25
80
180
Lagos
Enugu
Kaduna
Capacity
Step 3:
To/From
Akure
3
75
8
1
90
4
20
3
70
15
Aba
2
6
10
Kano
10
5
7
70
Requirement
75
25
80
180
23
The transportation cost is 3(75) + 8(15) + 6(10) + 4(10) + 3(10) = N695,000.
(b) Determination of initial feasible solution using the Least-Cost method.
Steps 1 and 2 are the same as that of the North-West Corner method.
Step 3:
To/From
Lagos
Akure
Enugu
Kaduna
3
8
2
5
Capacity
1
90
6
4
20
7
3
70
10
80
Aba
20
Kano
45
25
Requirement
75
25
80
180
The transportation cost is 3(10) + 2(20) + 5(45) + 7(25) + 1(80) = N550,000 which is N105,000
reduction in cost relative to the solution of North-West Corner method.
(c) Determine the initial feasible solution using Vogel’s Approximation method.
Steps 1 and 2 are the same as that of the North-West Corner method.
Step 3:
To/From
Lagos
Enugu
Kaduna
Capacity
Akure
3
8
1
90
Aba
2
6
4
20
Kano
5
7
3
70
Penalties
2
2
2
Requirement
75
1
25
1
80
2
180
This table shows that 2 is the largest penalty and cell (1, 3) is the least cost cell among those with
penalty of 2, thus, maximum possible allocation should be given to this cell, so 80 units is placed
on it accordingly to satisfy requirement at Kaduna.
Step 4: Crossing out the column whose requirement has been satisfied to obtain first reduced
penalty matrix:
To/From
Lagos
Enugu
Capacity
Akure
3
8
10
Aba
2
6
20
Kano
5
7
70
Penalties
5
4
2
Requirement
75
1
25
1
100
24
This table shows that 5 is the largest penalty and cell (1, 1) is the least cost cell in this row thus,
maximum possible allocation should be given to this cell, so 10 units is placed on it accordingly
as partial requirement at Lagos and maximum available capacity at Akure
Step 5: Crossing out the row whose capacity has been fully utilized to obtain second reduced
penalty matrix:
To/From
Lagos
Enugu
Capacity
Aba
2
6
20
Kano
5
7
70
Penalties
4
2
Requirement
65
3
25
1
90
Maximum possible allocation to cell (1,1) in this table, so 20 units is placed on it accordingly as
partial requirement at Lagos and maximum available capacity at Aba
Step 6: Crossing out the row whose capacity has been fully utilized to obtain third reduced penalty
matrix:
To/From
Lagos
Kano
Enugu
5
Capacity
7
Penalties
70
2
Requirement
45
25
70
Allocations were made to the cells accordingly the remaining requirements at Lagos and Enugu as
well as available capacity at Kano. Thus, the basic solution is as follows;
To/From
Lagos
Akure
Enugu
Kaduna
3
8
2
5
Capacity
1
90
6
4
20
7
3
70
10
80
Aba
20
Kano
45
Requirement
25
75
25
80
The transportation cost is 3(10) + 2(20) + 5(45) + 7(25) + 1(80) = N550,000
(d) Optimality test for the Vogel’s solution
Step 1: m + n – 1= 3+ 3 – 1 = 5.
Since the solution has five allocations, the first optimality test is satisfied
180
25
Step 2: Initial value table with cost values of cells where allocation were made
To/From
Lagos
Akure
Aba
Kano
Column value
Enugu
3
2
5
Kaduna
Row value
1
7
Step 3(a): Assuming column value of Lagos to be zero; the total row and column value of cell
(1,1) is 3 + 0 = 3 , therefore row value of Akure is 3 since 3+ 0 = 3 the first cell value. Then the
row values of Aba is 2 + 0 = 2 and that of Kano 5 + 0 = 5
To/From
Lagos
Akure
Aba
Kano
Column value
Enugu
3
2
5
0
Kaduna
Row value
1
3
2
5
7
Step 3(b): Computation of column values of Enugu (7 – 5 = 2) and Kaduna (1- 3 = -1)
To/From
Lagos
Akure
Aba
Kano
Column value
Enugu
3
2
5
0
Kaduna
Row value
1
3
2
5
7
2
-2
Lagos
Enugu
Kaduna
Row value
3
2
5
0
5
4
7
2
1
0
3
-2
3
2
5
Step 4: Computation of cost values of other cells.
To/From
Akure
Aba
Kano
Column value
Step 5: Test plan; Cost values of step 4 – Actual cost values
To/From
Akure
Aba
Kano
Lagos
Enugu
Kaduna
3
-3
0
2
-2
0
5
5
0
5
-8
-3
4
-6
-2
7
7
0
1
-1
0
0
-4
-4
3
3
0
This solution is the optimal one since there is no positive value in the test plan.
 Formulation and Solution Procedures for Assignment Problem
Hungarian method is the widely known method of solving assignment problems. The steps
involved in this method are as follows;
26
1. Prepare a balanced initial matrix, where the number of jobs and capacities are not equal,
extra rows or columns with zero cost elements (dummy) are added to make the problem
amenable to assignment algorithm.
2. Obtain the reduce-cost matrix or reduce-profit matrix. This is done by subtracting each
element in a row from the largest element in that row if it is a profit maximization problem
or subtracting the smallest elements in each row from all the elements in that row if it is a
cost minimization.
3. Obtain the total-opportunity cost matrix. This is done by subtracting the smallest elements
in each column from all the elements in that column.
4. Draw the minimum number of horizontal and vertical lines necessary to cover all zeros at
least once.
5. Obtain the optimal solution. If the number of lines equals the number of elements in a row
or column, optimal solution is reached and an assignment based on the zeros gives the
optimal solution.
6. If the number of lines is not equal to the number of elements in a row or column, subtract
the smallest uncovered element from all uncovered elements and add the same to all
elements at the intersection of the lines and repeat step 4 until the optimal assignment is
attained.
