OPERATIONS RESEARCH

OPERATIONS RESEARCH
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UNIT 1: APPROXIMATIONS
Apply given Technologies in estimating solutions in business environment
1.1 Newton – Raphson iteration method for solving polynomial equations.
1.2 Trapezium Rule for approximating a definite integral
1.3 Simpson’ Rule for approximating a definite integral.
1.4 Maclaurin Series expansion
The Trapezium Rule
The Trapezium Rule is a method of finding the approximate value of an integral between two limits.
The area involved is divided up into a number of parallel strips of equal width.
Each area is considered to be a trapezium (trapezoid).
If there are n vertical strips then there is n+ 1 vertical line (ordinates) bounding them?
The limits of the integral are between a and b, and each vertical line has length y1 y2 y3... yn+1
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Therefore in terms of the all the vertical strips, the integral is given by:
approx. integral = (strip width) x (average of first and last y-values, plus the sum of all y values
between the second and second-last value)
The trapezium rule is a way of estimating the area under a curve. We know that the area under a
curve is given by integration, so the trapezium rule gives a method of estimating integrals. This is
useful when we come across integrals that we don't know how to evaluate.
The trapezium rule works by splitting the area under a curve into a number of trapeziums, which we
know the area of.

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If we want to find the area und
into smaller intervals, each of which has length h (see diagram above).
Then we find that:
under a curve between the points x0 and xn, we divide this interval up
Where y0 = f(x0) and y1 = f(x1) etc
If the original interval was split up into n smaller interv
Example
intervals, then h is given by: h = (x
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3
als, n - x0)/n

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Example #1
Example #2

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Maclaurin Series
The infinite series expansion for f(x) about x = 0 becomes:
f '(0) is the first derivative evaluated at x = 0, f ''(0) is the second derivative evaluated at x = 0, and so
on.
[Note: Some textbooks call the series on this page Taylor Series (which they are, too), or series
expansion or power series.]
Maclaurin’s Series. A series of the form
Such a series is also referred to as the expansion (or development) of the function f(x) in powers of x, or its
expansion in the neighborhood of zero. Maclaurin’s series is best suited for finding the value of f(x) for a
value of x in the neighborhood of zero. For values of x close to zero the successive terms in the
expansion grow small rapidly and the value of f(x) can often be approximated by summing only the
first few terms.
A function can be represented by a Maclaurin series only if the function and all its derivatives exist
for x = 0. Examples of functions that cannot be represented by a Maclaurin series: 1/x, ln x, cot x.

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Example 1 Expand ex in a Maclaurin Series and determine the interval of convergence.
Solution. f(x) = ex, f '(x) = ex, f ''(x) = ex, f '''(x) = ex, ........ , f(n)(x) = ex
and
f(0) = 1, f '(0) = 1, f ''(0) = 1, f '''(0) = 1, ....... ,f(n)(0) = 1
so
Example 2. Expand sin x in a Maclaurin Series and determine the interval of convergence.
Solution. f(x) = sin x, f'(x) = cos x, f''(x) = - sin x, f'''(x) = - cos x, ......
Since sin 0 = 0 and cos 0 = 1 the expansion is

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Simpson's rule
Simpson's rule can be derived by approximating the integrand
interpolant P (x) (in red).
In numerical analysis, Simpson's rule
approximation of definite integrals
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f (x) (in blue) by the quadratic
is a method for numerical integration, the numerical
integrals. Specifically, it is the following approximation:
.
7
) , .

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OPERATIONS RESEARCH
Newton's method
In numerical analysis, Newton's method
after Isaac Newton and Joseph R
to the roots (or zeroes) of a real
methods, succeeded by Halley's method
(also known as the Newton–Raphson method
Raphson, is a method for finding successively better approximations
eal-valued function. The algorithm is first in the class of
method.
The Newton-Raphson method in one variable:
Given a function ƒ(x) and its derivative
is reasonably well-behaved a better approximation
ƒ '(x), we begin with a first guess x0. Provided the function
x1 is
Geometrically, x1 is the intersection point of the
process is repeated until a sufficiently accurate value is reached:
A. Description
tangent line to the graph of f, with the x
The function ƒ is shown in blue and the tangent line is in red. We see that
approximation than xn for the root
The idea of the method is as follows: one starts with an initial guess which is reasonably close to the
true root, then the function is approximated by its
tools of calculus), and one computes the
elementary algebra). This x-intercept will typically be a better approximat
than the original guess, and the method can be
Suppose ƒ : [a, b] → R is a differentiable
real numbers R. The formula for converging on the root can be easily derived. Suppose we have
some current approximation xn
referring to the diagram on the right. We know from the definition o
that it is the slope of a tangent at that point.
That is
Here, f ' denotes the derivative
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xn+1
x of the function f.
tangent line (which can be computed using the
), x-intercept of this tangent line (which is easily done with
approximation to the function's root
iterated.
function defined on the interval [a, b
. . Then we can derive the formula for a better approximation,
of the derivative at a given point
of the function f. Then by simple algebra we can derive
10
method), named
, . Householder's
x-axis. The
+is a better
h ion b] with values in the
xn+1 by
f .

OPERATIONS RESEARCH
We start the process ss off with some arbitrary initial value
x0. (The closer to the zero, the better. But,
in the absence of any intuition about where the zero might lie, a guess and check method might
narrow the possibilities to a reasonably small interval by appealing to
theorem.) The method will usually converge, provided this initial guess is close enough to the
unknown zero, and that ƒ'(x0) ≠
least quadratic (see rate of convergence
that the number of correct digits roughly at least doubles in every step. More details can be found in
the analysis section below.
B. Examples
Square root of a number
Consider the problem of finding the square root of a number. There are many
computing square roots, and Newton's method is one.
For example, if one wishes to find the square root of 612, this is equivalent to finding the solution to
The function to use in Newton's method is then,
with derivative,
With an initial guess of 10, the sequence given by Newton's method is
Where the correct digits are underlined. With only a few iterations one can obtain a solution accurate
to many decimal places.
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the intermediate value
.) 0. Furthermore, for a zero of multiplicity 1, the convergence is at
convergence) in a neighbourhood of the zero, which intuitively means
, 11
methods of

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Solution of a non-polynomial equation
Consider the problem of finding the positive number
finding the zero of f(x) = cos(x
x3 1 for x 1, we know that our zero lies between 0 and 1. We try a starting value of
(Note that a starting value of 0 will lead to an undefined result, showing the importan
starting point that is close to the zero.)
sider x with cos(x) = x3. We can rephrase that as
x) − x3. We have f'(x) = −sin(x) − 3x2. Since cos(
The correct digits are underlined in the above example. In particular,
decimal places given. We see that the number of correct digits after the decimal point increase
2 (for x3) to 5 and 10, illustrating the quadratic convergence.
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x6 is correct to the number of
12
. . x) ≤ 1 for all x and
x0 = 0.5.
importance of using a
increases from

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Newton-Raphson Method
This uses a tangent to a curve near one of its roots and the fact that where the tangent meets the x-axis
gives an approximation to the root.
The iterative formula used is:
Example
Find correct to 3 d.p. a root of the equation
f(x) = 2x2 + x - 6
given that there is a solution near x = 1.4

OPERATIONS RESEARCH
UNIT 2: LINEAR PROGRAMMING
HOURS: 20
LINEAR PROGRAMMING
Is a technique used to determine how best to allocate personnel, equipment, materials,
finance, land, transport e.t.c. , So that profit are maximized or cost are minimized or other optimization
criterion is achieved.
Linear programming is so called because all equations involved are linear. The variables in the problem
are Constraints. It is these constraints, which gives rise to linear equations or Inequalities.
The expression to the optimized is called the Objective function usually represented by an equation.
Question 1:
A furniture factory makes two products: Chairs and tables. The products pass through 3 manufacturing
stages; Woodworking, Assembly and Finishing.
The Woodworking shop can make 12 chairs an hour or 6 tables an hour.
The Assembly shops can assembly 8 chairs an hour or 10 tables an hour.
The Finishing shop can finish 9 chairs or 7 tables an hour.
The workshop operates for 8 hours per day. If the contribution to profit from each Chair is $4 and
from each table is $5, determine by Graphical method the number of tables and chairs that should be
produced per day to maximize profits.
Solution:
Let number of chairs be X.
Let number of tables be Y.
Objective Function is:
P = 4X + 5Y.
Constraints:
WW: X/12 + Y/6 = 8
X + 2Y = 96 when X=0 Y= 48 when Y=0 X= 96
AW: X/8 + Y/10 = 8
5X + 4Y = 320 when X=0 Y=80 when Y=0 X=64
FNW: X/9 + Y/7 = 8
7X + 9Y = 504 when X=0 Y=56 when Y=0 X=72
X=0 Y=0

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100
90
80
70
60
50
40
30
20 P------ this gives the maximum point
10
0
10 20 30 40 50 60 70 80 90 100
Question 2:
Mr. Chabata is a manager of an office in Guruwe; he decides to buy some new desk and chairs for his
staff.
He decides that he need at least 5 desk and at least 10 chairs and does not wish to have more than 25
items of furniture altogether. Each desk will cost him $120 and each chair will cost him $80. He has a
maximum of $2400 to spend altogether.
Using the graphical method, obtain the maximum number of chairs and desk Mr. Chabata can buy.
Solution:
Let X represents number of Desk.
Let Y represents number of Chairs.
P = 120X + 80Y.
Constraints:
X = 5
Y= 10
X + Y = 25

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30
25
20
PPoint P (10,15) gives the maximum point
15
10
5
0
5 10 15 20 25 30
The Optimum Solution = 120 * 10 + 80 * 15
= 1200 + 1200
= 2400
Question 3:
A manufacturer produces two products Salt and Sugar. Salt has a contribution of $30 per unit and Sugar
has $40 per unit. The manufacturer wishes to establish the weekly production, which maximize the
contribution. The production data are shown below:
Production Unit
Machine Hours Labour Hours Materials in Kg
Salt 4 4 1
Sugar 2 6 1
Total available per unit 100 180 40
Because of the trade agreement sales of Salt are limited to a weekly maximum of 20 units and to honor
an agreement with an old established customer, at least 10 units of Sugar must be sold per week.
Solution:
Let X represents Salt.
Let Y represents Sugar.

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P = 30X + 40Y.
Constraints:
4X + 2Y = 100 {machine hours}
4X + 6Y = 180 {Labour hours}
X + Y = 40 {material}
Y= 10
X= 20
X=0; Y=0;
SIMPLEX METHOD:
The graphical outlined above can only be applied to problems containing 2 variables. When 3 or more
variables are involved we use the Simplex method. Simplex comprises of series of algebraic procedures
performed to determine the optimum solution.
In Simplex method we first convert inequalities to equations by introducing a Slack variable.
A Slack variable represents a spare capacity in the limitation.
Simplex Method for Standard Maximization Problem
To solve a standard maximization problem using the simplex method, we take the following steps:
Step 1. Convert to a system of equations by introducing slack variables to turn the constraints into
equations, and rewriting the objective function in standard form.
Step 2. Write down the initial tableau.
Step 3. Select the pivot column: Choose the negative number with the largest magnitude in the
bottom row (excluding the rightmost entry). Its column is the pivot column. (If there are two
candidates, choose either one.) If all the numbers in the bottom row are zero or positive (excluding
the rightmost entry), then you are done: the basic solution maximizes the objective function (see
below for the basic solution).
Step 4. Select the pivot in the pivot column: The pivot must always be a positive number. For each
positive entry b in the pivot column, compute the ratio a/b, where a is the number in the Answer
column in that row. Of these test ratios, choose the smallest one. The corresponding number b is
the pivot.
Step 5. Use the pivot to clear the column in the normal manner (taking care to follow the exact
prescription for formulating the row operations and then relabel the pivot row with the label from
the pivot column. The variable originally labeling the pivot row is the departing or exiting variable
and the variable labeling the column is the entering variable.
Step 6. Go to Step 3.

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Simplex Method for Minimization Problem
To solve a minimization problem using the simplex method, convert it into a maximization
problem. If you need to minimize c, instead maximize p = -c.
Example
The minimization LP problem:
Minimize C = 3x + 4y - 8z subject to the constraints
3x - 4y ≤ 12,
x + 2y + z ≥ 4
4x - 2y + 5z ≤ 20
x ≥ 0, y ≥ 0, z ≥ 0
can be replaced by the following maximization problem:
Maximize P = -3x - 4y + 8z subject to the constraints
3x - 4y ≤ 12,
x + 2y + z ≥ 4
4x - 2y + 5z ≤ 20
x ≥ 0, y ≥ 0, z ≥ 0.
OR
STEPS TO FOLLOW IN SIMPLEX METHOD:
i. Obtain the pivot column as the column with the most positive indicator row.
ii. Obtain pivot row by dividing elements in the solution column by their corresponding pivot
column entries to get the smallest ratio. Element at the intersection of the pivot column and
pivot row is known as pivot elements.
iii. Calculate the new pivot row entries by dividing pivot row by pivot element. This new row is
entered in new tableau and labeled with variables of new pivot column.
iv. Transfer other row into the new tableau by adding suitable multiplies of the pivot row (as it
appears in the new tableau) to the rows so that the remaining entries in the pivot column
becomes zeroes.
v. Determine whether or not this solution is optimum by checking the indicator row entries of the
newly completed tableau to see whether or not they are any positive entries. If they are positive
numbers in the indicator row, repeat the procedure as from step 1.
vi. If they are no positive numbers in the indicator row, this tableau represents an optimum
solution asked for, the values of the variables together with the objective function could then be
stated.

OPERATIONS RESEARCH
Question 1:
Maximize Z = 40X + 32Y
Subject to:
40X +20Y = 600
4X + 10Y = 100
2X + 3Y = 38 Using the Simplex method
Solution:
To obtain the initial tableau, we rewrite the Objective Function as:
Z = 40X + 32Y
Introducing Slack variables S1, S2, S3 in the 3 inequalities above we get:
40X + 20Y + S1 = 600
4X + 10Y + S2 = 100
2X + 3Y + S3 = 38
X = 0
Y =0.
TABLEAU 1:
Pivot Element
X Y S1 S2 S3 Solution
S1 40
20 1 0 0 600
S4 10 0 1 0 100
2 S2 3 0 0 1 38
3 Z 40 32 0 0 0 0 Indicator Row
Pivot Column
TABLEAU 2:
X 1
0.5 0.025 0 0 15
S0 8 -0.1 1 0 40
2 S0 2 -0.05 0 1 8
3 Z 0 12 -1 0 0 -600

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TABLEAU 3:
X 1 0 0.0375 0 -0.25 13
S0 0 0.1 1 -4 8
2 Y 0 1
0.025 0 0.5 4
Z 0 0 -0.7 0 -6 -648 Indicator Row
Pivot Column
Conclusion:
Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, X =
13 and Y = 4 giving Z = 648.
NB:
a) If when selecting a pivot column we have ties in the indicator row, we then select the
pivot column arbitrary.
b) If all the entries in the selected pivot column are negative then the objective function is
unbound and the maximum problem has no solution.
c) A minimization problem can be worked as a maximization problem after multiply the
objective function and the inequalities by –1.
d) Inequalities change their signs when multiplied by negative number.
Question 2:
Maximize Z = 5X1 + 4X2
Subject to:
2X1 +3X2 = 17
X1 + X2 = 7
3X1 + 2X2 = 18 Using the Simplex method
Solution:
Max Z = 5X1 + 4X2
Subject to:
2X1 + 3X2 + S1 = 17
X1 + X2 + S2 = 7
3X1 + 2X2 + S3 = 18
X1 = 0
X2=0.

OPERATIONS RESEARCH
TABLEAU 1:
X1 X2 S1 S2 S3 Solution
S2 3 1 0 0 17
1 S1 1 0 1 0 7
2 S3
2 0 0 1 18
3 Z 5 4 0 0 0 0 Indicator Row
Pivot Column
TABLEAU 2:
S0 5/3 1 0 -2/3 5
1 S0 1/3 0 1 -1/3 1
2 X1
2/3 0 0 1/3 6
1 Z 0 2/3 0 0 -5/3 -30
TABLEAU 3:
X0 2 1
3/5 0 -2/5 3
S0 0 -1/5 1 -1/5 0
2 X1 0 -2/5 0 9/15 4
1 Z 0 0 -2/5 0 -7/5 -32
Pivot Column
Conclusion:
Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, X1=
4 and X2 = 3 giving Z = 32.

OPERATIONS RESEARCH
Question 3:
A company can produce 3 products A, B, C. The products yield a contribution of $8, $5 and $10
respectively. The products use a machine, which has 400 hours capacity in the next period. Each
unit of the products uses 2, 3 and 1 hour respectively of the machine’s capacity.
There are only 150 units available in the period of a special component, which is used singly in
products A and C.
200 kgs only of a special Alloy is available in the period. Product A uses 2 kgs per unit and Product
C uses 4kgs per units. There is an agreement with a trade association to produce no more than 50
units of product in the period.
The Company wishes to find out the production plan which maximized contribution.
Solution:
Maximize Z = 8X1 + 5X2+ 10X3
Subject to:
2X1 + 3X2 + X3 = 400 {machine hour}
X1 + X3 = 150 {component}
2X1 + 4X3 = 200 {Alloy}
X2 =50 {Sales}
X1 = 0
X2 = 0
X3 = 0.
Introducing slack variables:
Subject to:
2X1 + 3X2 + X3 + S1= 400
X1 + X3 + S2 = 150
2X1 + 4X3 + S3 = 200
X2 + S4 =50
X1 = 0
X2 = 0
X3 = 0.

OPERATIONS RESEARCH
TABLEAU 1:
X1 X2 X3 S1 S2 S3 S4 Solution
S2 3 1 1 0 0 0 400
1 S1 0 1 0 1 0 0 150
2 S2 0 3 4
0 0 1 0 200
S0 1 0 0 0 0 1 50
4 Z 8 5 10 0 0 0 0 0 Indicator Row
Pivot Column
NB: Ignore S4 in finding pivot row.
TABLEAU 2:
S3/2 3 0 1 0 -1/4 0 350
1 S1/2 0 0 0 1 -1/4 0 100
2 X1/2 0 3 1
0 0 1/4 0 50
S0 1 0 0 0 0 1 50
4 Z 3 5 0 0 0 -5/2 0 -500
Pivot Column

OPERATIONS RESEARCH
TABLEAU 3:
S3/2 0 0 1 0 -1/4 -3 200
1 S1/2 0 0 0 1 -1/4 0 100
2 X1/2 0 4 0 0 1/4 0 50
3 X0 1
0 0 0 0 1 50
2 Z 3 0 0 0 0 -5/2 -5 -750
Pivot Column
TABLEAU 4:
S0 0 -3 1 0 -1 -3 50
1 S0 0 -1 0 1 -1/2 0 50
2 X1
0 2 0 0 1/2 0 100
1 X0 1 0 0 0 0 1 50
2 Z 0 0 -6 0 0 -4 -5 -1050
Pivot Column
Conclusion:
100 and X2 = 50 giving Z = 1050.
Two slack variable S1 =0 and S2 =0. This means that there is no value to be gained by altering the
machine hours and component constraints.

OPERATIONS RESEARCH
GENERAL RULE:
Constraints only have a valuation when they are fully utilized.
These valuations are known as the SHADOW Prices or Shadow Costs or Dual Prices or Simplex
Multipliers.
A constraint only has a Shadow price when it is binding i.e. fully utilized and the Objective function would
be increased if the constraint were increased by 1 unit.
When solving Linear Programming problems by Graphical means the Shadow price have to be calculated
separately. When using Simplex method they are an automatic by product.
MIXED CONSTRAINTS:
This involves constraints containing a mixture of = and = varieties. Using Maximization problem
we use “Less than or equal to” type. (=).
Faced with a problem which involves a mixture of = and = variety. The alternative solution to deal
with “Greater than or equal to” (=) type is to multiply both sides by –1 and change the inequality sign.
Question 4:
Subject to:
3X1 + 12X2 + 6X3 = 660
6X1 + 6X2 + 3X3 = 1230
6X1 + 9X2 + 9X3 = 900
X3 =10
Solution:
The only constraint that need to be changed is X3 =10 by multiply by –1 both sides and we get:
-X3 = -10
Subject to:
3X1 + 12X2 + 6X3 + S1 = 660
6X1 + 6X2 + 3X3 + S2 = 1230
6X1 + 9X2 + 9X3 + S3 = 900
-X3+ S4 = -10

OPERATIONS RESEARCH
TABLEAU 1:
X1 X2 X3 S1 S2 S3 Solution
S1 3 12 6 1 0 0 600
S2 6 6 3 0 1 0 1200
S3 9 9 0 0 1 900
Z 5 3 4 0 0 0 0 Indicator Row
Pivot Column
FINAL TABLEAU:
X1 X2 X3 S1 S2 S3 Solution
S1 0 15/2 3/2 1 0 -1/2 150
S2 0 -3 -6 0 1 -1 300
X1 3/2 3/2 0 0 1/6 150
Z 0 -9/2 -7/2 0 0 -5/6 -750
Pivot Column
Conclusion:
150 producing Z = $750. Plus production to satisfy constrain (d) 20 units of X3 producing $ 40
contribution.
Therefore Total solution is 150 units of X1 and 10 units of X3 giving $790.
NB: Maximize Z = 5(150) + 3(0)+ 4(10)
= $790.
6
1

OPERATIONS RESEARCH
Question 5:
Subject to:
4X1 + 2X2 = 100
4X1 + 6X2 = 180
X1 + X2 = 40
X1 = 20
X2 =10
Solution:
The only constraint that need to be changed is X2 =10 by multiply by –1 both sides and we get:
-X2 = -10
Subject to:
4X1 + 2X2 + S1 = 100 {1}
4X1 + 6X2 + S2 = 180 {2}
X1 + X2 + S3 = 40 {3}
X1 + S4 = 20 {4}
-X2 + S5 =-10 {5}
TABLEAU 1:
X1 X2 S1 S2 S3 S4 S5 Solution
S1 4 1 0 0 0 0 100
S2 4 6 0 1 0 0 0 180
S3 1 1 0 0 1 0 0 40
S4 1 0 0 0 0 1 0 20
S5 0 -1 0 0 0 0 1 -10
Z 3 4 0 0 0 0 0 0 Indicator
Row
2
Pivot Column
The problem is then solved by the usual Simplex iterations. Each iteration improves on the one
before and the process continues until optimum is reached.