This method is illustrated with the following examples.
Example 22: Four operators are to be assigned to four machines with assignment cost as tabulated
below;
Operators/Machines
A
B
C
D
I
2
6
5
3
II
1
4
6
3
III
3
5
7
6
IV
3
2
1
1
Determine the optimal assignment and its cost.
Solution
Step 1: The assignment is balanced
Step 2: Reduce-cost matrix-subtracting of smallest elements in each row from other element in
that row
Operators/Machines
A
B
C
D
I
1
4
4
2
II
0
2
5
2
III
2
3
6
5
IV
2
0
0
0
Step 3: Total-opportunity cost matrix-subtracting of smallest elements in each column from other
element in that column
Operators/Machines
A
B
C
D
I
0
3
3
1
II
0
2
5
2
III
0
1
4
3
IV
2
0
0
0
27
Step 4: Draw the minimum number of horizontal and vertical lines necessary to cover all zeros at
least once.
Operators/Machines
A
B
C
D
I
0
3
3
1
II
0
2
5
2
III
0
1
4
3
IV
2
0
0
0
Step 5: Since the number of lines in step 4 is 2 and not equal to the number of columns or rows
which is 4, the assignment is not optimal- Subtracting the smallest uncovered element from all
the uncovered elements and adding to all element at the intersection of the lines.
Operators/Machines
A
B
C
D
I
0
2
2
0
II
0
1
4
1
III
0
0
3
2
IV
3
0
0
0
Step 6: Draw the minimum number of horizontal and vertical lines necessary to cover all zeros at
least once.
Operators/Machines
A
B
C
D
I
0
2
2
0
II
0
1
4
1
III
0
0
3
2
IV
3
0
0
0
The number of lines is 4 and equal to the number of columns or rows which is 4, thus, the optimal
solution is reached. The optimal assignment is therefore gives as;
Operators/Machines
A
B
C
D
I
II
1
III
IV
5
1
3
The assignment cost is 1+5+1+3 = 10 units.
Example 23: Four operators are to be assigned to four machines with assignment profit as
tabulated below;
Operators/Machines
A
B
C
D
I
2
6
5
3
II
1
4
6
3
Determine the optimal assignment and its profit.
III
3
5
7
6
IV
3
2
1
1
28
Solution
Step 1: The assignment is balanced
Step 2: Reduce-profit matrix-subtracting of each row element from the largest element in that row
Operators/Machines
A
B
C
D
I
1
0
2
3
II
2
2
1
3
III
0
1
0
0
IV
0
4
6
5
Step 3: Total-opportunity cost matrix-subtracting of smallest elements in each column from other
element in that column
Operators/Machines
A
B
C
D
I
1
0
2
3
II
1
1
0
2
III
0
1
0
0
IV
0
4
6
5
Step 4: Draw the minimum number of horizontal and vertical lines necessary to cover all zeros at
least once.
Operators/Machines
A
B
C
D
I
1
0
2
3
II
1
1
0
2
III
0
1
0
0
IV
0
4
6
5
The number of lines is 4 and equal to the number of columns or rows which is 4, thus, the optimal
solution is reached. The optimal assignment is therefore gives as;
Operators/Machines
A
B
C
D
I
II
1
III
IV
3
6
6
6
The profit of the assignment is 3+6+6+6 = 21 units.
Example 24: Four operators are to be assigned to four machines with assignment cost as tabulated
below. The constraints are that worker I cannot be assigned to machine III and also worker III
cannot be assigned to machine IV.
Operators/Machines
A
B
C
D
I
5
7
9
7
II
5
4
3
2
Determine the optimal assignment and its cost.
III
2
5
6
IV
3
5
7
29
Solution
Step 1: The assignment is balanced
Step 2: Reduce-cost matrix-subtracting of smallest elements in each row from other element in
that row
Operators/Machines
A
B
C
D
I
3
5
6
5
II
3
2
0
0
III
0
2
4
IV
0
3
5
Step 3: Total-opportunity cost matrix-subtracting of smallest elements in each column from other
element in that column
Operators/Machines
A
B
C
D
I
0
2
3
2
II
3
2
0
0
III
0
2
4
IV
0
3
5
Step 4: Draw the minimum number of horizontal and vertical lines necessary to cover all zeros at
least once.
Operators/Machines
A
B
C
D
I
0
2
3
2
II
3
2
0
0
III
0
2
4
IV
0
3
5
Step 5 : Since the number of lines in step 4 is 3 and not equal to the number of columns or rows
which is 4, the assignment is not optimal- Subtracting the smallest uncovered element from all
the uncovered elements and adding to all element at the intersection of the lines.
Operators/Machines
A
B
C
D
I
0
0
1
0
II
5
2
0
0
III
0
2
4
IV
0
1
3
Step 6: Draw the minimum number of horizontal and vertical lines necessary to cover all zeros at
least once.
Operators/Machines
A
B
C
D
I
0
0
1
0
II
5
2
0
0
III
0
2
4
IV
0
1
3
The number of lines is 4 and equal to the number of columns or rows which is 4, thus, the optimal
solution is reached. The optimal assignment is therefore gives as;
Operators/Machines
A
B
I
II
III
2
IV
2
30
C
D
3
7
The assignment cost is 7+3+2+2 = 14 units.
Example 25: Suppose that in example 24, a fifth machine is made available and its respective
assignment costs to the four workers are 2, 1, 2 and 8. Reformulate the problem as an assignment
model and determine the optimal solution. Will it be economical to replace one of the existing
machines? If yes, state the one.
Solution
Step 1(a): Reformulation of the assignment model
Operators/Machines
A
B
C
D
I
5
7
9
7
II
5
4
3
2
III
2
5
6
IV
3
5
7
V
2
1
2
8
The assignment model is not balanced because the number of workers and machine are not equal
Step 1(b): Add an extra row (dummy operator) to balance the model
Operators/Machines
A
B
C
D
E
I
5
7
9
7
0
II
5
4
3
2
0
III
2
5
6
0
IV
3
5
7
0
V
2
1
2
8
0
Step 2: Reduce-cost matrix-subtracting of smallest elements in each row from other element in
that row
Operators/Machines
A
B
C
D
E
I
3
6
7
5
0
II
3
3
1
0
0
III
1
3
4
0
IV
0
2
5
0
V
0
0
0
6
0
Step 3: Total-opportunity cost matrix-subtracting of smallest elements in each column from other
element in that column
Operators/Machines
A
B
C
D
E
I
3
6
7
5
0
II
3
3
1
0
0
III
1
3
4
0
IV
0
2
5
0
V
0
0
0
6
0
31
Step 4: Draw the minimum number of horizontal and vertical lines necessary to cover all zeros at
least once.
Operators/Machines
A
B
C
D
E
I
3
6
7
5
0
II
3
3
1
0
0
III
1
3
4
0
IV
0
2
5
0
V
0
0
0
6
0
Step 5 : Since the number of lines in step 4 is 4 and not equal to the number of columns or rows
which is 5, the assignment is not optimal- Subtracting the smallest uncovered element from all
the uncovered elements and adding to all element at the intersection of the lines.
Operators/Machines
A
B
C
D
E
I
2
5
6
4
0
II
3
3
1
0
1
III
0
2
3
0
IV
0
2
5
1
V
0
0
0
6
1
Step 6: Draw the minimum number of horizontal and vertical lines necessary to cover all zeros at
least once.
Operators/Machines
A
B
C
D
E
I
2
5
6
4
0
II
3
3
1
0
1
III
0
2
3
0
IV
0
2
5
1
V
0
0
0
6
1
The number of lines is 5 and equal to the number of columns or rows which is 4, thus, the optimal
solution is reached. The optimal assignment is therefore gives as;
Operators/Machines
A
B
C
D
E
I
II
III
IV
2
V
2
2
2
0
The assignment cost is 2+2+2+2 = 8 units. Since this yields savings of 6units (14-8units) when
compared with example 24, it will be economical to replace one of the machines. Machine one
will be replaced because the solution assigns dummy operator to it. Recall that this machine has
excessive assignment cost relative to others.
3.0 INTEGER PROGRAMMING
Integer programming is a linear programming with decision variables restricted to integer nonnegative values only. However, an integer programming problem could be pure or mixed. In pure
32
integer programming, all the decision variables take only integer values while Mixed program
contains variables restricted to only integer values and others that could have fractional values as
the case may be in real life applications. Approximation involves in integer programming
technique limits its application especially when non discrete number of variables such as exact
values of money is been handled. Integer programming model formulation uses the same
mathematical notation as in the ordinary LP problem and is generally written as;
Optimize (Max. or Min.): 𝑍 = ∑𝑘𝑖=1 𝑐𝑖 𝑥𝑖
Subject to the constraints: ∑𝑘𝑖=1 𝑎𝑖𝑗 𝑥𝑖 ≤ 𝑏𝑖 or ∑𝑘𝑖=1 𝑎𝑖𝑗 𝑥𝑖 ≥ 𝑏𝑖
𝑥𝑖 ≥ 0
Where; 𝑖&𝑗 = 1, 2, 3….,n 𝑥𝑖 = integer value.
The models can be solved using two major variant methods, Cutting Plane and Search (Branch
and Bound) methods. Both methods are applicable to pure as well as mixed IP problems.
Cutting Plane Method: In this method, certain “secondary conditions” are added to the model in
such a manner that the ultimate result satisfies the conditions of only integer solutions. These
secondary conditions cuts or eliminate certain aspects of the solution which are not feasible
integers. The steps involved are;
1. Minimization problem is converted into maximization problem.
2. Solve the maximization problem without considering the condition of integer values.
3. Test for the integerality of the solution, if the values of the variables in the solution are
integers, the solution is optimal.
4. If the solution is not optimal determine the highest fraction value in the solution value
column and convert the row with the largest value into an equation. If there is a negative
fraction, convert this into the sum of negative integer and non-negative fraction and obtain
the equations with fractional part of all coefficients by ignoring integral part and replacing
the whole number by zero. Here the technical coefficient is equal to the sum of the
fractional part of resource availability and some integer. Hence, it is equal to or greater
than the fractional part of resource availability. So, fractional part is taken to the R.H.S.
and the inequality is formed as greater than or equal to (≥) type. If this is to be converted
into (≤) type, it is multiplied with -1 to make it an inequality and a slack is introduced.
5. Add the new constraint to the simplex table of solution found in step 2 and solve the
problem by dual simplex method.
6. If the solution is not optimal repeat step 4 and 5 till an optimum solution with all the integer
values is obtained.
Branch and Bound Techniques: This method which considers all possible feasible integers only
as solution is used to solve integer programming problems with constraints which have both upper
and lower bounds (limits). Basically, this method involves dividing the feasible region into smaller
sub-sets, each sub-set is considered sequentially until a feasible solution giving the optimal value
33
of objective function is arrived at. Its special case is when all the integer variables are binary in
nature. The procedure involves the following steps;
1. Obtain the optimal solution of the linear programming problem without considering the
restriction of the integer.
2. Test for the integerality of the solution, if the values of the variables in the solution are
integers, the solution is optimal.
3. If the solution is not optimal, consider the upper bound values of the objective function and
determine the lower bound values by rounding off to the integer values of the decision
variables.
4. Sub divide the given LP problem into two problems as follows;
Sub-Problem I- Given LP problem with additional constraint xj ≤ [xj *]
Sub-Problem I- Given LP problem with additional constraint xj ≥ [xj *] + 1
Where xj * is the optimum value of xj (not an integer) and [xj *] is the largest integer contained in x
j
5. Solve the two LP problems. If the solution of ;
(a) The two sub-problems are integral, it is optimal, the select the solution which the
optimum value of the objective function.