OPERATIONS RESEARCH
1
TABLEAU 2:
X0 2 1
0 0 0 0 -1 10
S4 0 1 0 0 0 2 80
2 S4 0 0 1 0 0 6 120
3 S1 0 0 0 1 0 1 30
4 S1 0 0 0 0 1 0 20
5 Z 4 0 0 0 0 0 4 -40
This shows 10X2 being produced and $40 contribution. The first four constraints have surpluses of
80, 120, 30 and 20 respectively. Not optimums as there are still positive values in Z row.
TABLEAU 3:
X2 0.667 1 0 0.167 0 0 0 30
S2 2.667 0 1 -0.333 0 0 0 40
S3 -0.333 0 0 -0.167 0 0 0 10
S4 1 0 0 0 1 1 0 20
S5 0.333 0 0 0.167 0 0 20
Z 0.333 0 0 -0.667 0 0 0 -120
This shows 30X2 being produced and $120 contribution. All constraints have surpluses except
Labour hours. Not optimum as there is a positive value in Z row.

OPERATIONS RESEARCH
TABLEAU 4:
X1 1
0 0.375 0.125 0 0 0 15
X0 1 -0.25 0.250 0 0 2 0
2 S0 0 -0.125 -0.125 1 0 0 5
3 S0 0 -0.375 0.125 0 1 0 5
4 S0 0 -0.25 0.25 0 0 1 10
5 Z 0 0 -0.125 -0.625 0 0 0 -125
Conclusion:
Since the indicator row is negative the solution is optimum with 15X1 and 20X2 giving $125
contribution.
Shadow prices are X1 = $0.125 and X2 = $0.625.
Non-binding constraints are {3}, {4}, {5} with 5, 5 and 10 spare respectively.
DUALITY:
There is a dual or inverse for every Linear Programming problem. Because solving Simplex problem in
Maximization is quite simple and straightforward, it is usually to convert a Minimization problem into
Maximization problem using dual.
The dual or inverse of Linear Programming problem is obtained by making the constraints in the
inequalities coefficient of the new objective function.
The cofficiences of the original inequalities are combined with the cofficiences of the original objective
function as the constraints.
Question 6:
Minimize Z = 40X1 + 50X2
Subject to:
3X1 + 5X2 = 150
5X1 + 5X2 = 200
3X1 + X2 = 60
X1, X2 =0

OPERATIONS RESEARCH
Solution:
The Dual Linear Programming problem is as follows:
Maximize P = 150Y1 + 200Y2 +60 Y3
Subject to:
3Y1 + 5Y2 + 3Y3 = 40
5Y1 + 5Y2 + Y3 = 50
Y1=0, Y2=0, Y3=0.
Y1 Y2 Y3 S1 S2 Solution
S1 3 3 1 0 40
S2 5 5 1 0 1 50
P 150 200 60 0 0 0 Indicator Row
Pivot Column
Y2 3/5 1 3/5 1/5 0 8
S2 0 -2 -1 1 10
P 30 0 -60 -40 0 -1600
Y2 0 1 1.2 0.5 -0.3 5
Y1 0 -1 -0.5 0.5 5
P 0 0 -30 -25 -15 -1750
Conclusion:
Since the indicator row is negative the solution is optimum with 5Y1 and 5Y2 giving $1750
contribution.
5
2
1

OPERATIONS RESEARCH
1
Question 7:
Minimize Z = 16X1 + 11X2
Subject to:
2X1 + 3X2 = 3
5X1 + X2 = 8
X1, X2 =0 Using the Dual problem.
Solution:
The Dual Linear Programming problem is as follows:
Maximize P = 3Y1 + 8Y2
Subject to:
2Y1 + 5Y2 = 16
3Y1 + Y2 = 11
Y1=0, Y2=0.
Y1 Y2 S1 S2 Solution
S2 1 5
1 0 16
S3 1 0 1 11
2 Z 3 8 0 0 0 Indicator Row
Pivot Column
Y1 Y2 S1 S2 Solution
Y1 0.4 0.2 0 3.2
S2 2.6 0 -0.2 1 7.8
Z -0.2 0 -1.6 0 -25.6
Conclusion:
Since the indicator row is negative the solution is optimum. Hence P = 3Y1 + 8Y2 is
maximum when Y1 = 3.2 and Y2= 0 and P = 25.6
In the primary problem, the solution correspond the slack variable values in the final tableau.
i.e.
X1= S1 = 1.6
X2= S2 = 0.
Hence Z = 16*1.6 + 11*0 = 25.6

OPERATIONS RESEARCH
TERMS USED WITH LINEAR PROGRAMMING:
FEASIBLE REGION:
Represents all combinations of values of the decision variables that satisfy every restriction
simultaneous.
The corner point of the feasible region gives what is known as BASIC FEASIBLE
SOLUTION i.e. the solution that is given by the coordinates at the intersection of any two
binding constraints.
BINDING CONSTRAINTS:
Is an inequality whose graph forms the bounder of the feasible region.
NON BINDING CONSTRAINTS:
Is an inequality, which does not conform to the feasible region.
DUAL PRICE / SHADOW PRICES:
It is important that management information to value the scarce resources. These are known
as Dual price / Shadow price. Derived from the amount of increase (or decrease) in
contribution that would arise if one more (or one less) unit of scare resource was available.

OPERATIONS RESEARCH
ASSIGNMENT PROBLEM:
This is the problem of assigning any worker to any job in such a way that only one worker is assigned to
each job, every job has one worker assigned to it and the cost of completing all jobs is minimized.
STEPS TO BE FOLLOWED IN ASSIGNMENT PROBLEM:
a. Layout a two way table containing the cost for assigning a worker to a job.
b. In each row subtract the smallest cost in the row from every cost in the row. Make a new
table.
c. In each column of the new table, subtract the smallest cost from every cost in the column.
Make a new table.
d. Draw horizontal and vertical lines only through zeroes in the table in such a way that the
minimum number of lines is used.
e. If the minimum number of lines that covers zeroes is equal to the number of rows in the
table the problem is finished.
f. If the minimum number of lines that covers zeroes is less than the number of rows in the
table the problem is not finished go to step g.
g. Find the smallest number in the table not covered by a line.
i. Subtract that number from every number that is not covered by a line.
ii. Add that number to every number that is covered by two lines.
iii. Bring other numbers unchanged. Make a new table.
h. Repeat step d through step g until the problem is finished.

OPERATIONS RESEARCH
Question 1:
Use the assignment method to find the minimum distance assignment of Sales representative to
Customer given the table below: What is the round trip distance of the assignment?
Sales Representative Customer Distance (km)
A 1 200
A 2 400
A 3 100
A 4 500
B 1 1000
B 2 800
B 3 300
B 4 400
C 1 100
C 2 50
C 3 600
C 4 200
D 1 700
D 2 300
D 3 100
D 4 250
TABLEAU 1:
1 2 3 4
A 200 400 100 500
B 1000 800 300 400
C 100 50 600 200
D 700 300 100 250
TABLEAU 2:
1 2 3 4
A 100 300 0 400
B 700 500 0 100
C 50 0 550 150
D 600 200 0 150

OPERATIONS RESEARCH
TABLEAU 3:
1 2 3 4
A 50 300 0 300
B 650 500 0 0
C 0 0 550 50
D 550 200 0 50
TABLEAU 4:
1 2 3 4
A 0 250 0 300
B 600 450 0 0
C 0 0 600 100
D 500 150 0 50
Conclusion:
Since the number of lines is now equal to number of rows, the problem is finished with the
following assignment:
SALES REP CUSTOMER DISTANCE
A 1 200
B 4 400
C 2 50
D 3 100
750 km
Therefore total round Trip distance = 750 km * 2 = 1500 km
Question 2:
A foreman has 4 fitters and has been asked to deal with 5 jobs. The times for each job are estimated as
follows.
A B C D
1 6 12 20 12
2 22 18 15 20
3 12 16 18 15
4 16 8 12 20
5 18 14 10 17
Allocate the men to the jobs so as to minimize the total time taken.

OPERATIONS RESEARCH
Solution:
Insert a Dummy fitter so that number of rows will be equal to number of column.
TABLEAU 1:
A B C D DUMMY
1 6 12 20 12 0
2 22 18 15 20 0
3 12 16 18 15 0
4 16 8 12 20 0
5 18 14 10 17 0
TABLEAU 2:
A B C D DUMMY
1 0 4 10 0 0
2 16 10 5 8 0
3 6 8 8 3 0
4 10 0 2 8 0
5 12 6 0 5 0
TABLEAU 3:
A B C D DUMMY
1 0 4 10 0 3
2 13 7 2 5 0
3 3 5 5 0 0
4 10 0 2 8 3
5 12 6 0 5 3
Conclusion:
FITTERS JOBS TOTALS
A 1 6
B 4 8
C 5 10
D 2 15
Dummy 2 0
39

OPERATIONS RESEARCH
THE ASSIGNMENT TECHNIQUE FOR MAXIMIZING PROBLEMS:
Maximizing assignment problem typically involves making assignments so as to maximize
contributions.
STEPS INVOLVED:
a) Reduce each row by largest figure in that row and ignore the resulting minus
signs.
b) The other procedures are the same as applied to minimization problems.
Question 3:
A foreman has 4 fitters and has been asked to deal with 4 jobs. The times for each job are estimated as
follows.
W X Y Z
A 25 18 23 14
B 38 15 53 23
C 15 17 41 30
D 26 28 36 29
Allocate the men to the jobs so as to maximize the total time taken.
Solution:
TABLEAU 1:
W X Y Z
A 0 7 2 7
B 15 38 0 30
C 26 24 0 11
D 10 8 0 7
TABLEAU 2:
W X Y Z
A 0 0 2 0
B 15 31 0 23
C 26 17 0 4
D 10 1 0 0

OPERATIONS RESEARCH
TABLEAU 3:
W X Y Z
A 0 0 3 1
B 14 30 0 23
C 25 16 0 4
D 9 0 0 0
TABLEAU 4:
W X Y Z
A 0 0 7 1
B 10 26 0 19
C 21 12 0 0
D 9 0 4 0
Conclusion:
A W 25
B Y 53
C Z 30
D X 28
$136
Question 4:
A Company has four salesmen who have to visit four clients. The profit records from previous
visits are shown in the table and it is required to Maximize profits by the best assignment.
A B C D
1 6 12 20 12
2 22 18 15 20
3 12 16 18 15
4 16 8 12 20
Solution:
TABLEAU 1:
W X Y Z
1 6 12 20 12
2 22 18 15 20
3 12 16 18 15
4 16 8 12 20

OPERATIONS RESEARCH
TABLEAU 2:
W X Y Z
1 14 8 0 8
2 0 4 7 2
3 6 2 0 3
4 4 10 8 0
TABLEAU 3:
W X Y Z
1 14 6 0 8
2 0 2 7 2
3 6 0 0 3
4 4 10 8 0
Conclusion:
4 D 20
2 A 22
1 C 20
3 B 16
$78

OPERATIONS RESEARCH
TRANSPORTATION PROBLEM:
This is the problem of determining routes to minimize the cost of shipping commodities from one
point to another.
The unit cost of transporting the products from any origin to any destination is given. Further more,
the quantity available at each origin and quantity required at each destination is known.
⇒ STEPS TO BE FOLLOWED IN ASSIGNMENT PROBLEM:
Arrange the problem in a table with row requirements on the right and column
requirements at the bottom. Each cell should contain the unit cost approximates to
the shipment.
Obtain an initial solution by using the North West Corner rule. By this method one
begins at the up left corner cell and works up to the lower right corner. Place the
quantity of goods in the first cell equal to the smallest of the rows or column totals in
the table. Balance the row and column respectively until you reach the lower right
hand cell.
Find cell values for every empty cell by adding and subtract around the closing loop.
If all empty cell have + values the problem is finished. If not pick the cell with most
– (negative) value. Allocate a quantity of goods to that cell by adding and subtract the
small value of the column or row entries in the closed loop. The closed loop
techniques involves the following steps:
Pick an empty cell, which has no quantity of goods in it.
Place a + sign in the empty cell.
Use only occupied cells for the rest of the closed loop.
Find an occupied cell that has occupied values in the same row or same
column and place a – (negative) sign in this cell.
Go to the next occupied cell and place + sign in it.
Continue in this manner until you return to the unoccupied cell in which you
started.
A closed loop exists for every empty cell as long as they are occupied cell
equal to number of rows + number of column – 1.

OPERATIONS RESEARCH
3 5
6
5
- 4 - 3
7 4 9
6
5 - 1
2 7
5 12 10 8
Question 1:
A firm has 3 factory (A, B, C) and 4 warehouses (1, 2, 3, 4). The capacities of the factories and the
requirements of the warehouse are in the table below.
FACTORY CAPACITY WAREHOUSE REQUIREMENTS
A 220 1 160
B 300 2 260
C 380 3 300
4 180
The cost of shipping one unit from each factory to each warehouse is given below.
FACTORY WAREHOUSE COST $
o A 1 3
o A 2 5
o A 3 6
o A 4 5
o B 1 7
o B 2 4
o B 3 9
o B 4 6
o C 1 5
o C 2 12
o C 3 10
o C 4 8
Using the Transportation method, find the least cost shipping schedule and state what it is ?.
Solution:
TABLEAU 1:
1 2 3 4 Capacity
A 160 60 220
B 200 100 300
C 200 180 380
Req 160 260 300 180 900
This is the initial solution, which costs
(160 * 3) + (60 * 5) + (200 * 4) + (100 * 9) + (200 * 10) + (180 * 8)
= $5920.00

OPERATIONS RESEARCH
3 5
6
4 1
7 4 6
5 12 10 8
6
5
2 4 1
7 4 9
6
3 - 1
6
5
2 3 1
4 1
7 4 9
6
TABLEAU 2:
1 2 3 4 Capacity
A 160 60 220
B 260 40 300
C 200 180 380
Req 160 260 300 180 900
The costs = (160 * 3) + (60 * 6) + (260 * 4) + (40 * 9) + (200 * 10) + (180 * 8)
= $5680.00
TABLEAU 3:
1 2 3 4 Capacity
A 220 220
B 260 40 300
C 160 40 180 380
Req 160 260 300 180 900
The costs = (160 * 5) + (40 * 10) + (260 * 4) + (40 * 9) + (220 * 6) + (180 * 8)
= $5360.00
TABLEAU 4:
1 2 3 4 Capacity
A 220 220
B 260 40 300
C 160 80 140 380
Req 160 260 300 180 900
The costs = (160 * 5) + (80 * 10) + (260 * 4) + (40 * 6) + (220 * 6) + (140 * 8)
= $5320.00
9
5
5 - 1
-2 7
3 5
5 12 10 8
7
3 5
5 12 10 8
6

OPERATIONS RESEARCH
Conclusion:
Since all the cell values are positive the solution is optimum with the following allocations:
A Supplies 220 to 3
B Supplies 260 to 2
C Supplies 160 to 1
C Supplies 80 to 3
C Supplies 140 to 4
With a minimum cost of $ 5320.00
QUESTION
Below is a transportation problem where costs are in thousand of dollars.
SOURCES DESTINATIONS
A B C CAPACITIES
X 14 13 15 500
Y 16 15 12 400
Z 20 15 16 600
REQUIREMENTS 700 300 500
i. Solve this problem fully indicating the optimum delivery allocations and the corresponding
total delivery cost. [6 marks]
ii. There are two optimum solutions. Find the second one [4 marks].
iii. Solve the same problem considering XA is an infeasible (prohibited / impossible) route
and find the new total transportation cost [7 marks].
iv. If under consideration is a road network in a war zone, what is the simple economic effect of
bombing a bridge between X and A? [3 marks].
3 5
6
9
0
7 4
1
-4
5 12 10
Solution: PART (i)
TABLEAU 1:
A B C Capacity
X 500 500
Y 200 200 400
Z 4
100 500 600
Req 700 300 500 1500
The initial solution = (500 * 14) + (200 * 16) + (15 * 200) + (100 * 15) + (500 * 16)

OPERATIONS RESEARCH
= $227000000
3 5
6
9
4
7 4
4
5
5 12 10
3 5
6
9
4
7 4
5
TABLEAU 2:
A B C Capacity
X 500 500
Y 200 200 400 400
Z 0
300 300 600
Req 700 300 500 1500
Hence delivery allocations are:
X Supplies 500 to A
Y Supplies 200 to A
Y Supplies 200 to C
Z Supplies 300 to B
Z Supplies 300 to C
With a minimum cost of (500 * 14) + (200 * 16) + (15 * 300) + (200 * 12) + (300 * 16)
= $21900 0000
PART (ii)
The existence of an alternative least cost solution is indicated by a value of zero in an unoccupied
cell in the final table. We add and subtract the smallest quantity in the column or row of the zero to
get the alternative.
TABLEAU 1:
A B C Capacity
X 500 500
Y 0 4
400 400 400
Z 200 5 300 12 100 600
10
Req 700 300 500 1500

OPERATIONS RESEARCH
3 5
6
5 0
5 12 10
X Supplies 500 to A
Y Supplies 400 to C
Z Supplies 200 to A
Z Supplies 300 to B
Z Supplies 100 to C
= $21900 0000
PART (iii)
TABLEAU 1:
A B C Capacity
X --- 300 200 500
Y 400 400 400
Z 300 300 600
Req 700 300 500 1500
X Supplies 300 to B
X Supplies 200 to C
Y Supplies 400 to A
Z Supplies 300 to A
Z Supplies 300 to C
= $24100 0000
PART (iv)
The simple economic effect of bombing the bridge between X and A
= 24100 0000 – 21900 0000
= 2200 000
9
7 4
1

OPERATIONS RESEARCH
DUMMIES:
This is an extra row or column in a transportation table with zero cost in each cell and with a
total equal to the difference between total capacity and total demand.
In an unbalance transportation problem a dummy source or destination is introduced.
12 23
43
3
10 - 51
23 0
63 33
53
51
21 -22
30 -40 0
33 1 63 13
0
0
QUESTION:
The transport manager of a company has 3 factories A, B and C and four warehouses I, II, III and
IV is faced with a problem of determining the way in which factories should supply warehouses so
as to minimize the total transportation costs.
In a given month the supply requirements of each warehouse, the production capacities of the
factories and the cost of shipping one unit of product from each factory to each warehouse in $ are
shown below.
FACTORY WAREHOUSES
I II III IV PRO AVAIL
A 12 23 43 3 6
B 63 23 33 53 53
C 33 1 63 13 17
REQUIREMENTS 4 7 6 14 31
You are required to determine the minimum cost transportation plan [20 marks].
Solution:
TABLEAU 1:
I II III IV Dummy Capacity
A 4 2 6
B 5 6 14 28 53
C 17 17
Req 4 7 6 14 45 76
This is the initial solution, which costs
(4 * 12) + (2 * 23) + (5 * 23) + (6 * 33) + (14 * 53) + (28 * 0) + (17 * 0)
= $1149.00

OPERATIONS RESEARCH
12 23
43
3
50 60
23 0
63 33
53
1
-29 -22
30 -40 0
33 1 63 13
50
0
12 23
43
3
16 20
23 0
40
63 33
53
41
11 -22
30 0
33 1 63 13
10
0
12 23
43
3
32 42
23 0
18
63 33
53
19
11 22
52 0
33 1 63 13
32
0
TABLEAU 2:
A 4 2 6
B 7 6 12 28 53
C 17 17
Req 4 7 6 14 45 76
The costs=
(4 * 12) + (2 * 3) + (7 * 23) + (6 * 33) + (12 * 53) + (28 * 0) + (17 * 0)
= $1049.00
TABLEAU 3:
A 4 2 6
B 7 6 40 53
C 12 5 17
Req 4 7 6 14 45 76
The costs=
(4 * 12) + (2 * 3) + (7 * 23) + (6 * 33) + (40 * 0) + (12 * 13) + (5 * 0)
= $569.00
TABLEAU4:
A 4 2 6
B 2 6 45 53
C 5 12 17
Req 4 7 6 14 45 76
The costs=
(4 * 12) + (2 * 3) + (2 * 23) + (6 * 33) + (45 * 0) + (12 * 13) + (5 * 1)
= $459.00

OPERATIONS RESEARCH
Factory A Supplies Warehouse I
Factory A Supplies Warehouse IV
Factory B Supplies Warehouse II
Factory B Supplies Warehouse III
Factory B Supplies Warehouse Dummy
Factory C Supplies Warehouse II
Factory C Supplies Warehouse IV
With a minimum cost of (4 * 12) + (2 * 3) + (2 * 23) + (6 * 33) + (45 * 0) + (12 * 13) + (5 * 1)
= $459.00
QUESTION:
A well-known organization has 3 warehouse and 4 Shops. It requires transporting its goods from the
warehouse to the shops. The cost of transporting a unit item from a warehouse to a shop and the
quantity to be supplied are shown below.
DESTINATION
I II III IV TOTAL SUPPLY
SOURCE A 10 0 20 11 15
SOURCE B 12 7 9 20 25
SOURCE C 0 14 16 18 5
TOTAL DEMAND 5 15 15 10
Use any method to find the optimum transportation schedule and indicate the cost [14marks].
DEGENERATE SOLUTION:
It involves working a transportation problem if the number of used routes is equal to:
Number of rows + Number of column – 1.
However if the number of used routes can be less than the required figure we pretend that an empty
route is really used by allocating a zero quantity to that route.
MAXIMIZATION PROBLEMS:
Transportation algorithm assumes that the objective is to minimize cost. However it is possible to
use the method to solve maximization problem by either:
Multiply all the units’ contribution by – 1.
Or by subtracting each unit contribution from the maximum contribution in the
table.