(b) One is integral and the other non feasible, integral solution is the optimum.
(c) One is integral and the other non-integral, repeat step 3 to 5 till solutions with integral
values are obtained.
Example 26: An investment consultant has four projects with different investments and present
value of expected returns. Funds available for investment during the three proposals are also
available. The detailed information regarding the project is as follows;
Project
Investment During the Year
Present Value of Expected Returns
1
II
III
P-1
1,000,000
600,000
500,000
800,000
P-2
500,000
200,000
400,000
700,000
P-3
300,000
250,000
350,000
400,000
P-4
400,000
300,000
260,000
300,000
Funds of Investment 1,800,000 1,000,000
800,000
Formulate an integer programming model for the consultant to make a decision as to which project
should be accepted in order to maximize present value of expected return.
Solution: Let 𝑥1 ,𝑥2 , 𝑥3 and 𝑥4 are respective investment on projects P-1, P-2, P-3 and P-4
Maximize:
Z = 800000𝑥1 + 700000𝑥2 + 400000𝑥3 + 300000𝑥4
Subject to: 1000000𝑥1 + 500000𝑥2 + 300000𝑥3 + 400000𝑥4 ≤ 1800000
600000𝑥1 + 200000𝑥2 + 250000𝑥3 + 300000𝑥4 ≤ 1000000
500000𝑥1 + 400000𝑥2 + 350000𝑥3 + 260000𝑥4 ≤ 800000
x1  0; 𝑥2 ≥ 0;
x3  0; 𝑥4 ≥ 0
And all are integers
34
Example 27: A multinational company is planning to invest in four different projects in Business
Process Outsourcing (BPO) industry in an important town of North. The details of the investment
are provided below;
Project
Capital Requirement for three years
Present Value of Expected Returns
1
II
III
800
600
500
550
550
900
400
400
300
200
400
250
400
150
100
Funds of Investment
1500
1200
700
500
Formulate an integer programming model for the consultant to make a decision as to which project
should be accepted in order to maximize present value of expected return.
Solution: Let𝑥1 ,𝑥2 , 𝑥3 and 𝑥4 are respective investment on projects A, B, C and D; also let 𝑥𝑖 = 1
(if project i is accepted) and 𝑥𝑖 = 0 (if project i is rejected)
Maximize:
Z = 800𝑥1 + 550𝑥2 + 400𝑥3 + 250𝑥4
Subject to: 600𝑥1 + 900𝑥2 + 300𝑥3 + 400𝑥4 ≤ 1200
A
B
C
D
500𝑥1 + 400𝑥2 + 200𝑥3 + 150𝑥4 ≤ 700
550𝑥1 + 400𝑥3 + 100𝑥4 ≤ 500
𝑥1 + 𝑥2 ≥ 1
−𝑥3 + 𝑥4 ≤ 1
𝑥𝑖 = 0 or 1
4.0 GOAL PROGRAMMING
All the LP techniques treated above were restricted the goal of either maximizing benefits or
minimizing costs but in real life situations organizations are faced with multiple and conflicting
objectives, and it is not possible to solve this type of problem with the previously discussed LP
algorithms. Goal programming is the appropriate techniques for solving this type of problem. This
technique aims at full or partial achievement of the goals in the order of priority set by the
management or decision makers and treats the constraints of linear programming problems as goals
in the objective function. Low priority goals are considered only after the high priority goals have
been considered. This is very difficult to decide as contribution of a particular goal to the overall
well being of an organization is very difficult to determine, thus selected goal continues to remain
in the problem unless and until the achievement of a lower priority goal would cause the
management to fail to achieve a higher priority goal. Hence, the concept of underachievement and
overachievement of goals. Generally, goal programming models are expressed mathematically as;
Minimize:
𝑍 = ∑𝑘𝑖=1 𝑊𝑖 (𝑑𝑖 + + 𝑑𝑖 − )
Subject to the constraints: ∑𝑘𝑖=1 𝑎𝑖𝑗 𝑥𝑖 + 𝑑𝑖 + + 𝑑𝑖 − = 𝑏𝑖
+
−
𝑥𝑖 ≥ 0
Where; 𝑖 = 1, 2, 3….,n; 𝑥𝑖 , , 𝑑𝑖 , 𝑑𝑖 , 𝑖, 𝑗 ≥ 0
𝑖 is the decision variable
𝑊𝑖 is the weightage of goal i
𝑑𝑖 − is the degree of under achievement of goal i
35
𝑑𝑖 + is the degree of overachievement of goal i
The steps involved in the formulation of goal programming model are as follows;
1. Identify the decision variables and constraints.
2. Formulation of goals or objectives of the problem.
3. Formulation of the constraints.
4. Identify the least important and redundant goals.
The goal models can be solved by graphical method similar to that of LP problems described in
section 4. The only difference is that goal programming model goes with a number of goals and
total deviations from these goals are required to be minimized. Deviations minimization is done in
the order of priority. Steps involved are as follows;
1. Formulate the linear goal programming model.
2. Construct the graph of all the goals in relation with decision variables as in normal LP
problem.
3. Indicate the deviations with arrows
4. Identify the feasible region of the goal with the highest priority and determine the optimal
solution with respect to this goal.
5. Proceed to the next highest priority goal and determine its best solution until the goals are
exhausted.