OPERATIONS RESEARCH
UNIT 3: NON-LINEAR FUNCTIONS:
HOURS: 20.
NON-LINEAR FUNCTIONS:
o MARGINAL DISTRIBUTION:
PARTIAL INTEGRATION:
Partial integration is a function with more than one variable or finding the probability of a function
with more than one variables i.e. f(X1, X2, X3, ….Xn) and is just the rate at which the values of a
function change as one of the independent variables change and all others are held constant.
Question 1:
If f(x, y) = 2(x + y –2xy) given the intervals 0= x=1, 0=y=1.
Find the marginal distribution of x = f(x).
Find the marginal distribution of y = f(x).
Solution:
Pr {0=x=1} = 0∫1 2(x + y –2xy)dx
= 2 0∫1 (x + y –2xy)dx
= 2 [x2/2 + xy + x2y]0
1
= 2 [½ + y – y]
= 2[½]
= 1
Pr {0=y=1} = 0∫1 2(x + y –2xy)dy
= 2 ∫1 (x + y –2xy)dy
0= 2 [xy +y2/+ xy2]1
2 0
= 2 [x + ½ – x]
= 2[½]
= 1
Question 1:
If f(X1, X2) =(X2
1X2 + X3
1X2
2 + X1) given the intervals 0= X1=2, 1=X2 =3.
Find the marginal distribution of x = f(x).
Find the marginal distribution of y = f(x).
Find the Expected value of X1 (E(X1)).
Find the variance of X1 (Var (X1)).

OPERATIONS RESEARCH
Solution:
Pr {0=X1=2} = 0∫2 (X2
1X2 + X3
1X2
2 + X1)dx
= [X3
1X2 /3+ X4
1X2
2 /4+ X2
1/2]0
2
= [8X2 /3+ 4X2
2 + 2] – [0]
= 8X2 /3+ 4X2
2 + 2
Pr {1=X2=3} = 1∫3 (X2
1X2 + X3
1X2
2 + X1)dy
= [X2
1X2
2 /2+ X3
1X3
2 /3+ X1X2]1
3
= [9X2
1 /2+ 27X3
1 /3+ 3X2]1
3 –[X2
1 /2+ X3
2 /3+ X1]
= 9X2
1 /2+ 27X3
1 /3+ 3X2 – X2
1 /2- X3
2 /3 - X1
= 8X2
1 /2+ 26X3
1 /3+ 2X2
Expected value of E(X1) =0∫2 X. f(X1)dx
= 0∫2 X(X2
1X2 + X3
1X2
2 + X1)dx
= 0∫2 (X3
1X2 + X4
1X2
2 + X2
1)dx
= [X4
1X2 /4+ X5
1X2
2 /5+ X3
1/3]0
2
= [4X2 + 32X2
2 /5 + 8 /3] – [0]
= 4X2 + 32X2
2 /5 + 8 /3
Variance of X1 = Var (X1) = 0∫2 ([X1 - E(X1)]2 . f(X1)dx
= 0∫2 ([X1 - 4X2 + 32X2
2 /5 + 8 /3]2 * (X2
1X2 + X3
1X2
2 + X1)dx.
Question 2:
A manufacturing company produces two products bicycles and roller skates. Its fixed costs
production is: $1200 per week. Its variables costs of production are: $40 for each bicycle produced
and $15 for each pair of roller skates. Its total weekly costs in producing x bicycles and y pairs of
roller skates are therefore c= cost.
C(x, y) = 1200 + 40x + 15y for example; in producing x = 20 bicycles and y = 30 pairs of roller
skates/ week.
The manufacture experiences total cost of:
C(20, 30) = 1200 + 40(20) + 15(30)
= 1200 + 800 + 450
= 2450.
Question 3:
A manufacturing of Automobile tyres produces 3 different types: regular, green and blue tyres. If the
regular tyres sell for $60 each, the green tyres for $50 each and the blue tyres for $100 each. Find a
function giving the manufacture’s total receipts or revenue from the of x regular tyres and y green
tyres and z blue tyres.
R(x, y, z) = 60x + 50y +100z.

OPERATIONS RESEARCH
Solution:
Since the receipts of the sale of any tyre type is the price per tyre times the number of tyres sold:
The total receipts are:
R(x, y, z) = 60x + 50y + 100z
For example receipts from the sell of 10 tyres of each type would be:
R(10,10,10) = 60(10) + 50(10) + 100(10)
= 600 + 500 + 1000
= $2100
PARTIAL DIFFERENTIATION:
For a function “f” of a single variable, the derivative f measures the rate at which the values of f(x)
change as the independent variable x change.
A partial derivative of a function i.e. f(X1, X2, X3.. Xn) of several variables is just the rate at which the
values of the function change as one of the independent variable changes and all others are held
constant.
Question 3:
For the function f(x, y) = X3 + 4X2Y3
+ Y2
Find
df / dx
df / dy
f(-2; 3)
Solution:
df / dx = 3X2 + 8XY3
df / dy = 12X2Y2
+ 2Y
f(-2; 3) = X3 + 4X2Y3
+ Y2
= (-2)3 + 4(-2)2(3)3
+ (3)2
= -8 + 16(27) +9
= 433
Question 4:
A company produces electronic typewriters and word processors, it sells the electronic typewriters
for $100 each and word processors for $300 each. The company has determined that its weekly sales
in producing x electronic writers and y word processors are given by the following joint cost
function.
C(x, y) = 200 + 50x +8y + X2 + 2Y2
Find the numbers of x and y of machines that the company should manufacture and sell weekly in
order to maximize profits.

OPERATIONS RESEARCH
Solution:
Revenue function is given by:
R(x, y) = 100x + 300y
Profit = Revenue – Cost.
Then Profit function is given by:
P(x, y) = R(x, y) – C(x, y)
= (100x + 300y) – (200 + 50x +8y + X2 + 2Y2)
= 50x + 292y – 200 - X2 - 2Y2
To find the critical points of turning points of x and y. We set the partial derivative = 0.
Thus
dp / dx = 50 – 2x = 0
50 = 2x
x = 25
dp / dy = 292 – 4y =0
292 = 4y
y = 73
The production schedule for maximum profit is therefore x = 25 type writers and y = 73 word
processors which yields a profit of
P = 50(25) + 292(73) – 200 – 625 – 2(73)2
= 1250 + 21316 – 200 – 625 – 1065
= 22566 – 11493
= $11083
NECESSARY AND SUFFICIENT CONDITIONS FOR EXTREMA:
The necessary condition or the GRADIENT VECTOR of the extrema determines the turning
points or critical points of a function.
Let X0 be a variable representing the turning point and represented mathematically as:
X0 = (A0, B0, … N0).
In general form; a necessary condition or gradient vector for X0 to be an extrema point of f(x) is that
the gradient (Ñ) º Ñf (X0) = 0.

OPERATIONS RESEARCH
Question 1:
Given f(X1, X2, X3) =(X1 + 2X3 + X2X3 – X2
1 - X2
2 - X2
3)
Find the gradient vector for X0 i.e. Ñf (X0) = 0.
Solution:
The necessary condition (gradient vector) Ñf (X0) = 0 is given by:
df / dx1 = 1 - 2X1 = 0.
1 - 2X1 = 0. [1]
df / dx2 = X3 - 2X2 = 0.
X3 - 2X2 = 0. [2]
df / dx3 = 2 + X2 - 2X3 = 0.
2 + X2 - 2X3 = 0. [3]
(a) Finding X1 is given by 1 = 2X1
X1 = ½
(b) Equation 2 is given by X3 - 2X2 = 0.
X3 = 2X2.
(c) On equation 3 where therefore substitute X3 with 2X2.
Thus 2 + X2 - 2X3 = 0.
2 + X2 – 2(2X2) = 0.
2 + X2 – 4X2 = 0.
2 – 3X2 = 0.
X2 = 2/3.
Therefore X3 = 2X2.
X3 = 2(2 / 3)
X3 = 4/3.
Therefore X0 = (½, 2/3, 4/3)

OPERATIONS RESEARCH
Hessian matrix
In mathematics, the Hessian matrix
partial derivatives of a function
variables.
Given the real-valued function
function; that is, it describes the local curvature of a function of many
if all second partial derivatives
where x = (x1, x2, ..., xn) and Di
the Hessian becomes
(or simply the Hessian) is the square matrix
of f exist, then the Hessian matrix of f is the matrix
is the differentiation operator with respect to the
Some mathematicians define the Hessian as the
Bordered Hessian
A bordered Hessian is used for the second
problems. Given the function as before:
second-derivative test in certain constrained optimiza
but adding a constraint function such that:
the bordered Hessian appears as
determinant of the above matrix.
rmmakaha@gmail.com
54
of second-order
; ith argument and
optimization

OPERATIONS RESEARCH
If there are, say, m constraints then the zero in the north-north
west corner is an m
and there are m border rows at the top and
m border columns at the left.
The above rules of positive definite and negative definite can not apply here since a bordered
Hessian can not be definite: we have
z'Hz = 0 if vector z has a non-zero as its first element,
followed by zeroes.
The second derivative test consists here of sign restrictions of the determinants of a certain set of
m submatrices of the bordered Hessian. Intuitively, think of the
problem to one with n - m free variables. (For example,
constraint x+ x+ x= 1 can be reduced to the maximization of
1 2 3 constraint.)
A sufficient condition for X
HESSIAN matrix (denoted by H) eval
i. Positive definite when X
ii. Negative definite when X
The Hessian matrix is achieved by finding the 2
each equation with respect to all variables
Thus the Hessian matrix is evaluated at the point X
H/X= d2f/2
d2f/ 2
0 dX
1,
dX1 dX
2,
d2f/dX2 dX
2
1,
d2f/ dX
2
2,
d2f/dX3 dX
2
1,
d2f/ dX3 dX
2
2,
rmmakaha@gmail.com
he m constraints as reducing the
the maximization of f
f(x1,x2,1 − x
X0 a point to be extremism is that the
evaluate at X0 is:
X0 is a Minimum point.
X0 is a Maximum point.
2nd Partial derivation of the first Partial derivative of
defined.
0
,
d2f/ dX1 dX
2
3
d2f/ dX2 dX
2
3
d2f/ dX
2
3
55
× m block of zeroes,
n -
f(x1,x2,x3) subject to the
1 − x2) without

OPERATIONS RESEARCH
df / dx1 = 1 - 2X1 = 0.
df / dx2 = X3 - 2X2 = 0.
df / dx3 = 2 + X2 - 2X3 = 0.
To establish the sufficiency the function has to have:
H/X0 = -2 0 0
0 -2 1
0 1 -2
Since the Hessian matrix is 3 by 3 matrix then:
Find the 1st Principal Minor determinant of 1 by 1 matrix in the Hessian matrix.
Find the 2nd Principal Minor determinant of 2 by 2 matrix in the Hessian matrix.
Find the 3rd Principal Minor determinant of 3 by 3 matrix in the Hessian matrix.
The Positive definite when X0 is a Minimum point is evaluated as:
When 1st PMD = + ve.
When 2nd PMD = +ve.
When 3rd PMD = +ve.
Or
When 1st PMD = - ve.
When 2nd PMD = +ve.
When 3rd PMD = +ve.
Thus 3 by 3 Hessian matrix the number of positive number should be greater than one. (Should be
two or more).
The Negative definite when X0 is a Maximum point is evaluated as:
When 1st PMD = - ve.
When 2nd PMD = - ve.
When 3rd PMD = - ve.
Or
When 1st PMD = + ve.
When 2nd PMD = - ve.
When 3rd PMD = - ve.
Thus 3 by 3 Hessian matrix the number of negative number should be greater than two. (Should be
two or more).
H/X0 = -2 0 0
0 -2 1
0 1 -2
Thus the 1st PMD of (-2) = -2

OPERATIONS RESEARCH
Thus the 2nd PMD of –2 0
0 -2
= (-2 * -2) – (0 * 0)
= 4
The 3rd PMD = -2 0 0
0 -2 1
0 1 -2
= -2 –2 1 - 0 0 1 + 0 0 -2
1 -2 0 -2 0 1
= -2 {(-2 * -2) – (1 * 1)} – 0 (0 – 0) + 0 (0 – 0)
= - 2 (3)
= - 6
Thus the PMD is equal to –2, 4 and –6 and H/X0 is negative definite and X0 = (½, 2/3, 4/3) represents a
Maximum point.
Question 2:
Given f(X1, X2, X3) =(-X1 + 2X3 - X2X3 + X2
1+ X2
2 - X2
3)
i. Find the gradient vector for X0 i.e. Ñf (X0) = 0.
ii. Determine the nature of the turning points using Hessian Matrix.
Solution:
The necessary condition (gradient vector) Ñf (X0) = 0 is given by:
df / dx1 = -1 + 2X1 = 0.
-1 + 2X1 = 0. [1]
df / dx2 = -X3 + 2X2 = 0.
-X3 + 2X2 = 0. [2]
df / dx3 = 2 - X2 - 2X3 = 0.
2 - X2 - 2X3 = 0. [3]
(b) Finding X1 is given by -1 = 2X1
X1 = ½
(b) Equation 2 is given by -X3 + 2X2 = 0.
X3 = 2X2.
(c) On equation 3 where therefore substitute X3 with 2X2.
Thus 2 - X2 - 2X3 = 0.

OPERATIONS RESEARCH
2 - X2 – 2(2X2) = 0.
2 - X2 – 4X2 = 0.
2 – 5X2 = 0.
X2 = 2/5.
Therefore X3 = 2X2.
X3 = 2(2 / 5)
X3 = 4/5.
Therefore X0 = (½, 2/5, 4/5)
H/X0 = d2f/ dX
2
1,
d2f/ dX1 dX
2
2,
d2f/ dX1 dX
2
3
d2f/ dX2 dX
2
1,
d2f/ dX
2
2,
d2f/ dX2 dX
2
3
d2f/ dX3 dX
2
1,
d2f/ dX3 dX
2
2,
d2f/ dX
2
3
H/X0 = 2 0 0
0 2 -1
0 -1 -2
Thus the 1st PMD of (2) = 2
Thus the 2nd PMD of 2 0
0 2
= (2 * 2) – (0 * 0)
= 4
The 3rd PMD = 2 0 0
0 2 -1
0 -1 -2
= -2 2 -1 - 0 0 -1 + 0 0 2
-1 -2 0 -2 0 -1
= 2 {(2 * -2) – (-1 * -1)} – 0 (0 – 0) + 0 (0 – 0)
= 2 (-4) - (1)
= 2 (-5)
= - 10
Thus the PMD is equal to 2, 4 and –10 and H/X0 is positive definite and X0 = (½, 2/5, 4/5) represents a
Minimum point.

OPERATIONS RESEARCH
NON LINEAR ALGORITHMS: (COMPUTATIONS)
1 THE GRADIENT METHODS:
The general idea is to generate successive iterative points, starting from a given initial point, in the
direction of the fast and increase (maximization of the function).
The method is based on solving the simultaneous equations representing the necessary conditions
for optimality namely Ñf (X0) = 0.
Termination of the gradient method occurs at the point where the gradient vector becomes null.
This is only a necessary condition for optimality suppose that f(x) is maximized.
Let X0 be the initial point from which the procedure starts and define Ñf (Xk) as the gradient of f at
Kth point Xk.
This result is achieved if successive point Xk and Xk+1 are selected such that
Xk+1 = Xk + rk Ñf (Xk) where rk is a parameter called Optimal Step Size.
The parameter rk is determined such that Xk+1 results in the largest improvement in f. In other
words, if a function h(r) is defined such that h(r) = f(Xk )+ rk Ñf (Xk). This function is then
differentiated and equate zero to the differentiatable function to obtain the value of rk.
Question 3:
Consider maximizing f(X1, X2) =(4X1 + 6X2 - 2X2
1- 2X2X1 - 2X2
2)
And let the initial point be given by X0(1, 1).
Hint in X0(1, 1) X1 =1 and X2 = 1
Solution:
Find Ñf (X0) = (df/dx1, df/dx2)
= (4 – 4X1 – 2X2; 6 – 2X1 - 4X2)
1st iteration
Step 1: Find Ñf (X0) = (4 – 4 – 2; 6 – 2 - 4)
= (-2; 0)
Step 2: Find Xk+1 = Xk + rk Ñf (Xk)
X0+1 = X0 + rk Ñf (X0)
X1 = (1, 1) + r(-2; 0)
(1, 1) + (-2r, 0)
(1 + -2r, 1)
(1 – 2r, 1)
Thus h(r) = f(Xk )+ rk Ñf (Xk)
= f(X0 )+ r Ñf (X0)
= f(1 – 2r; 1)
= 4(1 – 2r) + 6(1) – 2(1 – 2r)2 – 2(1)(1 – 2r) - 2(1)2.

OPERATIONS RESEARCH
= 4(1 – 2r) + 6 – 2(1 – 2r) 2 – 2(1 – 2r) - 2.
= 4(1 – 2r) – 2(1 – 2r) + 6 – 2 – 2(1 – 2r)2.
= (4 – 2)(1 – 2r) + 4 – 2(1 – 2r)2.
= 2(1 – 2r) – 2(1 – 2r)2 + 4.
= – 2(1 – 2r)2 +2(1 – 2r) + 4.
= – 2(1 – 2r)2 + 2 – 4r + 4.
h1(r) = 0
– 2(1 – 2r)2 + 2 – 4r + 4 = 0.
- 4 * -2(1 – 2r) – 4 = 0.
8(1 – 2r) + - 4 = 0.
8 – 16r – 4 = 0
4 –16r = 0.
r = ¼
The optimum step size yielding the maximum value of h(r) is h1 = ¼.
This gives X1= (1 –2(¼); 1)
= (1 - ½; 1)
= (½; 1)

OPERATIONS RESEARCH
UNIT 4: PROJECT MANAGEMENT WITH
PERT/CPM
HOURS: 20
PROJECT MANAGEMENT:
TERMS USED IN PROJECT MANAGEMENT:
PROJECT:
Is a combination of interrelated activities that must be executed in a certain order
before the entire task can be completed?
ACTIVITY:
Is a job requiring time and resource for its completion?
ARROW:
Represents a point in time signifying the completion of some activities and the
beginning of others.
NETWORK:
Is a graphic representation of a project’s operation and is composed of activities and
nodes.
Benefits of PERT
PERT is useful because it provides the following information:
· Expected project completion time.
· Probability of completion before a specified date.
· The critical path activities that directly impact the completion time.
· The activities that have slack time and that can lend resources to critical path activities.
· Activity starts and end dates.
Limitations
The following are some of PERT's weaknesses:
· The activity time estimates are somewhat subjective and depend on judgement. In cases
where there is little experience in performing an activity, the numbers may be only a guess.
In other cases, if the person or group performing the activity estimates the time there may be
bias in the estimate.
· Even if the activity times are well-estimated, PERT assumes a beta distribution for these
time estimates, but the actual distribution may be different.

OPERATIONS RESEARCH
· Even if the beta distribution assumption holds, PERT assumes that the probability
distribution of the project completion time is the same as the that of the critical path.
Because other paths can become the critical path if their associated activities are delayed,
PERT consistently underestimates the expected project completion time.
Critical Path Analysis CPA (Network Analysis)
Critical Path Analysis (CPA) is a project management tool that:
· Sets out all the individual activities that make up a larger project.
· Shows the order in which activities have to be undertaken.
· Shows which activities can only taken place once other activities have been completed.
· Shows which activities can be undertaken simultaneously, thereby reducing the overall time
taken to complete the whole project.
· Shows when certain resources will be needed – for example, a crane to be hired for a
building site.
In order to construct a CPA, it is necessary to estimate the elapsed time for each activity – that is the
time taken from commencement to completion.
Then the CPA is drawn up a based on dependencies such as:
· The availability of labour and other resources
· Lead times for delivery of materials and other services
· Seasonal factors – such as dry weather required in a building project
Once the CPA is drawn up, it is possible to see the CRITICAL PATH itself – this is a route
through the CPA, which has no spare time (called ‘FLOAT’ or ‘slack’) in any of the activities. In
other words, if there is any delay to any of the activities on the critical path, the whole project will be
delayed unless the firm makes other changes to bring the project back on track.
The total time along this critical path is also the minimum time in which the whole project can be
completed.
Some branches on the CPA may have FLOAT, which means that there is some spare time available
for these activities.
What can a business do if a project is delayed?
· Firstly, the CPA is helpful because it shows the likely impact on the whole project if no
action were taken.
· Secondly, if there is float elsewhere, it might be possible to switch staff from another activity
to help catch up on the delayed activity.
· As a rule, most projects can be brought back on track by using extra labour – either by hiring
additional people or overtime. Note, there will be usually be an extra cost. Alternative
suppliers can usually be found – but again, it might cost more to get urgent help.

OPERATIONS RESEARCH
The key rules of a CPA
· Nodes are numbered to identify each one and show the Earliest Start Time (EST) of the
activities that immediately follow the node, and the Latest Finish Time (LFT) of the
immediately preceding activities
· The CPA must begin and end on one ‘node’ – see below
· There must be no crossing activities in the CPA
· East activity is labelled with its name eg ‘print brochure’, or it may be given a label, such as
‘D’, below.
· The activities on the critical path are usually marked with a ‘//’
In the example below
· The Node is number 3
· The EST for the following activities is 14 days
· The LFT for the preceding activities is 16 days
· There is 2 days’ float in this case (difference between EST and LFT)
· The activity that follows the node is labelled ‘D’ and will take 6 days
OR

OPERATIONS RESEARCH
A simple example – baking a loaf of bread
Here is a simple example, in which some activities depend on others having been undertaken in
order, whereas others can be done independently.
Activity Preceded by Elapsed time (minutes)
A weigh ingredients - 1
B mix ingredients A 3
C dough rising time B 60
D prepare tins - 1
E pre-heat oven - 10
F knock back dough and place in tins CD 2
G 2 nd dough rising time F 15
H cooking time E G 40
In this example, there is a clear sequence of events that have to happen in the right order. If any of
the events on the critical path is delayed, then the bread will not be ready as soon. However, tasks D
(prepare tins) and E (heat the oven) can be started at any time as long as they are done by the latest
finish time in the following node.
So, we can see that the oven could be switched on as early as time 0, but we can work out that it
could be switched on at any time before 71 – any later than this and it won’t be hot enough when
the dough is ready for cooking. There is some ‘float’ available for tasks D and E as neither is on the
critical path.
This is a fairly simple example, and we can see the LST and LFT are the same in each node. In a
more complex CPA, this will not necessarily be the case, and if so, will indicate that there is some
‘float’ in at least one activity leading to the node. However, nodes on the critical path will always
have the same EST and LFT.

OPERATIONS RESEARCH
HOW TO CONSTRUCT A CRITICAL PATH NETWORK DIAGRAM
Here is the data:
Activity Preceded by Duration (days)
A - 2
B - 3
C A 4
D B 5
E C 8
F E 3
G D,F 4
A
Here is what to do:
1. Draw the first ‘node’ and number it ‘1’.
2. Draw the line to show any ‘activities’ that are not preceded by any other activities.
3. Tick these activities off to show you have done them
4. Look at the next activity and see which it is preceded by. In this case it is activity C and it is
preceded by activity A
5. Draw this on the diagram and again tick off the activity on the list.
6. Do the same for activity D, E F. Look carefully at which activity they are preceded by
7. Now do activity G. This one is a little trickier, as it is preceded by more than one activity.
However, all you have to do is make them meet at one node – easy!
8. Next, draw a node on the end of the network diagram
9. Now number the nodes, following through the activities. If there are 2 activities starting at
the time, you need to number the shortest activity first.

OPERATIONS RESEARCH
The diagram is now ready for you to work out (make sure you understand these terms by reading
further).
· Earliest Start Time (EST) – top segment or left segment.
· Latest Finish Time (LFT) – bottom segment or right segment.
· Float Time
· The critical path
There are many ways to do the above, but the method below is the
simplest, so learn it and follow it!
i. Work out which ‘route’ takes longest (which is the critical path)
ii. In this case : A C E F G takes 21 days and B D G takes 12 days
iii. Consequently, A C E F G is the critical path
iv. As these activities take 21 days, you can write ‘21’ days in both the top or left and bottom or right
segment of the end node.
v. Now work backwards through the nodes on the critical path and enter the LFT in the
bottom or right segment and the EST in the top or left segment.
vi. Taking off the length of time it takes to complete the activity. So from node 6 would have
‘17’ in both segments.
vii. Now work backwards through any other nodes and enter the LFT in the bottom or right
segment. You MUST do this backwards for ALL other ‘routes’. You then do the same for
the EST for each route, but go forwards this time!
viii. Note: Backwards for LFT and Forwards for EST
ix. Here it is below!

OPERATIONS RESEARCH
Activity EST LFT Duration Float(LFT-D-EST)
A 0 2 2 0
B 0 12 3 9
C 2 6 4 0
D 3 17 5 9
E 6 14 8 0
F 14 17 3 0
G 17 21 4 0
RULES FOR CONSTRUCTING NETWORK DIAGRAM:
Each activity is represented by one and only one arrow in the network.
No two activities can be identified by the same head and tail events. If activities A and B
can be executed simultaneously, then a dummy activity is introduced either between A
and one end event or between B and one end event. Dummy activities do not consume
time or resources. Another use of the dummy activity: suppose activities A and B must
precede C while activity E is preceded by B only.
To ensure the correct precedence relationships in the network diagram, the following
questions must be answered as every activity is added to the network:
What activities must be completed immediately before this activity can start.
What activities must follow this activity?
What activities must occur concurrently with this activity?

OPERATIONS RESEARCH
9
29 35
Question 1:
ACTIVITY PRECEDED BY DURATION (Weeks)
A Initial activity 10
B A 9
C A 7
D B 6
E B 12
F C 6
G C 8
H F 8
I D 4
J G, H 11
K E 5
L I 7
Find the critical path and the time for completing the project.
Solution:
D I
6 4
B 9 E 12 L 7
K
A 5
10
Dummy J 11
C 7
G 8
F H
6 8
EARLISET START TIME:
Represents all the activities emanating from i. Thus ESi represent the earliest occurrence time of
event i.
Earliest finish time is given by:
EF = Max {ESi + D}
0
0 0
1
10 10
2
19 25
3
17 17
4
25 31
5
31 37
7
25 31
6
23 23
8
31 31
10
42 42

OPERATIONS RESEARCH
LATEST COMPLETION TIME:
It initiates the backward pass. Where calculations from the “end” node and moves to the “start”
node.
Latest start time is given by:
LSi = Min {LF – D}
DETERMINATION OF THE CRITICAL PATH:
A Critical path defines a chain of critical that connects the start and end of the arrow diagram.
An activity is said to be critical if the delay in its start will cause a delay in the completion date of
the entire project. Or it is the longest route, which the project should follow until its completion
date of the entire project.
The critical path calculations include two phases:
FORWARD PASS:
Is where calculations begin from the “start” node and move to the “end” node. At each
node a number is computed representing the earliest occurrence time of the corresponding
event.
BACKWARD PASS:
Begins calculations from the “end” node and moves to the “start” node. The number
computed at each node represents the latest occurrence time of the corresponding event.
DETERMINITION OF THE FLOATS:
A Float or Spare time can only be associated with activities which are non critical. By definition
activities on the critical path cannot have floats.
There are 3 types of floats.
1 TOTAL FLOAT:
This is the amount of time a path of activities could be delayed without affecting the overall
project duration.
Total Float = Latest Head Time – Earliest Tail time – duration.
= LS – ES.
= LF – ES – D
= LF – EF or EC.
1 FREE FLOAT:
This is the amount of time an activity can be delayed without affecting the commencement
of a subsequent activity at its earliest start time.
Free Float = Earliest Head Time – Earliest Tail Time – Duration.
= LF – ES – D
= ESj – ESi – D.

OPERATIONS RESEARCH
1 INDEPENDENT FLOAT:
This is the amount of time an activity can be delayed when all preceding activities are
completed as late as possible and all succeeding activities completed as early as possible.
Independent Float = EF – LS – D.
NORMAL EARLIEST TIME LATEST TIME TOTAL FLOAT
ACTIVITY TIME ES EF = ES + D LS = LF – D LF =LS – ES
A 10 0 10 0 10 0
B 9 10 19 16 25 6
C 7 10 17 10 17 0
D 6 19 25 25 31 6
E 12 19 31 25 37 6
F 6 17 23 17 23 0
G 8 17 25 23 31 6
H 8 23 31 23 31 0
I 4 25 29 31 35 6
J 11 31 42 31 42 0
K 5 31 36 37 42 6
L 7 29 36 35 42 6
Question 2:
Draw the network for the data given below then find the critical path as well total float and free
float.
ACTIVITY (I, J) DURATION
(0, 1) 2
(0, 2) 3
(1, 3) 2
(2, 3) 3
(2, 4) 2
(3, 4) 0
(3,5) 3
(3, 6) 2
(4, 5) 7
(4, 6) 5
(5, 6) 6

OPERATIONS RESEARCH
3
6 6
6
19 19
4
6 6
5
13 13
Solution:
0
0 0
2
2
1
2 4
2 3
3 6
Dummy
3
7 5
2
2
3 3
ACTIVITY D ES EF=ES+D LS=LF-D LF TOTAL FREE
Float Float
(0, 1) 2 0 2 2 4 2 2
(0, 2) 3 0 3 0 3 0 0
(1, 3) 2 2 4 4 6 2 2
(2, 3) 3 3 6 3 6 0 0
(2, 4) 2 3 5 4 6 1 1
(3, 4) 0 6 6 6 6 0 0
(3,5) 3 6 9 10 13 4 4
(3, 6) 2 6 8 17 19 11 11
(4, 5) 7 6 13 6 13 0 0
(4, 6) 5 6 11 14 19 8 8
(5, 6) 6 13 19 13 19 0 0

OPERATIONS RESEARCH
PERT ALGORITHM:
PROBABILISTIC TIME DURATION OF ACTIVITIES.
The following are steps involved in the development of probabilistic time duration of activities.
Make a list of activities that make up the project including immediate
predecessors.
Make use of step 1 sketch the required network.
Denote the Most Likely Time by Tm, the Optimistic Time by To and
Pessimistic time by Tp.
Using beta distribution for the activity duration the Expected Time Te for each
activity is computed by using the formula:
Te = (To + 4Tm + Tp) / 6.
Tabulate various times i.e. Expected activity times, Earliest and Latest times and
the EST and LFT on the arrow diagram.
Determine the total float for each activity by taking the difference between EST
and LFT.
Identify the critical activities and the expected date of completion of the project.
Using the values of Tp and To compute the variance (d2) of each activity’s time
estimates by using the formula: d2 = {{Tp – To} / 6}2.
Compute the standard normal deviate by:
Zo = (Due date – Expected date of Completion) / ÖProject variance.
Use Standard normal tables to find the probability P (Z = Zo) of completing
the project within the scheduled time, where Z ~ N(0,1).
Question 3:
A project schedule has the following characteristics:
Activity Most Likely Time Optimistic Time Pessimistic Time
1 – 2 2 1 3
2 – 3 2 1 3
2 – 4 3 1 5
3 – 5 4 3 5
4 – 5 3 2 4
4 – 6 5 3 7
5 – 7 5 4 6
6 – 7 7 6 8
7 – 8 4 2 6
7 – 9 6 4 8
8 – 10 2 1 3
9 – 10 5 3 7
I. Construct the project network.
II. Find expected duration and variance for each activity.

OPERATIONS RESEARCH
III. Find the critical path and expected project length.
IV. What is the probability of completing the project in 30 days.
Solution:
4 5 6
2
3 7 4 5
2
3 2
5
Expected job Time Te = (To + 4Tm + Tp) / 6.
Variance d2 = {{Tp – To} / 6}2.
Activity Tm To Tp Te d2
1 – 2 2 1 3 2 0.111
2 – 3 2 1 3 2 0.111
2 – 4 3 1 5 3 0.445
3 – 5 4 3 5 4 0.111
4 – 5 3 2 4 3 0.111
4 – 6 5 3 7 5 0.445
5 – 7 5 4 6 5 0.111
6 – 7 7 6 8 7 0.111
7 – 8 4 2 6 4 0.445
7 – 9 6 4 8 6 0.445
8 – 10 2 1 3 2 0.111
9 – 10 5 3 7 5 0.445
Critical path (*) comprises of activities (1 –2), (2 - 4), (4 –6), (6 –7), (7 –9) and (9 –10)
Expected project length is = 28 days.
Variance d2 = 0.111 + 0.445 + 0.445 + 0.111 + 0.445 + 0.445
= 2.00 (on critical path only)
1
0 0
2
2 2
3
4 8
5
8 12
7
17 17
9
23 23
10
28 28
4
5 5
6
10 10
8
21 26

OPERATIONS RESEARCH
(iv) Probability of completing the project in 30 days is obtained by:
Zo = (Due date – Expected date of Completion) / ÖProject variance.
= (30 – 28) / Ö2.
= 1.414 (Look this from Normal tables)
Now from Standard Normal tables Z= 0.4207.
P (t = 30) = P (Z = 1.414)
= 0.5 + 0.4207
= 0.9207
This shows that the probability of meeting the scheduled time will be 0.9207
COST CONSIDERATIONS IN PERT / CPM:
The cost of a project includes direct costs and indirect costs. The direct costs are associated with the
individual activities and the indirect costs are associated with the overhead costs such as
administration or supervision cost. The direct cost increase if the job duration is to be reduced
whereas the indirect costs increase if the job duration is to be increased.
1 TIME COST OPTIMIZATION PROCEDURE:
The process of shortening a project is called Crashing and is usually achieved by adding extra
resources to an activity. Project crashing involves the following steps:
Critical Path: Find the normal critical path and identify the critical activities.
Cost Slope: Calculate the cost slope for the different activities by using the
Formula: COST SLOPE = Crash cost – Normal cost.
Normal Time – Crash Time.
Ranking: Rank the activities in the ascending order of cost slope.
Crashing: Crash the activities in the critical path as per the ranking i.e. activities having
lower cost slope would be crashed first to the maximum extent possible. Calculate
the new direct cost by cumulatively adding the cost of crashing to the normal cost.
Parallel Crashing: As the critical path duration is reduced by the crash in step 3 other
paths become critical i.e. we get parallel critical paths. This means that project
duration can be reduced by simultaneous crashing of activities in the parallel critical
paths.
Optimal Duration. Crashing as per Step 3 and step 4 an optimal project is determined.
It would be the time duration corresponding to which the total cost (i.e. Direct cost
plus Indirect cost) is a minimum.

OPERATIONS RESEARCH
Question 4:
For the network given below find the optimum cost schedule for the completion of the project:
JOB NORMAL CRASH
TIME COST $ TIME COST $
1 – 2 10 60 8 120
2 – 3 9 75 6 150
2 – 4 7 90 4 150
3 – 4 6 100 5 140
3 – 5 9 50 7 80
3 – 6 10 40 8 70
4 – 5 6 50 4 70
5 – 6 7 70 5 110
Solution:
JOB COST SLOPE
*1 – 2 30 --- (4) = (120 –60) / (10 – 8)
*2 – 3 25 --- (3)
2 – 4 20
⇒ 3 – 4 40 ---(5)
3 – 5 15
3 – 6 15
4 – 5 10 --- (1)
5 – 6 20 --- (2)
10
9
6 9 7
10
7
6
The critical path = 1 –2, 2 –3, 3 –4, 4 –5 and 5 –7.
Expected project Length = 38 days.
Associated with 38 days the minimum direct project cost
= 60 + 75 + 90 + 100 + 50 + 40 + 50 + 70
= $535
1
0 0
3
19 19
5
31 31
6
38 38
4
25 25
2
10 10

OPERATIONS RESEARCH
In order to reduce the project duration we have to crash at least one of the jobs on the critical path.
This is being done because crashing of the job not on the critical path does not reduce the project
length.
1st Crashing:
On critical path the minimum cost slope is job 4 –5 and is to be crashed at extra cost of $10 per day.
10
9
3
19 19
6 9 7
5
29 29
4
25 25
10
7
4
1
0 0
2
10 10
Duration of project = 36 days and Total cost = $535 + $10 * 2 = $555.
2nd Crashing:
Now crash job 5 –6 and is to be crashed at extra cost of $20 per day.
10
9
6 9 5
10
7
4
6
36 36
1
0 0
5
29 29
4
25 25
2
10 10
3
19 19
6
34 34

OPERATIONS RESEARCH
3rd Crashing:
Now crash job 2 –3 for 3 days and is to be crashed at extra cost of $25.
10
6
3
16 16
6 9 5
5
26 26
4
22 22
10
7
4
1
0 0
2
10 10
4th Crashing:
Now crash job 1 –2 for 2 days and is to be crashed at extra cost of $30.
10
6
6 9 5
8
7
4
6
31 31
1
0 0
5
24 24
4
20 20
2
8 8
3
14 14
6
29 29

OPERATIONS RESEARCH
5th Crashing:
Final crash job 3 –4 for 1 day and it is to be crashed at extra cost of $40 and two critical paths
occurs.
10
6
3
14 14
5 9 5
5
23 23
4
19 19
8
7
4
1
0 0
2
8 8
Duration of project = 28 days and Total cost = $730+ $40 * 1 = $770.
Optimum Duration of project = 28 days and Total Cost = $770.
6
28 28

OPERATIONS RESEARCH
UNIT 5: RANDOM VARIABLES AND THEIR
PROBABILITY:
HOURS: 20.
RANDOM VARIABLES AND PROBABILITY FUNCTIONS (DISCRETE):
Given a Sample space S = {1,2,3,4,5,6} we may therefore use the variable such as X to represent
an outcome in the sample space such a variable is called Random variable.
When the outcome in a sample space are represented by values in a random variable the assignment
of probabilities to the outcome can be thought of as a function for which the domain is the sample
space, we refer to this as the probability function written as Pr.
We use the following notation with probability function Pr {X = a} which means the probability
associated with the outcome a while Pr {X in E} means the probability associated with event E.
Given S = {1,2,3,4,5,6}
a. Find the probability of S = 3.
b. Find the probability of S = 5.
c. Find the probability of X in E when E = 1,2.3.
Solution:
i. P (S = 3) = 1/6.
ii. P (S = 5) = 1/6.
iii. P (X in E) = ½.
PROBABILITY DENSITY FUNCTIONS:
2 PROPERTIES OF DISCRETE RANDOM VARIABLE:
a. It is a discrete variable.
b. It can only assume values x1, x2. …xn.
c. The probabilities associated with these values are p1, p2. …pn.
Where P(X = x1) = p1.
P(X = x2) = p2.
.
.
P(X = xn) = pn.
Then X is a discrete random variable if p1 + p2. …pn = 1.
This can be written as Σ P(X = x) =1.
all x

OPERATIONS RESEARCH
Question 1:
The P.d.f. of a discrete random variable Y is given by P (Y=y) = cy2, for y = 0,1,2,3,4.
Given that c is a constant, find the value of c.
Solution:
Y 0 1 2 3 4
P(Y =y) 0 c 4c 9c 16c
= Σ P(X = x) =1.
all x
1 = c + 4c + 9c + 16c
1 = 30c
c = 1/30.
Question 2:
The Pdf. of a discrete random variable X is given by P (X=x) = a(¾)x , for x = 0,1,2,3...
Find the value of the constant a.
Solution:
= ΣP(X = x) =1.
all x
P(X = 0) = a(¾)0.
P(X = 1) = a(¾)1.
P(X = 2) = a(¾)2.
P(X = 3) = a(¾)3 and so on.
So ΣP(X = x) =a + a(¾) + a(¾)2 + a(¾)3 + …
all x
= a( 1 + ¾ + (¾)2 + (¾)3 + …)
= a ( 1/1- ¾) - (sum of an infinite G.P with first term 1 and common ratio ¾)
= a(4)
4a = 1
a = ¼

OPERATIONS RESEARCH
EXPECTED VALUE/ MEAN / AVERAGE:
For a random variable X associated with a sample space {x1, x2. …xn} the concept of expected
value is the generalization of the average of numbers {x1, x2. …xn}.
2 EXPECTED VALUE WITH SAME PROBABILITIES:
The expected value of X with same probabilities is given by:
E(x) = X1 + X2 + …Xn/n.
Question 3:
Given that an die is thrown 6 times and the recordings are as follows then calculate the expected
mean or mean score
Score x 1 2 3 4 5 6
P(X = x) 1/6 1/6 1/6 1/6 1/6 1/6
Solution:
E(x) = X1 + X2 + …Xn/n.
= 1 + 2 + 3 + 4 + 5 + 6/6
= 21/6
= 7/2
= 3.5
2 EXPECTED VALUE WITH DIFFERENT PROBABILITIES:
The expected value of X with different probabilities is given by:
E(x) = P1 * X1 + P2 * X2 + …Pn * Xn.
Question 4:
Given a random variable X which has a Pdf shown below. Calculate the expected mean.
X -2 -1 0 1 2
P(X = x) 0.3 0.1 0.15 0.4 0.05
Solution:
E(x) = P1 * X1 + P2 * X2 + …Pn * Xn
= -2 * 0.3 + -1 * 0.1 + 0 * 0.15 + 1 * 0.4 + 2 * 0.05
= -0.2

OPERATIONS RESEARCH
Question 5:
A venture capital firm is determined based on the past experience that for each $100 invested in
a high technology startup company; a return of $400 is experienced 20% of time. A return of
$100 is experienced 40% of the time and zero (0) total loss is experienced 40% of the time.
What is the firm’s expected return based on this data?.
Solution:
S = {400, 100, 0}
E(x) = P1 * X1 + P2 * X2 + …Pn * Xn
= 400 * 0.2 + 100 * 0.4 + 0 * 0.4.
= 80 + 40 + 0
= $120
Question 6:
Given that an unbiased die was thrown 120 times and the recordings are as follows then
calculate the expected mean or mean score.
Score x 1 2 3 4 5 6
Frequency f 15 22 23 19 23 18 Total = 120
Solution:
E(x) = Σfx/Σf.
= (15 + 44 + 69 + 76 + 115 +108)/120.
= 3.558
Question 7:
The random variable X has Pdf P(X=x) for x = 1,2,3.
X 1 2 3
P(X = x) 0.1 0.6 0.3
Calculate:
a. E(3).
b. E(x).
c. E(5x).
d. E(5x + 3).
e. 5E(x) + 3.
f. E(x2).
g. E(4x2 - 3).
h. 4E(x2) – 3.