Example 28: A and B Ltd produces two types of products A and B using common production
facilities which are considered a scarce resource by the company. The scarce production facilities
are in two departments of machining and assembling. The company is in position to sell whatever
number it produces as their brand enjoys the market confidence. However, the production capacity
is limited because of the availability of the scarce resources. The company wants to set a goal of
maximum daily profit and would be satisfied with N2000 daily profit because of its other
constraints. The details of processing time, capacities of each of the departments and unit profit
combinations of product A and B are as follows;
Type of Product
Processing Time of the products Profit per Unit Product
Machining
Assembling
A
3
1
200
B
2
1
300
Available Time (hr)
100
50
If the company wishes to know the product mix that would get them the desired profit of N2000
per day, formulate the problem as a goal programming model.
Solution: Let 𝑥1 and 𝑥2 are respective number of products A and B while 𝑑𝑖 − is the amount by
which the actual profit will fall short of N2000/day and 𝑑𝑖 + is the amount by which the actual
profit exceed the desired profit of N2000/day.
36
Minimize:
Subject to:
𝑍 = 𝑑𝑖 + + 𝑑𝑖 −
3𝑥1 + 2𝑥2 ≤ 100
𝑥1 + 𝑥2 ≤ 50
200𝑥1 + 300𝑥2 + 𝑑𝑖 + + 𝑑𝑖 − = 2000
𝑥1 , 𝑥2 , 𝑑𝑖 + , 𝑑𝑖 − ≥ 0;
Example 29: A manufacturer produces two types of products X and Y. The plant has production
capacity of 500 hours per month and production of product X and Y or an average requires one
hour in the plant. The number of products X and Y sold per month and net profit from the sales of
these products is as follows;
Type of Product
Number of Products sold per Month
X
2503
Y
300
If the manufacturer set the following objectives arranged in the order of priority.
1. No under utilization of plant production capacity.
2. Sell maximum possible numbers of the products and also has twice as much desire to sell
product X than Y because the net profit from the sale of product X is twice the amount
from Y.
3. Minimize overtime operation of the plant.
Formulate the goal programming model of this problem.
Solution: Let 𝑥1 and 𝑥2 are respective number of products X and Y. The model is formulated as
follows
1. Since overtime operations are not allowed.
𝑥1 + 𝑥2 + 𝑑1 + − 𝑑1 − = 500
−
Where 𝑑1 is the degree of under utilization of production capacity and 𝑑1 + is the degree of
overtime production operation capacity?
2. Since maximization of sale is one of the goals, positive deviation is not allowed in the sales
related constraints. Hence,
𝑥1 + 𝑑2 − = 250
𝑥2 + 𝑑3 − = 300
Where 𝑑2 − is the degree of under achievement of sales goal for product X while 𝑑3 − is the degree
of under achievement of sales goal for product Y.
Minimize:
𝑍 = 𝑝1 𝑑1 − + 2𝑝2 𝑑2 − + 𝑝2 𝑑3 − + 𝑝3 𝑑1 +
Subject to:
𝑥1 + 𝑥2 + 𝑑1 + − 𝑑1 − = 500
𝑥1 + 𝑑2 − = 250
𝑥2 + 𝑑3 − = 300
𝑥1 , 𝑥2 , 𝑑1 − , 𝑑2 − , 𝑑3 + ≥ 0;
Example 30: A manufacturing plant of an electronic firm produces two types of television sets,
both colour and black-white. Past records shows that production of either a colour or a black-white
set requires on an average of one hour in the plant. The plant has a normal production capacity of
40 hours per week. The marketing department reports that because of the limited sales opportunity.
The maximum numbers of colour and black-white sets that can be sold are 24 and 30 respectively
37
for the week. The gross margin from the sale of a colour set is N80, where as it is N 40 from black
–white set. The following is the scale of preference of the organization desired goals;
1. No under utilization of plant production capacity and production workers lay off..
2. Sell maximum possible numbers of the products and also has twice as much desire to sell
colour television than black-white because the gross margin from the sale of colour
television is twice the amount from black-white sets.
3. Minimize overtime operation of the plant as much as possible.
Formulate the model of this goal programming problem.
Solution: Let 𝑥1 and 𝑥2 are respective number of colour and black-white televisions produced. The
model is formulated as follows;
1. Since under utilization and over utilization of production of capacity are not desired.
−
𝑥1 + 𝑥2 + 𝑑1 + − 𝑑1 − = 40
Where 𝑑1 is the degree of under utilization of production capacity and 𝑑1 + is the degree of
overtime production operation capacity.
2. Sales constraints;
𝑥1 + 𝑑2 − − 𝑑2 + = 24
𝑥2 + 𝑑3 − − 𝑑3 + = 30
Where 𝑑2 − and𝑑3 − are the degree of under achievement of sales goal while 𝑑2 + and 𝑑3 + are the
degree of under achievement of sales goal.
Minimize:
𝑍 = 𝑝1 𝑑1 − + 2𝑝2 𝑑2 − + 𝑝2 𝑑3 − + 𝑝3 𝑑1 +
Subject to:
𝑥1 + 𝑥2 + 𝑑1 + − 𝑑1 − = 40
𝑥1 + 𝑑2 − − 𝑑2 + = 24
𝑥2 + 𝑑3 − − 𝑑3 + = 30
𝑥1 , 𝑥2 , 𝑑1 − , 𝑑2 − , 𝑑3 − , 𝑑1 + , 𝑑2 + , 𝑑3 + ≥ 0;
5.0 DYNAMIC PROGRAMMING
All the programming techniques studied above assumed that values of decision variables remain
constant over the planning period, hence the problems are considered as static and their solutions
are applicable to specific situations for a particular period of time only. There are many dynamic
situations in which decision variables will change with time, dynamic programming techniques is
the most adequate to solve this type of problem. Dynamic programming (DP) is mathematical
procedure designed primarily to improve the computational efficiency of solving select
mathematical programming problems by decomposing them in to smaller and computationally
simpler sub-problems. The technique involves decision making over time and attempts to optimize
the allocation of multi-stage resources (decision variables). Dynamic programming typically
solves the problems in stages with each stage involving optimizing one variable. The computations
at different stages are linked through recursive computations in a manner that yields a feasible
optimal solution to the entire problems when the last stage is reached.
The following steps are involved in the formulation and solving dynamic programming model;
38
1. Define decision variables, objective function and constraints of the problems.
2. Specifically define stages of a multi stage decision problem. This involves finding out
different variables and their values for each state and specifying the relationship by which
the state is determined at one stage with the state and decisions at the next stage.
3. Developing optimal return function through recursion relationship.
4. Construct tabular representation clearly showing values and computation at each stage of
the solution.
5. Determine the optimal solution after sequential solving of all stages of the problem
Example 31: A corporation is entertaining proposals from its three plants for possible expansions
of facilities. The corporation is budgeting N5 million for allocation to all the three plants. Each
plant is requested to submit its proposals given total costs (C) and total revenue (R) for each
proposal. Table 1 summarizes the costs and revenues in millions of naira. The goal of the
cooperation is to maximize the total revenue resulting from the allocation of the N5 million to the
three plants.
Proposal
1
2
3
C1
1
2
-
PLANT I
R1
5
6
-
C2
2
3
4
PLANT II
R2
8
9
12
C3
1
-
PLANT III
R3
3
-
Formulate and solve dynamic programming model of this problem.
Solution:
Zero-cost proposals are introduced to allow for the possibility is not allocating funds to individual
plants.
Proposal
1
2
3
4
C1
0
1
2
-
PLANT I
R1
0
5
6
-
C2
0
2
3
4
PLANT II
R2
0
8
9
12
C3
0
1
-
PLANT III
R3
0
3
-
To construct the network, we first define the stages of the problem. In this example, each plant
represents a stage, since our goal is to determine an optimum proposal for each plant. Next, we
know that these stages are linked together via the capital restriction. We need to include this
linkage in the model in manner that allows us to optimize each stage separately without sacrificing
feasibility. Thus let;
X1 = amount of capital allocation to stage 1
X2 = amount of capital allocated to stages 1 and 2
X3 = amount of capital allocated to stages 1, 2 and 3.
These notations are used to decompose the problem as follows.
First, note that we do not know in advance what the specific values of X1 and X2 but we do know
that they must lie somewhere between 0 and 5. In fact, because the cost of different proposals are
discrete, X1 and X2 can each assume one of the values of 0, 1, 2, 3, 4 and 5. On the other hand, X3
39
which is the amount of capital allocated to all three stages equals 5. The dynamic network model
of problem is as follows.
Each possible value of X1, X2 and X3 is represented by node associated with the respective stages
j = 1, j = 2, and j = 3. The lengths of the arcs joining the nodes of successive stages represent the
revenues resulting from using the best feasible proposal between the two nodes. Consider the arcs
from node 0 at j = 0 to node X1 (0, 1, 2, 3, 4, 5) at j = 1. Arc (0, 0) joining nodes 0 at j = 0 and 0 at
j = 0) means that no funds are assigned to stage 1. Proposal 1 with R1 = 0 is the only feasible one
in this case, and hence arc (0, 0) is assigned the length R1 (0, 0) = 0. On the other hand, arc (0, 1)
admits proposals 1 and 2 as feasible; hence we choose proposal 2 with the higher revenue R 1 = 5
and assign arc (0, 1) length of 5. Following the same reasoning, arc (0, 2) is assigned length of 6
because all three proposals are feasible and proposal 3 has the highest revenue, R1 = 6. The values
X1 = 3, 4 and 5 represents “overspending” and cannot yield more than R1 = 6. The arcs between
stages 1 and 2 are defined by the nodes X1 and X2. By definitions, X1 and X2 represents the amount
allocated to stage 2 alone. Thus arc (x1, x2) is feasible for proposals whose cost does not exceed
X2-X1 and the length of this arc is the largest R2 among all feasible proposals. For example, for X1
= 1 and X2 = 5, the amount X2-X1 = 4 allocated to state 2 renders all of its four proposals feasible.
Since proposals 4 has the highest revenue (R2 = 12), arc (1, 5) is assigned length of 12. On the
other hand, arc (1, 3) has X2-X1 = 2 allowing only proposals 1 and 2 as feasible. Consequently, the
length of arc (1, 3) is 8.
Note that when X1>X2, the associated arc is not defined since naturally X2 must at least be equal
to X1 and also that each arc is uniquely associated with a proposal. Our objective is to find a
connected chain of arcs from node 0 at j = 0 to node 5 at j = 3 that has the longest length (largest
revenue). From the special definition of X1, X2, and X3 any node that connected path from node 0
at j = 0 to node 5 at j = 3 represents a feasible combination of proposals. If you examine the network
carefully, you will notice that some arc can be deleted as obviously non-optimal. For example, arc
(0, 1) between j = 1 and j = 2 can be deleted because assigning X2-X1 = 1 million niara to stage 2
will still result in zero revenue, since it does not provide sufficient capital to undertake proposal 2.
Deletion of non-optimal arcs will reduce the amount of computations. However, to minimize
confusion we shall not try to introduce such simplification at this point.
This example has three alternative optimal solutions as tabulated below;
Optimal Path
Optimal Proposals
(0, 1); (1, 4); (4, 5)
2, 3, 2
(0, 1); (1, 5); (5, 5)
2, 4, 1
(0, 2); (2, 4); (4, 5)
3, 2, 2
Note that three solutions yielded N17 million each. Detail of the computation and interpretation
will be discussed in the class.
40
Example 32: Repeat the exercise if the total capital available for all three plants is;
(a) N4 million.
(b) N6 million.
6.0 NETWORK ANALYSIS
Network analysis is a generic name for several types of network models such critical path method
(CPM) and Programmed Evaluation and Review Technique (PERT) used planning scheduling and
executing large time-bound projects which involve careful co-ordination of a variety of complex
and inter-related activities and resources. Many production/operations management problems lend
themselves to network modeling because of its merits over the traditional Gantt or Bar Chart which
equally could be used in project scheduling. The bar chart or the Gantt chart is graphical
representations where bars, indicate processes operations, jobs or activities on a time scale. The
limitations of this form of representation for production planning and control processes are as
follows:
(a)
It is not flexible, particularly to great variables.