OPERATIONS RESEARCH
Solution:
X 1 2 3
5x 5 10 15
5x + 3 8 13 18
x2 1 4 9
4x2 – 3 1 13 33
P(X = x) 0.1 0.6 0.3
a. E(3) = ΣP(X = x).
all x
= Σ3P(X = x).
all x
= 3(0.1) + 3(0.6) + 3(0.3).
= 3.
b. E(x) =ΣxP(X = x).
all x
=1(0.1) +2(0.6) +3(0.3)
= 2.2
c. E(5x) = Σ5xP(X = x).
all x
= 5(0.1) + 10(0.6) +15(0.3)
= 11
d. E(5x + 3) = Σ(5x + 3)P(X = x).
all x
= 8(0.1) + 13(0.6) + 18(0.3)
= 14
e. 5E(x) + 3 = 5(2.2) + 3
= 14
f. E(x2) = Σ x2P(X = x).
all x
= 1(0.1) + 4(0.6) + 9(0.3)
= 5.2

OPERATIONS RESEARCH
g. E(4x2 - 3) = Σ(4x2 – 3)P(X = x).
all x
= 1(0.3) + 13(0.6) + 33(0.3)
= 17.8
h. 4E(x2) – 3 = 4(5.2) – 3
=20.8 – 3
= 17.8
VARIANCE:
The expected value of a random variable is a measure of central tendency i.e. what values are
mostly likely to occur while Variance is a measure of how far apart the possible values are spread
again weighted by their respective probabilities.
The formula for variance is given by:
Var (x) = E(x – μ)2 this can be reduced to
Var (x) = E(x2) - μ2
Question 8:
The random variable X has probability distribution shown below.
x 1 2 3 4 5
P(X =x) 0.1 0.3 0.2 0.3 0.1
Find:
i. μ = E(x).
ii. Var(x) using the formula E(x – μ)2
iii. E(x2)
iv. Var(x) using the formula E(x2) - μ2
Solution:
i. E(x) =μ = ΣxP(X = x).
all x
=1(0.1) + 2(0.3) + 3(0.2) + 4(0.3) + 5(0.1)
= 3
ii. Var (x) = E(x – μ)2
=Σ(x – 3)2P(X = x).
all x

OPERATIONS RESEARCH
X 1 2 3 4 5
(x – 3) -2 -1 0 1 2
(x – 3)2 4 1 0 1 4
P(X = x) 0.1 0.3 0.2 0.3 0.1
= 4(0.1) + 1(0.3) + 0(0.2) + 1(0.3) + 4(0.1)
= 1.4
iii. E(x2) = Σx2P(X = x).
all x
= 1(0.1) + 4(0.3) + 9(0.2) + 16(0.3) + 25(0.1)
= 10.4
iv. Var(x) = E(x2) - μ2
= 10.4 – 9
= 1.4
STANDARD DEVIATION:
Is the square root of its variance given by the following formula:
δ = √Var (x).
Question 9:
From the question given above find the standard deviation for part (iv).
δ = √Var (x).
δ = √1.4
= 1.183215957
= 1.18
CUMULATIVE DISTRIBUTION FUNCTION:
When we had a frequency distribution, the corresponding Cumulative frequencies were obtained
by summing all the frequencies up to a particular value.
In the same way if X is a discrete random variable, the corresponding Cumulative Probabilities
are obtained by summing all the probabilities up to a particular value.
If X is a discrete random variable with Pdf P(X = x) for x = x1, x2. …xn then the Cumulative
distribution function is given by:
F(t) = P(X = t)
= Σt P(X = x).
x = x1
The Cumulative Distribution is sometimes called Distribution function.

OPERATIONS RESEARCH
Question 10:
The probability distribution for the random variable X is given below then constructs the
Cumulative distribution table.
X 0 1 2 3 4 5 6
P(X =x) 0.03 0.04 0.06 0.12 0.4 0.15 0.2
Solution:
F(t) = Σt P(X = x).
x = x1
So
F(0) = P(X = 0) = 0.03
F(1) = P(X = 1) = 0.03 + 0.04 = 0.07
F(2) = P(X = 2) = 0.03 + 0.04 + 0.06 = 0.13 and so on.
The Cumulative Distribution table will be as follows:
X 0 1 2 3 4 5 6
F(x) 0.03 0.07 0.13 0.25 0.65 0.8 1
Question 11:
For a discrete random variable X the Cumulative distribution function F(x) is given below:
X 1 2 3 4 5
F(x) 0.2 0.32 0.67 0.9 1
Find:
a) P(x = 3).
b) P(x 2).
Solution:
a. F(3) = P(x = 3) = P(x = 1) + P(x = 2) + P(x = 3)
= 0.67
F(2) = P(x = 2) = P(x = 1) + P(x = 2)
= 0.32
Therefore P(x = 3) = 0.67 – 0.32
= 0.35
b. P(x 2) = 1 – P(x = 2)
= 1 – F(2)
= 1 – 0.32
= 0.68

OPERATIONS RESEARCH
PROBABILITY DISTRIBUTION (CONTINUOUS RANDOM VARIABLES):
A random variable X that can be equal to any number in an interval, which can be either finite
or infinite length, is called a Continuous Random Variable.
PROBABILITY DENSITY FUNCTIONS:
There are 2 essential properties of Pdf:
Because probabilities cannot be negative. The integral of a function must be non-negative
for all choices of interval [a, b] i.e. f(x) = 0 for all values in the sample space for the
random variable X.
Since the probability associated with the entire sample space is always 1. The integral of
f(x) of the entire sample space = 1.
Question 1:
A continuous random variable has Pdf f(x) where f(x) = kx, 0= x = 4.
i. Find the value of constant k.
ii. Sketch y = f(x).
iii. Find P(1 = X = 2½).
2½[⅛x]∂x.
Solution:
i. ∫ f(x) ∂x = 1.
all x
∫4 kx ∂x = 1.
0
[kx2/]4 = 1.
20
8k = 1
k = ⅛
ii. Sketch of y = f(x).
½ y = ⅛x
0 4
P(1= x = 2½) = ∫1
= [x2/16]1
2½
= 0.328

OPERATIONS RESEARCH
Question 2:
A continuous random variable has Pdf f(x) where
Kx 0= x = 4.
f(x)= k(4 – x) 2= x = 4
0 otherwise
a) Find the value of constant k.
b) Sketch y = f(x).
Solution:
D =∫a
b P ∂x + ∫a
b Q ∂x = 1.
2 kx ∂x + ∫2
=∫0
4 k(4 – x) ∂x = 1.
= [kx2/2]0
2 + [4xk - kx2/2]2
4 = 1.
= [4k/2] – [0] + {[16k - 16k/2] – [8k - 4k/2]} = 1.
[2k] + {[8k] – [6k]} = 1.
4k = 1
k = ¼
c. Sketch y = f(x).
X 0 1 2 3 4
Y 0 ¼ ½ ¾ 1
F(x) = kx.
X 2 3 4
Y ½ ¼ 0
F(x) = k(4 – x)
1
¾
½
¼
0 1 2 3 4

OPERATIONS RESEARCH
EXPECTED VALUE OR MEAN (CONTINUOUS RANDOM VARIABLE):
For a continuous random variable X defined within finite interval [a, b] with continuous Pdf f(x)
then the expected value or mean is given by:
1 6/7 x2 ∂x + ∫1
1 + 6/7[2/3x3 – x4/4]1
E(x) = ∫a
b x. f(x) ∂x
Question 3:
6/7x 0= x = 1.
f(x)= 6/7x(2 – x) 1= x = 2
0 otherwise
i. Find E(x).
ii. Find E(x2).
Solution:
D =∫a
b P ∂x + ∫a
b Q ∂x = 1.
E(x) = ∫a
b x. f(x) ∂x
E(x) = ∫0
2 6/7 x2(2 – x)∂x
= 6/7[x3/3]0
2
= 6/7[⅓] + 6/7{16/3 – 4 – (⅔ - ¼)}
= 6/7[5/4]
= 15/14
E(x2) = ∫a
b x2. f(x) ∂x
E(x2) = ∫0
1 6/7 x3 ∂x + ∫1
2 6/7 x3(2 – x)∂x
= 6/7[x4/4]0
1 + 6/7[x4/2 – x5/5]1
2
= 6/7[¼] + 6/7{8 - 32/5 – (½ - 1/5)}
= 6/7[31/20]
= 93/70

OPERATIONS RESEARCH
VARIANCE AND STANDARD DEVIATION:
The variance and Standard Deviation associated with a continuous random variable X on the
sample space [a, b] is given by:
4 ⅛x2 ∂x
= ⅛[x3/3]0
b x2. f(x) ∂x
=∫0
Var (x) = ∫a
b [x - E(x)]2 . f(x) ∂x or Var(x) = E(x2) - μ2 ∂x
and Standard deviation = √Var (x).
E(x) = μ.
Question 4:
A continuous random variable has Pdf f(x) where f(x) = ⅛x, 0= x= 4.
Find:
a) E(x).
b) E(x2).
c) Var (x).
d) The standard deviation of x.
e) Var(3x +2).
Solution:
i. E(x) = ∫a
b x. f(x) ∂x
E(x) = ∫0
4
= 8/3
ii. E(x2) = ∫a
4 ⅛x3 ∂x
= ⅛[x4/4]0
4
= ⅛(64)
= 8
iii. Var(x) = E(x2) - μ2 ∂x
= E(x2) - E2(x) ∂x
= 8 – (8/3)2
= 8/9

OPERATIONS RESEARCH
iv. Standard Deviation = d = √Var (x).
= √8/9
= 2√2/3
v. Var(3x + 2) = 9 Var(x) this has been obtained form the concept
Var (ax)= Var a2(x).
= 9 (8/9)
= 8
MODE:
The Mode is the value of X for which f(x) is greatest in the given range of X. It is usually to
draw a sketch of y = f(x) and this will give an idea of the location of the Mode.
For some Probability Density functions it is possible to determine the mode by finding the
maximum point of the curve y = f(x) from the relationship f1(x) = 0.
f1(x) = d/∂x * f(x).
Question 5:
A continuous random variable has Pdf f(x) where f(x) = 3/80(2 + x)(4 – x), 0= x= 4.
a) Sketch y = f(x).
b) Find the mode.
Solution:
X 0 1 2 3 4
Y 24/80 27/80 24/80 15/80 0
a)
27/80 Mode
24/80 f(x) = 3/80(2 + x)(4 – x)
15/80
0 1 2 3 4

OPERATIONS RESEARCH
b) The mode = f(x) = 3/80(2 + x)(4 – x)
= 3/80(8 + 2x – x2)
f1(x) = (2+ 2x)
f1(x) = 0.
0 = 2+ 2x
2x = 2
x = 1
MEDIAN:
The median splits the area under the curve y = f(x) into 2 halves so if the value of the Median is
m. Therefore the formula for the median is given by:
∫m f(x) ∂x = 0.5.
a
F(m) = 0.5
Question 6:
Find:
a. The median m.
Solution:
m = ∫a
m f(x) ∂x = 0.5.
F(m) = 0.5
f(x) = ⅛x ∂x
0.5 = m2/16
m2 = 8
m = 2.83
CUMULATIVE DISTRIBUTION FUNCTION: F(x)
When considering a frequency distribution the corresponding cumulative frequencies were
obtained by summing all the frequencies up to a particular value.
In the same way if X is a continuous random variable with Pdf f(x) defined for a=x=b then
the Cumulative Distribution Function is given by F(t):
F(t) = P(X = t) = ∫a
t f(x) ∂x

OPERATIONS RESEARCH
2 PROPERTIES OF CDF:
F(b) = ∫a
b f(x) ∂x = 1.
If f(x) is valid for -¥ = x = ¥ then F(t) = ∫-¥
t f(x) ∂x where the interval is taken
over all values of x = t.
The Cumulative distribution function is sometimes known as just as the distribution
function.
Question 6:
Find:
i. The Cumulative distribution function F(x).
ii. Sketch y = F(x).
iii. Find P(0.3 =x= 1.8).
Solution:
i. F(t) = ∫a
t f(x) ∂x
t ⅛x ∂x
= ⅛[x2/2]0
F(t) = ∫0
t
= t2/16
F (t) = t2/16 0=t=4
NB: (1) F(4) = 42/16 = 1
0 x = 0.
F(x) = x2/16 0= x = 4
1 x = 4
ii. Sketch y = F(x).
X 0 1 2 3 4
Y 0 1/16 1/2 9/16 1

OPERATIONS RESEARCH
1
F(x) = 1
9/16
1/2 F(x)= x2/16
1/16
0 1 2 3 4
iii. P(0.3 = x = 1.8) = F(1.8) – F(0.3)
F(1.8) = (1.8)2/16
= 0.2025
F(0.3) = (0.3)2/16
= 0.005625
Therefore P(0.3 = x = 1.8) = F(1.8) – F(0.3)
= 0.2025 – 0.005625
= 0.196875
= 0.197
Question 7:
x/3 0= x = 2.
f(x)= -2x/3 +2 2= x = 3
0 otherwise
a. Sketch y = f(x).
b. Find the Cumulative distribution function F(x).
c. Sketch y = F(x).
d. Find P(1 = X = 2.5)
e. Find the median m.

OPERATIONS RESEARCH
t x/3∂x
t
Solution:
i. Sketch y = f(x).
X 0 1 2
Y 0
X 2 3
Y 0

y = x/3
y =-2x/3 +2
0 1 2 3
ii. CDF = F(t) = ∫0
= [x2/6]0
t
= t2/6
F (t) = x2/6 0=x=2
NB: F (2) = 22/6 =
F(t) = F(2) + (Area under the curve y = -2x/3 +2 between 2 and t)
So
F(t) = F(2) + ∫2
t (-2x/3 +2) ∂x
= F(2) + [-x2/3 + 2x]2
= + {-t2/3 +2t – ( -4/3 + 4)}
= -t2/3 +2t – 2 2= t = 3
NB: F(2) = -9/3 + 6 – 2 = 1

OPERATIONS RESEARCH
Therefore CDF =
x2/6 0= x = 2.
f(x)= - x2/3 +2x -2 2= x = 3.
1 x = 3.
iii. Sketch of y = F(x).
y = 1
1
y = - x2/3 +2x -2
2/3
y = x2/6
1/3
0 1 2 3
iv. P(1 = X = 2.5) = F(2.5) – F(1) as 2.5 is in the range 2= x =3.
F(2.5) = - x2/3 +2x –2
F(2.5) = - (2.5)2/3 +2(2.5) –2
= 11/12
F(1) = x2/6 as 1 is in the range 0 = x = 2.
F(1) = x2/6
F(1) = 12/6
= 1/6
Therefore P(1 = X = 2.5) = F(2.5) – F(1)
= 11/12 - 1/6
= 0.75
v. m = ∫a
m f(x) ∂x = 0.5 where m is the median.
F(2) = so the median must lie in the range 0 = x = 2.
F(m) = m2/6
m2/6 = 0.5
m2 = 3.
m = 1.73

OPERATIONS RESEARCH
OBTAINING THE PDF FROM THE CDF:
The Probability Density Function can be obtained from the Cumulative Distribution function as
follows:
Now F(t) = ∫a
t f(x) ∂x a= t = b.
So
f(x) = d/∂x * F(x).
= F1(x).
NB: The gradient of the F(x) curve gives the value of f(x).
Question 8:
0 x = 0.
F(x)= x3/27 0= x = 3
1 x = 3.
Find the Pdf of X, f(x) and sketch y = f(x).
Solution:
a. f(x) = d/∂x * F(x).
= d/∂x(x3/27).
= 3x2/27
= x2/9
Therefore the Pdf is equal to:
x2/9 0=x=3
f(x) =
0 otherwise.
b. Sketch of y = f(x).
1 y = x2/9
0 1 2 3

OPERATIONS RESEARCH
Question 9:
A continuous random variable X takes values in the interval 0 to 3.
It is given that P(X x) = a + bx3, 0 = x = 3.
i. Find the values of the constants a and b.
ii. Find the Cumulative distribution function F(x).
iii. Find the Probability density function f(x).
iv. Show that E(x) = 2.25.
v. Find the Standard deviation.
Solution:
a. P(X x) = a + bx3, 0 = x = 3.
So P(X 0) = 1 and P(X 3) = 0.
i.e. a + b(0) = 1 and a + b(27) = 0
Therefore a = 1 and 1 + 27b = 0.
B = -1/27.
So P(X x) = 1 - x3/27, 0 = x = 3.
b. Now P(X = x) = x3/27 (CDF)
X3/27 0=x=3
F(x) =
1 x 3.
c. f(x) = d/∂x * F(x).
= d/∂x(x3/27).
= 3x2/27
= x2/9
b x. f(x) ∂x
=∫0
d. E(x) = ∫a
3 x. x2/9∂x
3 x3/27∂x
=∫0
3
= [x4/36]0
= 2.25

OPERATIONS RESEARCH
3 x4/9∂x – 2.252.
1∂X [(XY + Y2/2 - XY2)]0
½∂X[(XY + Y2/2 - XY2)]0
½∂X[(¼X + 1/32 - 1/16X)]
e. Var(x) = ∫a
b [x - E(x)]2 . f(x) ∂x = ∫a
b x2.f(x)∂x - E2(X)
=∫0
=[x5/45]0
3 - 5.0625.
= 0.3375
f. δ = √ Var (x).
= √ 0.3375
= 0.581
RELATIONSHIPS AMONG PROBABILITY DISTRIBUTIONS:
JOINT PROBABILITY DISTRIBUTION:
Question 1:
2(X + Y - 2XY)0= X=1, 0= Y=1
Given f(X, Y) = 0 Otherwise
i. Show that this is a PDF.
ii. Find P(0 = X =½), (0 = Y =¼).
iii. Find CDF.
Solution:
a) =∫a
b∂X∫a
b∂Y
1∂X∫0
=∫0
1∂Y [2(X + Y - 2XY)]
= 2∫0
1
1∂X [(X + ½ - X)]
= 2∫0
= 2[(X2/2 + ½X - X2/2)]0
1
= 2[½ + ½ - ½]
= 2 * ½
= 1
b) =∫a
b∂X∫a
b∂Y
½∂X∫0
=∫0
¼∂Y [2(X + Y - 2XY)]
= 2∫0
¼
= 2∫0
½∂X[(3/16X + 1/32)]
= 2∫0
= 2[(3/32X2 + 1/32X)] 0
½

OPERATIONS RESEARCH
x∂U[(UV + V2/2 - UV2)]0
x∂U[(UY + Y2/2 - UY2)]
= 2[(3/128 + 1/64)]
= 2 * 5/128
= 5/64
c) F(u, v) =∫a
b∂u∫a
b∂v
x∂U∫0
=∫0
y∂V[2(U + V – 2UV)]
= 2∫0
y
= 2∫0
= 2[(YU2/2 + UY2/2 - U2X2/2)]0
X
= 2[(X2Y/2 + XY2/2 - X2X2/2)]
= X2Y + XY2 - X2X2
NB: Given CDF to get PDF we differentiate the function.
Given PDF to get CDF we integrate the function.
Given a PDF to prove that it is a PDF integrate until the answer is 1.
Given PDF the marginal distribution is found by integrating.
Given CDF the marginal distribution is found by differentiating.
Question 1:
Given the CDF = x2 + 3xyz + z.
Find the corresponding PDF.
Solution:
To get the corresponding PDF we differentiate with respect to x, y, z.
f(x, y, z) =x2 + 3xyz + z
= d/∂x(2x + 3yz).
= d/∂y(3z).
= d/∂z (3).
= 3