(b)
It cannot show the interdependence of activities.
(c)
For complicated projects it will be difficult to show all details.
(d)
Does not indicate in case of delays which jobs are more important than others such that
delivery dates or termination time could be met.
The network analysis was, therefore, developed in the late 1950s as solution to these problems of
Gantt/Bar chart. The techniques can be used in the planning and scheduling of the construction
works, ship and aero plane building, lunching of missiles, maintenance (especially plant overhaul),
assembly work, plant removal (such as changing factory site or department), marketing (i.e
launching a new product, or planning a sales drive) as well as research and development to;
(a)
find out what sequence is necessary for carrying out all the activities involved.
(b)
estimate with any degree of accuracy how long each activity would last.
(c)
have some idea of what resources are needed, and in what quantities, at any one time.
(d)
make some estimate of the cost of the project as a whole, before starting, and during work
(e)
keep adequate control of cost during the project and have some means of reducing them
where possible.
41
Although, CPM is a deterministic and activity oriented model while PERT is probabilistic and
event oriented one, both techniques are coverage on several aspects and often treated as twins.
Application of these techniques involves the following sequential steps;
1. Dividing the entire project into key activities by work break-down structure procedure.
2. Establishing the precedence order of carrying out the activities involved. This order can be
decided by stating the following:i.
activities that must be completed before starting a particular activities
(predecessor activities).
ii.
activities that must follow a particular activity (successor activities).
iii.
activities that can be performed simultaneously (concurrent activities).
3. Using the established precedence order to put the activities in a logical sequence by graphical
representation notations (network diagram). The symbols used in the network construction
include;
(a) Activities are represented by arrow (
) and events are represented by node (O).
(b) Each activity is represented by one and only one straight arrow which must start
and end in a node.
(c) Arrows should not cross each other and dangling is not permitted. Dummy activity
represented with doted arrow is used to prevent dangling.
(d) All events (nodes) should be numbered in an ascending order, and no events
number can be repeated.
4. Determination of each activity duration and inserting it in the network. Where multiple time
estimates are involved for a particular activities, the expected (average) duration, d of the
activities will be determined using Equation (2);
𝑑=
𝑡𝑜 +4𝑡𝑚 +𝑡𝑝
1
6
The variance, σe of this activity duration is computed as;
𝜎𝑒 =
𝑡𝑝 −𝑡𝑜
6
or 𝜎𝑒
2
=
(𝑡𝑝 −𝑡𝑜 )
2
36
where to = optimistic time, tp = pessimistic time and tm = most likely time
Probability of achieving scheduled date;
2
42
𝑥=
𝑠−𝑡
𝜎
3
where x = Standard value for ordinate required, i.e. scheduled date
s = Scheduled date of complete
t = Expected project duration
In using three activity duration estimates to calculate the probabilities of completing projects or
parts of projects on or before given scheduled completion dates, it is important to consider subcritical paths. This is particularly important when the length of such paths approaches the length
of the critical path, and also where the duration of the activities on such paths is subject to
comparatively large variance. Unless jobs have been done before, it is often difficult to obtain
accurate estimate of activity duration. One advantage of using multiple estimates is that it
encourages people to commit themselves to estimates when they might be reluctant to give a single
estimate. But if this method is used principally for this reason, then there is little to be said for
using these figure and subsequently calculating project duration and probabilities to several places
of decimals. In such cases it may be sufficient merely to take the average of the three estimates –
in fact many computer programmes provide this facility. Analyzing of the network model to
determine for all activities the Earliest Start, Latest Finish Earliest Finish and Latest Start dates,
Floats or Slacks as well as the project critical activities/path and the entire project duration.
i. Earliest Start date for an activity, ES is the date at which the activity can start without affecting
the total project duration. Earliest Start date of an activity is determined by working forward
through the network adding up the durations of all the activities preceding the one under study.
In any node where there is more than one possible figure the highest one is taken.
ii. The Earliest Finish date of any activity, EF is the earliest possible date at which an activity can
finish without affecting the total project duration. It is the sum of the activity’s earliest start
date and its duration (d).
𝐸𝐹 = 𝐸𝑆 + 𝑑
4
iii. Latest Finish date for an activity, LF is the latest date by which the activity must get completed
without delaying the project completion. This is obtained by working from the end of the
network, subtracting the durations of all succeeding activities to the one under study from the
project finish date. Where there is more than one possible figure at any node, the lowest one is
taken.
43
iv. The Latest Start date for an activity, LS is the latest possible date by which the activity must
start without delaying the project completion. It is the difference between activities latest finish
date and the duration.
𝐿𝑆 = 𝐿𝐹 − 𝑑
5
v. Float or Slack is the extra time a non critical activity can consume without delaying the project
completion date.
Float can be total, free, independent or interfering.
(a) The Total Float, TF is the difference between the total time available for any activity and
the activity duration.
𝑇𝐹 = 𝐿𝐹 − 𝐸𝑆 − 𝑑
6
Free Float, FF is the part of the total float that does not affect the subsequent activities
which is obtained when all the activities are started at the earliest. This is given as the
difference between the earliest date of the head event and the sum of the earliest date of the
preceding event and activity duration.
𝐹𝐹 = 𝑇𝐹 − 𝐻𝑆
7
Where HS is Head slack
(b) Independent Float, IF is the part of float that remains unaffected after utilizing the float of
the preceding activities which does not affect the succeeding activities. It is given as the
difference between the earliest date of the head event and the sum of the latest date of the
tail event and activity duration.
𝐼𝐹 = 𝐹𝐹 − 𝑇𝑆
8
Where TS is Tail slack
(c) Interfering Float is the part of float which affects the starting of subsequent activities. It is
given as the difference between the latest date of the head event and the earliest date of that
event.
Our treatment of network analysis so far has assumed that only time is important in executing
tasks. There are certainly many situations where time is indeed the only or the most important
factor but in the majority of cases other factors affect our ability to do the job. We have assumed,
44
for example, that the correct quantities. The availability of such facilities as manpower, plant, etc,
determines not only our ability to do the job but also the time required to do it. Estimates of activity
duration will rely implicitly upon our capability to undertake those activities; consequently it is a
little unrealistic to speak of activity durations in the abstract. An estimate of the duration of an
activity may differ substantially depending on the time at which the activity is undertaken. When
very little other work is being undertaken activity duration is likely to be shorter than when
facilities are already heavily loaded or committed. Each estimate of duration is based upon the
assumed use of a certain amount of resource, and consequently when the project duration is
initially calculated. The duration of an activity depends upon the quantity of resource allocated to
it. At additional cost more resource can usually be acquired and the activity duration decreased. In
many cases this additional expenditure can be justified by the earlier completion of the activity
and of the project. If a heavy penalty, clause applies for late completion of the project, or if the
project must be completed by a given date so that it can begin earning revenue (e.g. a hotel or
holiday camp ready for the beginning of the holiday season), then additional expenditure during
manufacture may be economically justifiable. In network analysis it is assumed that cost is linearly
related to activity duration, and that as duration decrease the cost increases
Example 33: A university of Technology is aspiring to introduce a new department called
Aeronautical Engineering Technology to the existing departments. The Dean of the school of
Engineering and Engineering Technology have been directed to prepare an overall project
schedule. As consultants in the area of project planning and control, you are required to assist the
Dean in doing so by constructing an activity network to represent the project and determine the
earliest finish date for the entire project, total float on each activity, critical path/activities, latest
start date for activity B, earliest finish date for the activity F, the effect on the project duration if
activity A is delayed for 3 weeks and the effect on the project duration if the activity F were to
take 7 weeks.
Solution:
The first thing to do should be the identification of all activities. The following constitute such
activities:
A
B.
C.
Perform pilot lectures for ten students in new facility.
Build the facility.
Install all equipment and furnishings.
45
D
Recruit Lecturers and other supportive staff.
E.
Train Staff
F.
Pass safety inspection of municipal authority.
These activities are very general, which is acceptable, but are they in the proper sequence? As
consultants, we encourage the dean to establish precedence relationships and also time estimates
for activity. And this is shown as follows:
Activity
Description
Predecessor
to
Time (week)
tm
tp
B
B
D
C, E, F
20
2
8
22
4
4
24
3
16
25
5
7
30
4
20
30
6
9
Build facility
Pass Safety Inspection
Install Equipment
Recruit Staff
Train Staff
Perform Pilot lecture
Example 34: The following table gives the activities and other relevant data for a project.
Activity
Duration (days)
Cost (N)
Normal
Crash
Normal
Crash
1-2
4
3
600
800
1-3
2
2
400
400
1-4
5
4
750
900
2-3
7
5
400
600
2-5
7
6
800
1000
3-5
2
1
500
650
4-5
5
4
600
850
If the indirect cost per day is N200, find the;
(a)
Normal duration and cost of the project
(b)
Minimum possible duration of the project and its associated cost.
(c)
Optimal duration and cost of the project
7.0 FLOW PRODUCTION LINES BALANCING
The fundamental problem of layout planning for assembly lines is to find the number of work
station (workers) and the tasks to be performed at each station so that the desired quantity of output
is achieved. Notice several important points in this definition. First, the design focuses on
46
achieving a desired level of productive capacity (output capacity). The desired capacity is an
important management decision. Secondly, if the tasks are to be assigned to work stations, the
sequence of task must be considered; which tasks must be done first, and which ones may follow.
Usually, some different sequences are possible and others are prohibited by the nature of the
product. Therefore, sequence restriction must be identified and compiled with the design. Finally,
the definition emphasis our concern with attaining desired output efficiency, without using
unnecessary input resources. In summary, the design of product layout should address the problem
of capacity, sequencing, equalization performance times of work centres and efficiency in
assembly or machine lines. The design problem of finding ways to equalize performance times at
all stations in order to reduce cost of idle time in any production line is called the line balancing
problem. How can the cost of idleness be reduced? The most common way is to redefine the work
content at each station. Perhaps elementary tasks can be reassigned in different way and we can
come up with work assignments that are more evenly distributed in terms of time. If productive
times required by stations were equal, we could have no idle time, and the line would be perfectly
balanced. The process of solving a line balancing problem involves identification and elimination
of unnecessary work stations. Our procedure for elemental task;
1.
Define elemental tasks
2.
Identify precedence requirements;
3.
Calculate the minimum number of work stations needed;
4.
Apply an assignment heuristic for specifying the work content at each station.
5.
Calculate effectiveness and efficiency, and
6.
Seek further improvement.
Example 35: The desired output of a window manufacturing company with the tabulated layout
design is 320 units per day.
47
Determine: (a) Bottle neck operation, production cycle time and whether this layout design is good
(b) If better design is possible; if yes; what is the optimal layout for the company
Example 36: The unsplitable tasks precedence relationship of MOUAU Fish Processing Limited
is tabulated as follows;
Task
A
B
C
D
E
F
G
H
i.
Time (Min)
3
6
8
5
2
4
5
5
Preceding Tasks
A
A
A
A
A, C & B
A, C
A, B, C, D. E, F & G
Identify the bottle neck operation, minimum cycle time and output per day of this layout at three
shifts of eight hours per day with one-hour break per shift.
ii.
Determine the optimal layout of this system and compare its output, efficiency and cost of idle time
at N5000 per hour.