OPERATIONS RESEARCH
UNIT 6 DECISION THEORY
HOURS: 20
DECISION THEORY:
There are four types of criteria that we will look at.
Expected Value (Realist)
Compute the expected value under each action and then pick the action with the largest
expected value. This is the only method of the four that incorporates the probabilities of the
states of nature. The expected value criterion is also called the Bayesian principle.
Maximax (Optimist)
The maximax looks at the best that could happen under each action and then chooses the
action with the largest value. They assume that they will get the most possible and then they
take the action with the best best case scenario. The maximum of the maximums or the
best of the best. This is the lotto player; they see large payoffs and ignore the probabilities.
Maximin (Pessimist)
The maximin person looks at the worst that could happen under each action and then
choose the action with the largest payoff. They assume that the worst that can happen will,
and then they take the action with the best worst case scenario. The maximum of the
minimums or the best of the worst. This is the person who puts their money into a
savings account because they could lose money at the stock market.
Minimax (Opportunist)
Minimax decision making is based on opportunistic loss. They are the kind that look back
after the state of nature has occurred and say Now that I know what happened, if I had
only picked this other action instead of the one I actually did, I could have done better. So,
to make their decision (before the event occurs), they create an opportunistic loss (or regret)
table. Then they take the minimum of the maximum. That sounds backwards, but
remember, this is a loss table. This similar to the maximin principle in theory; they want the
best of the worst losses.
Example: A bicycle shop
Zed and Adrian and run a small bicycle shop called Z to A Bicycles. They must order bicycles for
the coming season. Orders for the bicycles must be placed in quantities of twenty (20). The cost per
bicycle is $70 if they order 20, $67 if they order 40, $65 if they order 60, and $64 if they order 80.
The bicycles will be sold for $100 each. Any bicycles left over at the end of the season can be sold
(for certain) at $45 each. If Zed and Adrian run out of bicycles during the season, then they will
suffer a loss of goodwill among their customers. They estimate this goodwill loss to be $5 per

OPERATIONS RESEARCH
customer who was unable to buy a bicycle. Zed and Adrian estimate that the demand for bicycles
this season will be 10, 30, 50, or 70 bicycles with probabilities of 0.2, 0.4, 0.3, and 0.1 respectively.
Actions
There are four actions available to Zed and Adrian. They have to decide which of the actions is the
best one under each criteria.
1. Buy 20 bicycles
2. Buy 40 bicycles
3. Buy 60 bicycles
4. Buy 80 bicycles
Zed and Adrian have control over which action they choose. That is the whole point of decision
theory - deciding which action to take.
States of Nature
There are four possible states of nature. A state of nature is an outcome.
1. The demand is 10 bicycles
Zed and Adrian have no control over which state of nature will occur. They can only plan and make
the best decision based on the appropriate decision criteria.
Payoff Table
After deciding on each action and state of nature, create a payoff table. The numbers in parentheses
for each state of nature represent the probability of that state occurring.
Action
State of Nature Buy 20 Buy 40 Buy 60 Buy 80
Demand 10
(0.2)
50 -330 -650 -970
Demand 30
(0.4)
550 770 450 130
Demand 50
(0.3)
450 1270 1550 1230
Demand 70
(0.1)
350 1170 2050 2330

OPERATIONS RESEARCH
Ok, the question on your mind is probably How the [expletive deleted] did you come up with those
numbers?. Let's take a look at a couple of examples.
Demand is 50, buy 60:
They bought 60 at $65 each for $3900. That is -$3900 since that is money they spent. Now,
they sell 50 bicycles at $100 each for $5000. They had 10 bicycles left over at the end of the
season, and they sold those at $45 each of $450. That makes $5000 + 450 - 3900 = $1550.
Demand is 70, buy 40:
They bought 40 at $67 each for $2680. That is a negative $2680 since that is money they
spent. Now, they sell 40 bicycles (that's all they had) at $100 each for $4000. The other 30
customers that wanted a bicycle, but couldn't get one, left mad and Zed and Adrian lost $5
in goodwill for each of them. That's 30 customers at -$5 each or -$150. That makes $4000 -
2680 - 150 = $1170.
Opportunistic Loss Table
The opportunistic loss (regret) table is calculated from the payoff table. It is only needed for the
minimax criteria, but let's go ahead and calculate it now while we're thinking about it.
The maximum payoffs under each state of nature are shown in bold in the payoff table above. For
example, the best that Zed and Adrian could do if the demand was 30 bicycles is to make $770.
Each element in the opportunistic loss table is found taking each state of nature, one at a time, and
subtracting each payoff from the largest payoff for that state of nature. In the way we have the table
written above, we would subtract each number in the row from the largest number in the row.
Action
State of Nature Buy 20 Buy 40 Buy 60 Buy 80
Demand 10 0 380 700 1020
Demand 30 220 0 320 640
Demand 50 1100 280 0 320
Demand 70 1980 1160 280 0
Remember that the numbers in this table are losses and so the smaller the number, the better.
Expected Value Criterion
Compute the expected value for each action.
For each action, do the following: Multiply the payoff by the probability of that payoff occurring.
Then add those values together. Matrix multiplication works really well for this as it multiplied pairs
of numbers together and adds them. If you place the probabilities into a 1x4 matrix and use the 4x4
matrix shown above, then you can multiply the matrices to get a 1x4 matrix with the expected value
for each action.

OPERATIONS RESEARCH
Here is an example of the Buy 60 action if you wish to do it by hand.
0.2(-650) + 0.4(450) + 0.3(1550) + 0.1(2050) = 720
The expected values for buying 20, 40, 60, and 80 bicycles are $400, 740, 720, and 460 respectively.
Since the best that you could expect to do is $740, you would buy 40 bicycles.
Maximax Criterion
The maximax criterion is much easier to do than the expected value. You simply look at the best you
could do under each action (the largest number in each column). You then take the best (largest) of
these.
The largest payoff if you buy 20, 40, 60, and 80 bicycles are $550, 1270, 2050, and 2330 respectively.
Since the largest of those is $2330, you would buy 80 bicycles.
Maximin Criterion
The maximin criterion is as easy to do as the maximax. Except instead of taking the largest number
under each action, you take the smallest payoff under each action (smallest number in each column).
You then take the best (largest of these).
The smallest payoff if you buy 20, 40, 60, and 80 bicycles are $50, -330, -650, and -970 respectively.
Since the largest of those is $50, you would buy 20 bicycles.
Minimax Criterion
Be sure to use the opportunistic loss (regret) table for the minimax criterion. You take the largest
loss under each action (largest number in each column). You then take the smallest of these (it is
loss, afterall).
The largest losses if you buy 20, 40, 60, and 80 bicycles are $1980, 1160, 700, and 1020 respectively.
Since the smallest of those is $700, you would buy 60 bicycles.
Putting it all together.
Here is a table that summarizes each criteria and the best decision.
Action
Criterion Buy 20 Buy 40 Buy 60 Buy 80 Best Action
Expected Value 400 740 720 460 Buy 40
Maximax 550 1270 2050 2330 Buy 80
Maximin 50 -330 -650 -970 Buy 20
Minimax 1980 1160 700 1020 Buy 60

OPERATIONS RESEARCH
Practice Problem
Since there aren't any problems of this kind in the text, work this problem out, and then you can
check your answer with the instructor.
Finicky's Jewelers sells watches for $50 each. During the next month, they estimate that they will sell
15, 25, 35, or 45 watches with respective probabilities of 0.35, 0.25, 0.20, and ... (figure it out). They
can only buy watches in lots of ten from their dealer. 10, 20, 30, 40, and 50 watches cost $40, 39, 37,
36, and 34 per watch respectively. Every month, Finicky's has a clearance sale and will get rid of any
unsold watches for $24 (watches are only in style for a month and so they have to buy the latest
model each month). Any customer that comes in during the month to buy a watch, but is unable to,
costs Finicky's $6 in lost goodwill.
Find the best action under each of the four decision criteria.
DECISION UNDER RISK:
A. THE EXPECTED VALUE CRITERION:
Muchadura buys and sell tomatoes at $3 and $8 a case respectively. Because tomatoes are perishable,
they must be bought a day and sold after that, they become valuable.
A 90 days observation of a business reveals the following information in cases.
Daily Sales Number Of Days Sold Probability Of Demand
10 cases 18 (18 / 90) = 0.2
11 cases 36 (36 / 90) = 0.4
12 cases 27 (27 / 90) = 0.3
13 cases 9 (9 / 90) = 0.1
Total number of days 90 = 1
The Probability of demand = Number of days sold / Total number of days.
There are only 4 sales volumes but their sequence in UNKNOWN. The problem is how many cases
must Muchadura stock for selling the following day. If he stocks more or less than the demand on
any day he will suffer some losses.
Example:
If he stocks 13 cases on a particular day and the customer demand on that particular day is 10 cases,
then he gets a profit of 3 cases.
Let buying price = $3.
Let Selling price = $8.

OPERATIONS RESEARCH
NB: This method if getting Maximum profit = Selling Price – Buying Price per case = $5. Then 10 *
5 = $50 and perishable ones = 3 * 3 = $9 and then the Maximum profit = $50 - $9 = $41.
For the 3 unsold cases he suffer a loss for each equivalent to the Buying price. All such losses are
calculated in a conditional profit table. I.e. a table of profit on a given supply and demand levels as
illustrated below.
STOCK ACTION EXPECTED PROFIT
Possible Probability
Demand of Demand Profits
10 11 12 13 10 11 12 13
10 50 47 44 41 0.2 10 9.4 8.8 8.2
11 50 55 52 49 0.4 20 22 20.8 19.6
12 50 55 60 57 0.3 15 16.5 18 17.1
13 50 55 60 65 0.1 5 5.5 6 6.5
Total 1 50 53.4 53.6 51.4
Muchadura must stock 12 cases which give the highest Expected profit of $53.60 on a daily bases to
get the maximum profit over a long time.
B. EXPECTED PROFIT WITH PERFECT INFORMATION:
Suppose Muchadura had perfect information i.e. complete and accurate information about the
future. Sales demand would still vary from 10 to 13 cases in their respective probabilities. But
Muchadura would not know in advance how many cases would be needed each day.
The table below shows the conditional profits value in such situation.
10 11 12 13 10 11 12 13
10 50 0.2 10
11 55 0.4 22
12 60 0.3 18
13 65 0.1 6.5
Total 1 10 22 18 6.5
Maximum possible profit = 10 + 22 + 18 + 6.5.
= $56.50

OPERATIONS RESEARCH
C. MINIMISING EXPECTED LOSS:
There are two types of Losses:
Overstocking
Under stocking.
OVERSTOCKING:
Means you will loss the buying amount.
UNDER STOCKING:
It means you will loss the profit.
STOCK ACTION EXPECTED LOSS
Demand of Demand Loss
10 11 12 13 10 11 12 13
10 0 3 6 9 0.2 0 0.6 1.2 1.8
11 5 0 3 6 0.4 2 0 1.2 2.4
12 10 5 0 3 0.3 3 1.5 0 0.9
13 15 10 5 0 0.1 1.5 1 0.5 0
Total 1 6.5 3.1 2.9 5.1
The Expected Minimum Loss is $2.9. The values above the zero diagonal are due to overstocking
and those below are due to under stocking.
D. THE EXPECTED VALUE OF PERFECT INFORMATION:
The expected value of perfect information is calculated as:
The expected Profit with perfect information – The expected value Criterion.
= 56.50 – 53.60
= 2.9
In general the expected value of perfect information = Minimum Expected Loss.

OPERATIONS RESEARCH
E. ITEMS WITH A SALVAGE VALUE:
Main items do not become completely worthless after their prime deaths. They will have reduced
values (Salvage values). The Salvage values must be considered when computing conditional profits
or losses.
Consider the previous case when the cost price was $3 and selling price was $8. Any unsold cases
will be disposed / salvage of at $4 the profit will be $4 - $3 = $1.
10 11 12 13 10 11 12 13
10 50 51 52 53 0.2 10 10.2 10.4 10.6
11 50 55 56 57 0.4 20 22 22.4 22.8
12 50 55 60 61 0.3 15 16.5 18 18.3
13 50 55 60 65 0.1 5 5.5 6 6.8
Total 1 50 54.2 56.8 58.2
Muchadura must stock 13 cases, which give the maximum profit of $58.20.
DECISION MAKING UNDER UNCERTAINITY:
Under conditions of uncertainty, only payoffs are known and nothing is known about the
likelihood of each state of nature.
Different persons have suggested several decisions rules for making decisions under such
situations:
A. MAXIMIN (PESSIMISTIC):
It is based upon the consecutive approach to assume that the worst possible is going to
happen. The decision maker considers each alternative and locates the minimum payoff for
each and then selects that alternative which maximizes the minimum payoffs.
STEPS INVOLVED:
i. Determine the minimum assured payoffs for each alternative.

OPERATIONS RESEARCH
ii. Choose that alternative which corresponds to the Maximum of the above minimum
payoff.
B. MAXIMAX CRITERIA (OPTIMISTIC):
Is based upon extreme optimistic, the decision maker selects that particular strategy which
corresponds to the maximum of the maximum payoffs of each strategy.
STEPS INVOLVED:
i. Determine the maximum possible payoff for each alternative.
ii. Select that alternative which corresponds to the maximum of the above maximum
payoff.
C. HURWICZ:
In order to overcome the disadvantage of extreme pessimistic (Maximin) and extreme
Optimism (Maximax) criteria. Hurwicz introduced the concept of co-efficient of optimism
or pessimism as a.
Therefore the Hurwicz concept is given by:
a Maximin qj {V (ai, qj)} + (1 - a) Maximax qj {V(ai, qj)}
D. LAPLACE CRITERION:
This criterion is based on what is known as the principal if insufficient reason. Since the
probabilities associated with the occurrences of q1, q2….qn are unknown, we do not have
enough information to conclude that these probabilities will be different.
For if this is not the case, we should be able to determine these probabilities and the
situation will no longer be a decision under uncertainty. Thus because of insufficient reason
to believe otherwise; the states of q1, q2….qn are equally likely to occur.
The Laplace principal assumes that of q1, q2….qn are likely to occur therefore the
probability = 1/n where n = number of occurrences.

OPERATIONS RESEARCH
E. MAXIMIN REGRET CRITERION (SALVAGE):
This is a less conservative criterion. In this case a new loss matrix is created such that V(Ai,
Qj) is replaced by R(Ai, Qj) which is defined by:
Max {V (Ai, Qj) – V(Ai, Qj)} if V is profit
R(Ai, Qj)=
V(Ai, Qj) – Min {V(Ai, Qj)} if V is loss
R(Ai, Qj) is the difference between the best choice in column Qj and the values of V(Ai, Qj)
in the same column.
R(Ai, Qj) is a regret matrix.
Question 1:
Given the following data:
Customers Category
Q1 Q2 Q3 Q4
Supplier A1 5 10 18 25
Level A2 8 7 8 23
A3 21 18 12 21
A4 30 22 9 25
Using the above data:
Find
i. Maximin Criterion.
ii. Maximax Criterion.
iii. Hurwicz Criterion.
iv. Laplace Criterion.
v. (Salvage) Maximin Regret Criterion.
Solution:
a) Maximin Criterion.
Customers Category
Q1 Q2 Q3 Q4 Maximin
Supplier A1 5 10 18 25 5
Level A2 8 7 8 23 7
A3 21 18 12 21 12
A4 30 22 9 25 9
Maximin = 12
The optimal decision is to take supply level A3

OPERATIONS RESEARCH
b) Maximax Criterion.
Customers Category
Q1 Q2 Q3 Q4 Maximax
Supplier A1 5 10 18 25 25
Level A2 8 7 8 23 23
A3 21 18 12 21 21
A4 30 22 9 25 30
Maximax = 30
c) Minimax Criterion.
Customers Category
Q1 Q2 Q3 Q4 Minimax
Supplier A1 5 10 18 25 25
Level A2 8 7 8 23 23
A3 21 18 12 21 21
A4 30 22 19 25 30
Minimax = 21
d) Hurwicz Criterion.
Customers Category
Q1 Q2 Q3 Q4
Level A2 8 7 8 23
A3 21 18 12 21
A4 30 22 19 25
Hurwicz: = a Maximin qj {V (ai, qj)} + (1 - a) Maximax qj {V(ai, qj)}
Let assume that α = ½.
Maximin Maximax a Maximin + (1 - a) Maximax
5 25 15
7 23 15
12 21 16.5
9 30 19.5

OPERATIONS RESEARCH
e) Laplace Criterion.
Customers Category
Q1 Q2 Q3 Q4
Level A2 8 7 8 23
A3 21 18 12 21
A4 30 22 19 25
Laplace: = 1/n
Expected costs for different actions A1, A2, A3 and A4 are:
E(A1) = ¼(5 + 10 + 18 + 25) = 14.5
E(A2) = ¼(8 + 7 + 8 + 23) = 11.5
E(A3) = ¼(21 + 18 + 12 + 21) = 18
E(A1) = ¼(30 + 22 + 19 + 25) = 24
f) Savage Minimax Regret Criterion.
Customers Category
Q1 Q2 Q3 Q4
Level A2 8 7 8 23
A3 21 18 12 21
A4 30 22 19 25
Q1 Q2 Q3 Q4 Max{ R(Ai, Qj)}
A1 0 3 10 4 10
A2 3 0 0 2 3 Minimax
A3 16 11 4 0 16
A4 25 15 11 4 25
The optimal decision is to take supply level A2 which gives Minimax.

OPERATIONS RESEARCH
Question 2:
A firm has to decide on its advertising campaign. It has a choice between Tv, Newspaper,
Poster and Radio advertising. The return on the advertising medium is measured by the
number of potential customers who have the opportunity to see each medium. This will
depend on the type of the weather. The figures in the table are the numbers of potential
customers in thousands. The firm has funds for using only to one medium.
Poor Moderate Good Excellent
Tv 200 190 170 130
Newspaper 180 160 150 130
Poster 110 140 140 190
Radio 210 190 160 110
Decide which medium the firm should use based on the following information:
i. Maximin Criterion.
ii. Maximax Criterion.
iii. Hurwicz Criterion.
iv. Laplace Criterion.
v. (Salvage) Maximin Regret Criterion.
Solution:
a) Maximin Criterion.
Poor Mod Good Excellent Maximin
Tv 200 190 170 130 130
Newspaper 180 160 150 130 130
Poster 110 140 140 190 110
Radio 210 190 160 110 110
Maximin = 130
The best medium to use for advertising is Tv or Newspaper since they have the highest
number of (possible) potential customers of 130

OPERATIONS RESEARCH
b) Maximax Criterion.
Poor Mod Good Excellent Maximax
Tv 200 190 170 130 200
Newspaper 180 160 150 130 180
Poster 110 140 140 190 190
Radio 210 190 160 110 210
Maximax = 210
The best medium to use for advertising is the Radio since it has the highest number of
(possible) potential customers of 210.
c) Minimax Criterion.
Poor Mod Good Excellent Minimax
Tv 200 190 170 130 200
Newspaper 180 160 150 130 180
Poster 110 140 140 190 190
Radio 210 190 160 110 210
Minimax = 180
The best medium to use for advertising is the Newspaper with 180 potential customers.
d) Hurwicz Criterion.
Poor Mod Good Excellent
Tv 200 190 170 130
Newspaper 180 160 150 130
Poster 110 140 140 190
Radio 210 190 160 110

OPERATIONS RESEARCH
Hurwicz: = a Maximin qj {V (ai, qj)} + (1 - a) Maximax qj {V(ai, qj)}
Let assume that α = 0.7
Maximin Maximax a Maximin + (1 - a) Maximax
130 200 151
130 180 145
110 190 109.7
110 210 140
The best medium to use for advertising is the Tv since it has the highest number of
(possible) potential customers of 151
g) Laplace Criterion.
Tv 200 190 170 130
Newspaper 180 160 150 130
Poster 110 140 140 190
Radio 210 190 160 110
Laplace: = 1/n
Expected costs for different actions A1, A2, A3 and A4 are:
E(Tv) = ¼(200 + 190 + 170 + 130) = 172 500
E(Newspaper) = ¼(180 +160 + 150 + 130) = 155 000
E(Poster) = ¼(110 + 140 + 140 + 190) = 145 000
E(Radio) = ¼(210 + 190 + 160 + 110) = 167 500
The best medium to use for advertising is Tv since it has the highest number of (possible)
potential customers who amount to 172 500

OPERATIONS RESEARCH
h) Savage Minimax Regret Criterion.
Tv 200 190 170 130
Newspaper 180 160 150 130
Poster 110 140 140 190
Radio 210 190 160 110
Poor Mod Good Excellent Max {R(Ai, Qj)}
Tv 90 50 30 20 90
Newspaper 70 20 10 20 70 Minimax
Poster 0 0 0 80 80
Radio 100 50 20 0 100
The best medium to use for advertising is Newspaper, which minimize maximum view ship.

OPERATIONS RESEARCH
UNIT 7: THEORY OF GAMES:
HOURS: 20
CHARACTERISTICS OF A COMPETITIVE GAME

OPERATIONS RESEARCH

OPERATIONS RESEARCH
ALTERNATIVELY
GAME THEORY:
It deals with Decision under Uncertainty which involves 2 or more intelligent opponents in which
each opponent aim to optimize his / her own decision at the expense of the other opponent.
Typical examples include launching an advertisement campaign. (2) Competing for products or
planning war techniques for opposing enemy.
In game theory an opponent is referred to as a Player. Each player has the number of choices,
limited or unlimited called Strategies. The outcomes payoffs of a game are summarized as functions
of different strategies for each player.
A game with 2 players were a gain of one player is equal to the loss of another player is called a Two
Person Zero Sum game.
To illustrate two person zero sum game considers a coin-matching situation in which each of the
players A and B select a Head H or Tail T.
If the outcome match Head or Tail Player A wins $1 from Player B otherwise Player A loses $1 to
Player B.
In this game each player has 2 strategies Head or Tail which yields the following 2*2 game matrix
expressed in terms of payoff to Player A.
Player B
H T
1 -1
-1 1
Player A H
T
OPTIMUM SOLUTION OF TWO PERSON ZERO SUM GAME:
Is obtained by using minimax Maximin criterion according to which Player A (whose strategies
represents rows) select a strategy (mixed or Pure) which maximize his minimum gains, the
minimum being taken over all the strategies of Player B.
In similar way Player B selects his strategy that minimize his maximum loses.

OPERATIONS RESEARCH
1 VALUE OF THE GAME:
Is the maximum guaranteed gain to Player A or the minimum possible loss to Player B denoted
by V.
When Maximin possible value is equal to Minimax value of the corresponding pure strategies,
the game is said to have the Optimum Strategies and has a Saddle point.
Player B
1 2 3 4
Maximin
8 2 9 5
6 7 18
7 3 -4 10
Player A 1 2
2 5
5
3 -4
Minimax 8 5 9 18
The game has a Saddle point which exists at row 2 column 2 and the value of the game = 5.
Player B
1 2 3
Maximin
1 3 6
2 1 3
6 2 1
Player A 1 1
2 1
3 1
Minimax 6 3 6
NB: The optimum solution which shows that Maximin Minimax means it is a mixed strategy i.e.
there is no agreement (poor strategy). Below and above show that there is no Saddle point which
means Maximin Minimax.
Player B
1 2 3 4
Maximin
5 -10 9 0
6 7 8 1
8 7 15 2
3 4 -1 4
Player A 1 -10
2 1
3 2
4 -1
Minimax 8 7 15 4

OPERATIONS RESEARCH
2 6
-2 -1
2 4 5
10 7 q
4 p 8
l 6 2
-1 l -7
-2 4 l
Consider the following game G.
Player B
B1 B2 Maximin
Player A A1 2
A2 -1
Minimax 2 6
Maximin = Minimax which is a condition strictly determinable game hence the game is strictly
determinable whatever l maybe. The value of the game = 2 with the best strategy for Player A from
A1 and the best strategy for Player B from B1.
Question 1:
Find the range of values of P and Q, which will render the entry (2,2) a Saddle point of the game.
Player B
B1 B2 B3
Maximin
Player A A1 2
A2 7
A3 4
Minimax 10 7 8
Ignoring values of P and Q determine the Maximin and Minimax value of the payoff matrix. The
Maximin =f and Minimax = f thus there exists a Saddle point at position (2,2). This impose a
condition on P as P= f and on Q as Q = f.
Hence the required range for P and Q is = 7= Q, P= 7.
Question 2:
For what values of l is the game with the following payoff. Payoff is strictly determinable.
Player B
B1 B2 B3
Maximin
Player A A1 2
A2 -7
A3 -2
Minimax -1 6 2
This shows that the value of the game (V) lies between –1 and 2.
That is –1 = V = 2. For a strictly determinable we have –1 = l = 2.

OPERATIONS RESEARCH
GRAPHICAL SOLUTION OF 2 BY N M BY 2 GAMES:
The graphical solution of 2 by N and M by 2 is only applicable to games in which at least on of the
Players have 2 strategies only.
Consider the following 2 by N games.
Player B
Y1 Y2 Y3
Player A X1
X2
The expected payoff of the corresponding to the pure strategies of B are given below:
This shows that A’s average payoff varies linearly with X1.
According to the Minimax criterion for mixed strategies games Player A should select the value V
that minimize his maximum expected playoffs. This may be done by plotting a straight line as
function of X1.
Question 3:
Consider the following 2 by 4 games.
Player B
1 2 3 4
Player A 1
2
Solution:
A11 A12 A1n
A21 A22 A2n
B’s Pure Strategies A’s Expected Payoff
1 (A11 –A12) X1 + A21
2 (A21 – A22) X1 + A22
N (A2n –A2n) X1 + A2n
2 2 3 -1
4 3 2 6
1 (2 – 4) X1 + 4
2 (2 – 3) X1 + 3
3 (3 – 2) X1 + 2
4 (-1 –7) X1 + 6

OPERATIONS RESEARCH
6 6
5 5
4 4
3 3
2 2
1 1
X1 0 1/2 0 X1
- 1 -1
-2 -2
This is a point of integration of any 2 of the lines or more i.e. lines 2, 3 and 4.
A’s optimum strategy is X1 = ½ and X2 = 5/2 and the value of the game is obtained by substituting
X1 in the equations of any of the lines passing the intersection of the point.
The value of the game = 5/2.
Question 4:
Consider the following 2 by 3 game, find B’s pure strategies and A’s expected payoff using the
graphically method.
Player B
1 2 3
Player A 1
2
Solution:
1 - 2X1 + 4
2 -X1 + 3
3 X1 + 2
4 - 7X1 + 6
1 3 11
8 5 2
1 - 7X1 + 8
2 - 2X1 + 5
3 9X1 + 2

OPERATIONS RESEARCH
11 11
10 10
8 9
8 8
7 7
6 6
5 5
4 H 4
3 3
2 2
1 1
X1 0 0 X1
- 1 Maximin -1
-2 -2
Since Player A wishes to Maximize his Minimum expected payoff we consider the highest point of
intersection on the lower elevation of A’s expected payoff equations.
This point it represents the maximum value of the game for A.
Lines 2 and 3 passes through it defines the relative moves B2 and B3 it along B needs to play.
The solution of the original 2 * 3 game reduces to a 2 by 2 payoff matrix.
Player B
B1 B2
Player A A1 X1+ X2 = 1
3 11
A2 Y1 + Y2 = 1
5 2
Using the method of solution for a 2 * 2 games the optimum strategy can be easily obtained by:
X1 = 3 – 11 / (3 + 2) – (5 + 11) = 8 / 11
X2 = 1 - 8 / 11 = 3 / 11
The value of the game V = (2 * 3) – (5 * 11) / (3 + 2) – (5 + 11)
= 49 / 11

OPERATIONS RESEARCH
Question 4:
Consider the following 2 by 4 game, find B’s pure strategies and A’s expected payoff using the
graphically method.
Player B
1 2 3 4
Player A 1
2
1 3 -3 7
2 5 4 -6
1 - X1 + 2
2 - 2X1 + 5
3 - 7X1 + 4
4 13X1- 6
Solution:
11 11
10 10
9 9
8 8
7 7
6 6
5 5
4 4
3 3
2 2
1 1
X1 0 0 X1
-1 H -1
-2 -2
-3 Maximin -3
-4 -4
-5 -5
-6 -6

OPERATIONS RESEARCH
Player B
B1 B2
-3 7
A2 Y1 + Y2 = 1
4 -6
X1 = -3 – 7 / (-3 + -6) – (4 + 7) = 11 / 20
X2 = 1 - 11 / 20 = 9 / 20
Y1 = -6 – 7 / (-3 + -6) – (4 + 7) = 13 / 20
Y2 = 1 - 13 / 20 = 7 / 20
The value of the game V = (-2 * -6) – (7 * 4) / (3 + 2) – (5 + 11)
= 40 / 20 = 2
MINIMAX PROBLEMS:
Obtain the optimum strategy for both persons and the value of the game for zero person sum game
whose payoff matrix is as follows:
Player B
1 2
Player A 1
2
3
4
5
6
1 -3
3 5
-1 6
4 1
2 2
-5 0

OPERATIONS RESEARCH
B’s Expected Payoff
1 - 4Y1 - 3
2 - 2Y1 + 5
3 - 6Y1 + 6
4 5Y1+ 1
5 2Y1+ 2
6 -5Y1+ 0
3 5
4 1
Solution:
A’s Pure Strategies
11 11
10 10
9 Minimax 9
8 8
7 H 7
6 6
5 5
4 4
3 3
2 2
1 1
Y1 0 0 Y1
- 1 -1
-2 -2
-3 -3
-4 -4
-5 -5
-6 -6
-7 -7
Player B
B1 B2
A2 Y1 + Y2 = 1
X2 = 3 – 5 / (3 + 1) – (4 + 5) = 2 / 5

OPERATIONS RESEARCH
X1 = 1 - 2 / 5 = 3 / 5
Y1 = 1 – 5 / (3 + 1) – (4 + 5) = 4 / 5
Y2 = 1 - 4 / 5 = 1 / 5
The value of the game V = (1 * 3) – (4 * 5) / (3 + 1) – (4 + 5)
= 17 / 5
GENERAL SOLUTION OF M * N RECTANGULAR GAMES:
In a rectangular game with M * N payoff matrix if there does exists any saddle point and is not
possible to reduce the size of the game to 2 by 2 payoff matrix.
The following methods are generally used to solve the game:
Linear programming method.
Iterative Method.
Question 5:
Solve the following game; consider the game problem in which B’s linear programming problem.
Player B
1 2 3
Maximin
8 4 2
2 8 4
1 2 8
Player A 1 2
2 2
3 1
Minimax 8 8 8
Player B
Y1 Y2 Y3
Player A X1
X2
X3
8 4 2
2 8 4
2 2 8
B’s linear programming problem is as follows:
Maximize Z = Y1 + Y2 + Y3
Subject to:
8Y1 + 4Y2 + 2Y3 = 1 {1}
2Y1 + 8Y2 + 4Y3 = 1 {2}
Y1 + 2Y2 + 8Y3 = 1 {3}
Y1, Y2, Y3 = 0

OPERATIONS RESEARCH
TABLEAU 1:
Y1 Y2 Y3 S1 S2 S3 Solution
S1 8
4 2 1 0 0 1
S2 8 4 0 1 0 1
2 S1 2 8 0 0 1 1
3 Z 1 1 1 0 0 0 0
PC
TABLEAU 2:
Y1 0.5 0.25 0.125 0 0 0.125
S2 0 7 3.5 -0.25 1 0 0.75
S3 0 1.5 7.75 -0.125 0 1 0.875
Z 0 0.5 0.75 -0.125 0 0 -0.125
PC
1
TABLEAU 3:
Y1 0 0 0.107 -0.072 0 0.072
1 Y0 2 1
0.5 0.036 0.143 0 0.107
S0 0 7 -0.179 -0.215 1 0.715
3 Z 0 0 0.5 -0.143 -0.072 0 -0.1785

OPERATIONS RESEARCH
TABLEAU 3:
Y1 0 0 0.107 -0.072 0 0.072
1 Y0 1 0 0.049 0.159 -0.072 0.057
2 Y0 0 -0.026 -0.031 0.143 0.102
3 1
Z 0 0 0 -0.156 -0.087 -0.072 -0.2295
Conclusion:
Since they are no positive number in the Z row the solution is Optimum. Hence for maximum Z, Y1=
0.072, Y2= 0.057 and Y1= 0.102 producing Z = $0.2295.
Solving a Matrix Game Using the Simplex Method
A game may be solved using the simplex method as follows.
Before you start, check for saddle points. If you find one, you have already solved the game; the
optimal strategies are the pure strategies passing through a saddle point. Otherwise, continue with
the following steps.
1. Reduce the payoff matrix by dominance.
2. Add a fixed number k to each of the entries so that they all become non-negative.
3. Set up and solve the associated linear programming problem using the simplex method.
4. Obtain the optimal strategies and the expected value as follows:
Column Strategy
a. Express the solution to the linear programming problem as a column vector.
b. Normalize by dividing each entry of the solution vector by the sum of the entries, or
by the value of the objective variable p.
c. Insert zeros in positions corresponding to the columns deleted during reduction.

OPERATIONS RESEARCH
Row Strategy
d. Write down the row vector whose entries are the numbers appearing in the bottom
row of the final tableau underneath the slack variables.
e. Normalize by dividing each entry of the solution vector by the sum of the entries.
f. Insert zeros in positions corresponding to the rows deleted during reduction.
Value of the Game
e = 1/p - k.

OPERATIONS RESEARCH
UNIT 8: INVENTORY MODELLING:
HOURS: 20
INVENTORY SYSTEM MODEL:
Inventory deals with manufacturing sufficient stock of goods (parts and raw materials) to ensure
a smooth operation of a business activity.
1 GENERALIZED INVENTORY MODEL:
The ultimate objective of inventory model is to answer the following two questions:
⇒ How much to order.
⇒When to order.
The answer to the 1st question is expressed in terms of economic order quantity (EOQ), which is the
optimum amount that should be ordered every time an order is placed and may vary with time
depending on the situation under consideration.
The 2nd part depends on the type of the inventory type. If the inventory system requires periodic
review i.e. equal time intervals (e.g. every week or month). The time for acquiring a new order
usually coincides with the beginning of each time interval.
If a system is of continuous review type a reorder point is usually specified by an inventory level at
which a new order must be placed.
We can express the solution of the general inventory problems as follows:
PERIODIC REVIEW CASE:
Receive a new order of the amount specified by the order quantity at equal intervals of order
time, which can be weekly or monthly.
CONTINUOUS REVIEW CASE:
When the inventory level reaches the re-order point places an order whose size is equal to
the quantity.
The order quantity and re-order point are normally determined by minimizing the total
inventory cost, the point which can be expressed as a function of its principle components in
the following manner.
Total inventory cost = Purchasing cost + Setup cost + Holding cost + Storage cost.
1 PURCHASING COST:

OPERATIONS RESEARCH
Purchasing cost is the price of the quantity of goods to be supplied. The purchasing cost
becomes an important factor when the commodity unit prices become dependant on the
size of the order. This situation is normally expressed in terms of quantity discount or price
break where the unit price of an item decrease with the increase of order quantity.
1 SETUP COST / ORDERING COST: Represents a fixed charge incurred when an
order is placed. Thus to satisfy the demand for a given time period, ordering of
smaller quantities will result in a higher set up cost during the period than if the
demand is satisfied by placing large orders.
1 HOLDING COST / CARRYING COST: Represents the cost of carrying inventory
or represents the cost of maintaining inventory e.g. warehouse rental, security,
handling, depreciation, stock maintenance, insurance and loses of interest on
capital.
1 SHORTAGE COST: Is a penalty incurred when we run out of stock of the needed
commodity. It generally includes costs due to the loss of customer’s goodwill as
well as potential loss in income. We need to minimize the total cost involved in
stocking operations by determining the optimum quantity for the replenishment
order Q.
INVENTORY NOTATION:
2 Tc – Total inventory cost per unit time.
2 D – Total demand per unit time.
2 Ci – Cost of production or purchase per unit expressed as $ per unit.
2 Q – Quantity per order.
2 Q* - Optimum quantity = EOQ.
2 H – Holding cost or Carrying cost per unit of inventory per time.
2 T – Time between orders or length of inventory cycle Q/D.
2 K – Cost of processing an order or Ordering cost / Setup cost.
2 L – Leading time.
2 Cs – Shortage cost.
Total inventory cost = Purchasing cost + Setup cost + Holding cost + Storage cost.
Tc = CiD + QH/2 + DK/Q + Cs.

OPERATIONS RESEARCH
We need to minimize Tc depending on the quantity Q hence by differentiation we get
d(Tc)/dQ = d(QH/2)/ dQ + d(DK/Q)/dQ + d(CiD)/dQ + d(Cs)/dQ
0 = H/2 – DKQ2
0 = H/2 – DK/Q
2
Multiply both sides by 2Q2 we get 0 = HQ2 – 2DK
0 = HQ2 – 2DK
HQ2 = 2DK
Q2 = 2DK/H
Q* = Ö2DK/H (This is the formula for EOQ)
Economic Order Quantity is a deterministic model referred to as Wilson Economic Lot Size.
EOQ ASSUMPTIONS:
Tc = CiD + QH/2 + DK/Q + Cs.
The above model will hold for the following assumptions:
1) Demand should be constant.
2) Replenishment should be instantaneous.
3) Shortages are not allowed.
4) Fixed quantity order per inventory order level should be maintained.
Question 1:
The demand for an item is 18000 units per month and the holding cost per unit is $14.40 per year
and the cost of ordering is $400. No shortages are allowed and the replenishment rate is
instantaneous.
a) Determine the optimum order quantity.
b) The total cost per year of the inventory if the cost of one unit is $1.
c) The number of orders per year.
d) The time between orders.
Solution:
Since all the information asked for are years. Convert all information in months to years.
D = 18000 * 12 = 216000 per year.
H = $14.40 per year.
K = $400.
EOQ = Q* = Ö2DK/H
Ö2* 21600 * 400/14.40
Ö172800000/14.40

OPERATIONS RESEARCH
Ö12000000
3464.101615
3464
Tc = CiD + QH/2 + DK/Q.
= (1 * 216000) + (3464 * 14.40)/2 + (216000 * 400)/3464
= 216000 + 24940.80 + 24942.26
= 265883.06
The number of orders per year = D/Q
=216000/3464
= 62.36 orders
The time between orders = Q/D
= 3464/216000
= 0.016
= 0.02
Question 2:
The demand for a particular item is 18000 units per year. Holding cost per unit is $1.20 per year and
the cost of procurement is $400. No shortages are allowed and the replenishment rate is
instantaneous.
a. Determine the optimum order quantity.
b.The number of orders per year.
c. The time between orders.
d.The total cost per year of the inventory if the cost of one unit is $1.
Solution:
D = 18000 per year.
H = $1.20 per year.
K = $400.
EOQ = Q* = Ö2DK/H
Ö2* 18000 * 400/1.20
Ö14400000/1.20
Ö12000000
3464.101615
3464

OPERATIONS RESEARCH
The number of orders per year = D/Q
=18000/3464
= 5.196 orders
The time between orders = Q/D or 365/5.196 = 70.24 days
= 3464/18000
= 0.1924
= 0.19 days
= (1 * 18000) + (3464 * 1.20)/2 + (18000 * 400)/3461
= 18000 + 206.40 + 2078.52
= 22156.92
Question 3:
Suppose a company has soft drinks products. It has a constant annual demand rate of 3600 cases. A
case of soft drinks cost the Harare drinks company $3. Ordering cost is $20 per order and holding
cost is charged at 25% of the cost per unit. There are 250 working days per year and the lead-time is
5 days.
Identify the following aspects of inventory policy:
i. Economic order quantity.
ii. Re-order point.
iii. Cycle time (Length) = Q/D.
iv. The total annual cost.
Solution:
D = 3600.
H = 25% of $3.
K = $20 per order.
Ci = $3 per unit.
i. EOQ = Q* = Ö2DK/H
Ö2* 3600 * 20/0.75
Ö144000/0.75
Ö192000
438.178
438
ii. Re-order point = Daily Demand * Lead-time.
= 3600/250 * 5
= 72 crates

OPERATIONS RESEARCH
iii. Cycle time = Q/Daily demand.
Daily Demand is given by = 3600/250 = 14.40
Cycle Time = 438/14.40
= 30.42 days
iv. Tc = CiD + QH/2 + DK/Q.
= (3 * 3600) + (438 * 0.75)/2 + (20 * 3600)/438
= 10800 + 164.25 + 164.38356
= 11128.633
= 11128.63
RE-ORDER POINT:
Is a point in which a new order should be placed?
1 GENERAL RULES TO DETERMINE THE RE-ORDER POINT:
i. Determine the number of days the order should be placed (Cycle length) = Q/D.
ii. Determine the number of days which is the minimum level to re-order new entities and is
given by Lead-time – Cycle Length (Q/D) where Lead-time is the time between the placement
of an order and its receipt.
iii. Determine the level of inventory (stock) to re-order new items and is given by: Number of
days which gives minimum level to order new items * Demand.
Question 4:
The daily demand for a commodity is approximately 100 units. Every time an order is placed a fixed
cost of $100 is incurred and daily holding cost per unit inventory $0.02.
If the lead-time is 12 days.
Determine:
i. Economic order quantity.
ii. Re-order point.
iii. Cycle time.
iv. Minimum inventory level.
v. Level of inventory stock.

OPERATIONS RESEARCH
Solution:
D = 100.
H = $0.02.
K = $100.
i. EOQ = Q* = Ö2DK/H
Ö2* 100 * 100/0.02
Ö20000/0.02
Ö1000000
1000
ii. Re-order point = Demand * Lead-time
= 100 * 12
= 1200
iii. Cycle time = Q/D
= 1000/100
= 10
vi. Minimum inventory level = Lead-time – Cycle Length.
= 12 – 10
= 2 days
vii. Level of inventory stock = Minimum inventory level * Demand.
= 2 * 100
= 200 orders.
EOQ WITH GRADUAL REPLENISHMENT / ECONOMIC PRODUCTION LOT
SIZE:
Suppose a demand of an item in a company is D units per unit time and the company can
produce them at a rate of R units per unit time where RD. The cost of set-up is $K and the
holding cost per unit time is $H. We wish to determine the Optimum manufacturing quantity
and the total cost per year assuming the cost of one unit is $Ci. It is assumed that shortages are
not allowed.
EOQ = Q* = Ö2DK/H(1 –
D
/R)

OPERATIONS RESEARCH
Question 5:
A company uses 100000 units per year that cost $3 each. The carrying costs are 1% per month and
the ordering cost are $2.50 per year.
What is the economic order quantity made the items themselves on a machine with a potential
capacity of 600000 units per year.
Solution:
D = 100000.
K = $2.50.
H = 1% = 0.01 * 12 * 3 = 0.36
EOQ = Q* = Ö2DK/H(1 –
D
/R)
= Ö2* 100000 * 2.50/0.36(1 –
100000
/600000)
= Ö500000/0.36(1 –
1
/6)
=Ö1666666.667
= 1290.994449
=1291.
Question 6:
A company manufactures an item, which is also used in the company. The demand of this item is
18000 units per year and the production rate is 3000 per month. The cost of set up is $500. The
holding cost of 1 unit per month is $0.15.
Determine the optimum manufacturing quantity and the total cost per year assuming the cost of 1
unit is $2 and shortages are not allowed.
Solution:
D = 18000 per year.
H = $0.15 * 12 = $1.8 per year.
K = 500
Ci = $2
R = 3000 * 12 = 36000 per year.
EOQ = Q* = Ö2DK/H(1 –
D
/R)
= Ö2* 18000 * 500/1.8 *(1 –
18000
/36000)
= Ö18000000/0.36(1 – 0.5)
=Ö18000000/0.18
=3162.27766
= 3162

OPERATIONS RESEARCH
= (2 * 18000) + (3162 * 1.8)/2 + (18000 * 500)/3162
= 36000 + 2845.8 + 2846.29981
= 41692.09981
= 41692
INVENTORY CONTROL (TYPES OF CONTROL SYSTEMS):
There are 2 two basic inventory control systems:
Reorder Level
Periodic Review system.
1 PERIODIC REVIEW SYSTEM:
This is the reviewing of all stocks at fixed intervals periodically. Therefore absolute
stocks can be eliminated. Variable quantities can be ordered.
1 THE REORDER LEVEL SYSTEM:
The reorder level system results in fixed quantities being ordered at variable intervals
dependent upon demand. The reorder level system usually has three (3) control
levels namely:
The reorder level.
The minimum level.
The maximum level.
NB:
The Reorder Level is calculated so that if the worst anticipated
position occurs, Stock would be replenished in time.
The Minimum Level is calculated so that management will be
warned when demand is above average. There may be no danger
but the situation needs watching.
The Maximum Level is calculated so that management will be
warned when demand is the minimum anticipated and
consequently the stock level is likely to rise.

OPERATIONS RESEARCH
Question 7:
The following is an illustration of a company with a simple manual reorder level system.
Normal Usage = 110 per day.
Minimum Usage = 50 per day.
Maximum Usage = 140 per day.
Lead time = 25 – 30 days.
EOQ = 5000
Using the above data calculate the various control levels i.e.
Reorder level.
Minimum Usage.
Maximum Usage. [20 marks].
Solution:
a) Reorder Level = Maximum Usage * Maximum Lead time.
= 140 * 30
= 4200 units
b) Minimum Level = Reorder Level – (Average Lead time * Normal Usage)
= 4200 – (27.5 * 110)
= 4200 – 3025
= 1175 units
c) Maximum Level = Reorder Level + EOQ – (Min Lead time * Min Usage)
= 4200 + 5000 – (25 * 50)
= 7950
Question 8:
The following is an illustration of a company with a simple manual reorder level system.
Normal Usage = 560 per day.
Minimum Usage = 250 per day.
Maximum Usage = 750 per day.
Lead time = 15 – 20 days.
EOQ = 10000
Using the above data calculate the various control levels i.e.
Reorder level.
Minimum Usage.
Maximum Usage. [20 marks].
Solution:
a) Reorder Level = Maximum Usage * Maximum Lead time.
= 750 * 20
= 15000 units

OPERATIONS RESEARCH
b) Minimum Level = Reorder Level – (Average Lead time * Normal Usage)
= 15000 – (17.5 * 560)
= 15000 – 9800
= 5200 units
c) Maximum Level = Reorder Level + EOQ – (Min Lead time * Min Usage)
= 15000 + 10000 – (15 * 250)
= 21250

OPERATIONS RESEARCH
UNIT 9: REGRESSION ANALYSIS AND
CORRELATION:
HOURS: 20
CORRELATION:
Correlation is a measure of strength of a linear relationship between 2 sets of numbers. There are 2
measures of correlation namely:
The Pearson Product Moment Correlation Coefficient represented by r.
The Spearman’s Rank Correlation Coefficient represented by R.
A coefficient of +1 indicates a perfect Positive relationship whilst a coefficient of – 1 shows perfect
negative correlation.
Hence the correlation coefficient will have values ranging between – 1 and + 1 inclusive.
A coefficient of zero implies no correlation.
SCATTER DIAGRAM:
It is a way of representing a set of data by a scatter of plots or dots. One variable is plotted on the
X-axis and another on the Y-axis. Normally variable X is the one controlled over (independent
variable) and Y variable is the variable that you are interested in (dependent variable).
POSITIVE CORRELATION: NEGATIVE CORRELATION:
* * *
* * * *
Delivery * * * *
Time * * *
* * * *
* * * *
*
Number of deliveries Temperature
NO CORRELATION: NON-LINEAR CORRELATION:
* * * * * *
* * * * * * **
Salary * * * * * * **
* * * * * * **
* * * * * * * **
* * * * * * * **
Age *
Quantity produced
The first diagram is a positive because number of deliveries increase so apparently does the delivery
time. The second is a negative correlation because as air temperature increases the heating cost falls.

OPERATIONS RESEARCH
For the third correlation shows that there is no correlation existing between salary and age of an
employee. For non-linear correlation it suggests that quantity produced and efficiency are correlated but
not linearly.
Given two sets of data represented by the variables X and Y the Product Moment
Correlation Coefficient is given by:
r = Σ (X - r) (Y – Ý)
Ö Σ (X - r)2 * (Y – Ý)2
OR
n ΣXY - ΣX * ΣY
r =
2 2 2 2
[n Σ X - (ΣX)] [n Σ Y - (ΣY)]
Question 1:
Calculate r for the data given below:
X 15 24 25 30 35 40 45 65 70 75
Y 60 45 50 35 42 46 28 20 22 15
Solution:
X Y XY X2 Y2
15 60 900 225 3600
24 45 1080 576 2025
25 50 1250 625 2500
30 35 1050 900 1225
35 42 1470 1225 1764
40 46 1840 1600 2116
45 28 1260 2025 784
65 20 1300 4225 400
70 22 1540 4900 484
75 15 1125 5625 225
424 363 12835 21926 15123

OPERATIONS RESEARCH
n ΣXY - ΣX · ΣY
r =
2 2 2 2
[n Σ X - (ΣX)] [n Σ Y - (ΣY)]
10 * 12835 - 424 * 363
r =
10* 21926 - 179776 * 10 * 15123 - 131769
128350 - 153912
r =
39484 * 19461
- 25562
r =
27719.9959
r = - 0.922150238 r = - 0.92
Question 2:
Calculate r for the data given below:
X 1 2 3 6 5 6
Y 5 10.5 15.5 25 16 22.5
Solution:
X Y XY X2 Y2
1 5 5 1 25
2 10.5 21 4 110.25
3 15.5 46.5 9 240.25
4 25 100 16 625
5 16 80 25 256
6 22.5 135 36 506.25
21 94.5 387.5 91 1762.75

OPERATIONS RESEARCH
n ΣXY - ΣX * ΣY
r =
2 2 2 2
[n Σ X - (ΣX)] [n Σ Y - (ΣY)]
6 * 387.5 - 21 * 94.5
r =
6* 91 - 441 * 6 * 1762.75 – 8930.25
2325 – 1984.5
r =
546 - 441 * 10576 – 8930.25
340.5
r =
415.7598467
r = 0.819
(B) THE SPEARMAN’S RANK CORRELATION COEFFICIENT:
When two sets of data are given positions or ranked, we use Spearman’s Rank Correlation Coefficient
represented by R. It is given by:
R = 1 - 6 Σd2
n(n2 -1)
Where d = difference between pairs of ranked values.
n = number of pairs of ranking.
1 = is an independent number.
Question 2:
A group of eight students were tested in Maths and English and the ranking in the two subjects are
given below:
Student A B C D E F G H
Maths 2 7 6 1 4 3 5 8
English 3 6 4 2 5 1 8 7

OPERATIONS RESEARCH
Solution:
Students Maths English D D2
A 2 3 - 1 1
B 7 6 1 1
C 6 4 2 4
D 1 2 -1 1
E 4 5 -1 1
F 3 1 2 4
G 5 8 -3 9
H 8 7 1 1
TOTAL = 22
R = 1 - 6 Σd2
n(n2 -1)
R = 1 – 6 * 22
8(82 -1)
R = 1 - 132
8* 63
R = 1 – 0.2619
= 0.738
REGRESSION:
This is a statistical technique that can be used for short to medium term forecasting. It seeks to
establish the line of “best fit” for some observed data.
These are 2 ways of obtaining the trend of a set of data:
2 The graphical method.
2 The method of least squares.

OPERATIONS RESEARCH
THE GRAPHICAL METHOD:
In this method a scatter diagram is plotted and a straight line is produced through the middle of
the points representing the trend line.
This line can then be used (when extended) to predict a future value.
y
* Trend line or line of best fit.
* * *
* *
* *
* *
*
x
Least Squares Regression Lines
This is a method of finding a regression line without estimating where the line should go by eye.
If the equation of the regression line is y = ax + b, we need to find what a and b are. We find
these by solving the normal equations.
Normal Equations
The normal equations for the line of regression of y on x are:
Sy = aSx + nb and
Sxy = aSx2 + bSx
The values of a and b are found by solving these equations simultaneously.
For the line of regression of x on y, the normal equations are the same but with x and y
swapped.

OPERATIONS RESEARCH
METHOD OF LEAST SQUARES:
This method gives an equation of the trend line or regression line. The equation is given by:
Y = A + Bx
Where B = n ΣXY - ΣX · ΣY or
n ΣX2 - (ΣX)2
= ΣXY -n X Y
n X2 - nX2
and A = Y - BX
x is an independent variable.
n is the number of items in the list.
Question 3:
Consider the following data:
X 1 2 3 4 5 6
Y 6 4 3 5 4 2
i. Find the equation of the regression line.
ii. Find Y when X = 4 using the equation.
Solution:
X Y XY X2
1 6 6 1
2 4 8 4
3 3 9 9
4 5 20 16
4 4 20 25
6 2 21 36
21 24 75 91
B = n ΣXY - ΣX · ΣY
n ΣX2 - (ΣX)2

OPERATIONS RESEARCH
B = 6(75) – 21 * 24
6 (91) – (21)2
= 450 – 504
546 – 441
= -18/35
X = ΣX/n 21/6 = 3.5
Y = ΣY/n 24/6 = 4
A = Y - BX
= 4 – (7/2 * - 54/105)
= 4 + 9/5
=29/5
Y = A + Bx
= 29/5 – 18/35X
(ii) When X = 4
Y = A + Bx
= 29/5 – 18/35* 4
= 29/5 – 72/35
= 131/35 or 326/35

OPERATIONS RESEARCH
Question 4:
For data below calculate the equation of the regression Y upon X.
X 98 78 74 80 80 83 95 100 97 75
Y 14 9 10 11 10 11 12 13 11 9
Solution:
X Y XY X2
98 14 1372 9604
78 9 720 6084
74 10 740 5476
80 11 880 6400
80 10 800 6400
83 11 913 6889
95 12 1140 9025
100 13 1300 10000
97 11 1067 9409
75 9 675 5625
860 110 9589 74912
B = n ΣXY - ΣX · ΣY
n ΣX2 - (ΣX)2
B = 10(9589) – 860 * 110
10 (74912) – (860)2
= 95890 – 94600
749120 – 739600
= 0.1355
X = ΣX/n 860/10 = 86
Y = ΣY/n 110/10 = 11

OPERATIONS RESEARCH
A = Y - BX
= 11 – (0.1355 * 86)
= 11 – 11.653
= - 0.6533
Y = A + Bx
= - 0.6533 + 0.1355X

OPERATIONS RESEARCH
UNIT 10 INTRODUCTION TO QUEUEING
THEORY
HOURS: 20
Introduction to Queuing Theory
Queueing theory is a set of mathematical tools for the analysis of probabilistic systems of customers
and servers.
In everyday life, it is seen that a number of people arrive at a cinema ticket window. If the people
arrive “too frequently” they will have to wait for getting their tickets or sometimes do without it.
Under such circumstances, the only alternative is to form a queue, called the waiting line, in order
to maintain a proper discipline. Here the arriving people are called the customers and the person
issuing the tickets is called a server.
Specifically, the theory helps the manager to seek answer to the questions like the following when
actual/expected arrival and service rates are known and a set of assumption is made.
· What is the fraction of time the system providing service is expected to be busy (or, idle)?
· What are the chances of a certain number of customers present in the system at a random
point in time?
· On the average, how many customers are expected to be present in the system?
· What is the average length of queues formed from time to time?
· How much time a customer is expected to spend in the system – in waiting for getting the
service?
· How much time a customer typically spends in the queue?
· What is the expected total cost of providing service for various alternative service levels and,
accordingly, what is the optimal level of service to provide?
Definitions:
· Arrival Rate refers to the average number of customers who require service within a specific
period of time.
· A Capacitated Queue is limited as to the number of customers who are allowed to wait in
line.
· Customers can be people, work-in-process inventory, raw materials, incoming digital
messages, or any other entities that can be modeled as lining up to wait for some process to
take place.
· A Queue is a set of customers waiting for service.
· Queue Discipline refers to the priority system by which the next customer to receive
service is selected from a set of waiting customers. One common queue discipline is first-in-first-
out, or FIFO.

OPERATIONS RESEARCH
· A Server can be a human worker, a machine, or any other entity that can be modeled as
executing some process for waiting customers.
· Service Rate (or Service Capacity) refers to the overall average number of customers a
system can handle in a given time period.
· Stochastic Processes are systems of events in which the times between events are random
variables. In queueing models, the patterns of customer arrivals and service are modeled as
stochastic processes based on probability distributions.
· Utilization refers to the proportion of time that a server (or system of servers) is busy
handling customers.
Queueing Notation:
In the literature, queueing models are described by a series of symbols and slashes, such as
A/B/X/Y/Z, where A indicates the arrival pattern, B indicates the service pattern, X indicates the
number of parallel servers, Y indicates the queue’s capacity, and Z indicates the queue discipline. We
will be concerned primarily with the M/M/1 queue, in which the letter M indicates that times
between arrivals and times between services both can be modeled as being exponentially distributed.
The number 1 indicates that there is one server; we will also study some M/M/n queues, where n is
some number greater than 1.
Symbols:
Performance Measure Random
Variable
Expected
Value
Number of Customers in the
System
N L
Number of Customers in the
Queue
Nq Lq
Number of Customers in Service Ns Ls
Time Spent in the System T W
Time Spent in the Queue Tq Wq
Time Spent in Service Ts Ws
System Parameters
Number of Servers S
Arrival Rate (number per unit of time) l (Greek letter
lambda)
Service Rate (number per unit of time) m (Greek letter mu)
Utilization Factor r (Greek letter rho)

OPERATIONS RESEARCH
Formulas:
The utilization factor, or r (rho), also the probability that a server will be busy at any point in time:
(i)
Idle time, or the proportion of time servers are not busy, or the probability that a server will be idle
at any given time:
(ii)
The average time a customer spends in the system:
= + = 1
m -l
q s W W W
(iii)
The average number of customers in the queue:
r
l
2 2
=
( -
) r
m m l
-
=
1
Lq
(iv)
The average number of customers in the system:
L = Lq + Ls (v)
The average time spent waiting in the queue:
r
l
=
1 q W (vi)
( -
) m ( r )
m m l
-
=
Those who would commune more directly with the Arab mathematician al Jebr (whose name is the
source for our word algebra) will note that there is another formula for N:

  =
r
1 r
Therefore:
(vii)
The single most important formula in queueing theory is called Little’s Law:
(viii)
Little’s Law can be restated in several ways; for example, here is a formula for W:
(ix)
There are also some formulas that are based on the special characteristics of the exponential
distribution. These generally are beyond the scope of this course, but one useful example is offered
here. In an M/M/n system, the probability of a customer spending longer than some specific time
period t in the system is given by the following formula. e is the base of the natural logarithms
(roughly 2.718):
(x)
r
l
m
=
S
1- r
( )
N N N q s + =
-
+ =

 

 
-
+ =

 

 

 

 


+ -
 

 


-

 

-
=
-
=
r
r
r
l
m
l
m
l
m
l
m
m
l
m
l
m
l
m
l
m
l
m
l
m
l
m
r
r
2
1
1
1
1 1
1
L =
-
L =lW
W
L
=
l
P{T ³ t} = e(l-m )t

OPERATIONS RESEARCH
MEASURES OF THE MODEL
i. To find expected (average) number of units in the system, Ls i.e. the average number of customers in the
system both waiting and in service.
l m r r l m
l m r
= = =
S L ,where 1
- -
1 1
ii. To find expected (average) queue length, Lq i.e. average number of customers waiting in the queue.
l r
2
m r
= - =
q S L L
1
-
iii. To find mean (or expected) waiting time in the queue (excluding service time), Wq i.e. the
average time spent by a customer in the queue before the commencement of his service.
q W
l r
= =
m m - l m -
r
( ) (1 )
iv. To find expected waiting time in the system (including service time), Ws i.e. the total time
spent by a customer in the system.
W
S
1
-
m l
=
v. To find expected waiting time of a customer who has to wait, (W | W 0).
( 1
W|W 0
) =
-
m (1 r )
vi. To find expected length of non-empty queue, (L | L 0) or Lq’.
( 1
L | L 0
) =
-
1 r
vii. To find out the variance of queue length.
r
r
=
2 Var.{n}
-
(1 )
Inter-Relationship between Ls, Lq, Ws, Wq.
Ls = λWs,
Lq = λWq,
Wq = Ws – 1/v.
Lq = Ls – λ/v.

OPERATIONS RESEARCH
LIMITATIONS FOR APPLICATION OF QUEUEING MODEL:
1. The single channel queueing model can be fitted in situations where the following conditions
are satisfied.
2. The service time has exponential distribution. The average service rate is denoted by μ.
3. Arrivals are from infinite population.
4. The queue discipline is FCFS i.e. the customers are served on first come, first served basis.
5. There is only a single service station.
6. The mean arrival rate is less than the mean service rate, i.e. λ v.
7. The waiting space available for customers in the queue is infinite.
QUEUEING MODELS:
· It involves a customer’s standpoint where a waiting line is created. Customers arrive at the
facility and they join a waiting line (or Queue). The server chooses a customer from the
waiting line to begin service. Upon the completion of a service, the process of choosing a
new (waiting) customer is repeated.
· It is assumed that no time is lost between the completion of a service and the admission of
a new customer into the facility.
· The major actors in queuing situation are Customer and the Server. The interaction
between the Customer and the Serve are of interest only in as far as it relates to the period
of time the customer needs to complete a service. From the standpoint of customer
arrivals we will be interested in the time intervals that separate successive arrivals.
· The customer arrivals and service times are summarized in terms of probability
distributions normally referred to as: arrivals and service time distribution.
· These distributions may represent situations where customers arrive and are served
individually {banks or supermarkets}.
· Customers may arrive and or be served in groups are referred to as Bulk queues.
· In queue model the manner of choosing customer from the waiting line to start service is
referred to as Service discipline.
· The Service discipline occurs as FCFS rule (first come first service), LCFS rule (last come
first served) and SIRO (service in random order).
· Customers arriving at a facility may be put in Priority queues such that those with higher
priority will receive preference to start service first.
· The facility may include more than one server thus allowing as many customers as the
number of servers to be serviced simultaneously this is referred as Parallel service.
· Facilities which comprise a number of series stations through which the customer may pass
before service is completed (e.g. processing of a product on a sequence of machines) is
referred to as Queues in series or Tandem queues.

OPERATIONS RESEARCH
· A combination of Parallel and Series is called Network queues.
· Calling source is the source from which calls for service (arrivals of customer) are
generated.
· Queue size the number of limited customers that may be allowed at a facility may be
because of space limitation (e.g. car spaces in a drive-in bank).
PURE BIRTH MODEL / PROCESS:
Pure birth model is process situation where customer arrive and never leave. E.g. issuing of birth
certificates for newborn babies. These certificates are normally kept as permanent records in a
Central office administered by the state health department.
The birth of new babies and issuing of birth certificates is a completely random process that can be
described ay a Poisson distribution.
Assuming that l is the rate at which birth certificates are issued, the pure birth process of having n
arrivals (birth certificates) during the time period t is given as:
Pn(t) = (lt)n e-lt
n!
Where n = 0, 1, 2, 3… and l is the rate of arrival per unit time with the expected number of arrivals
during t being equal to lt.
Question 1:
Suppose that the birth in a state are spaced over time according to a an exponential distribution with
one birth occurring every 7 minutes on the average.
Solution:
Since the average inter arrival (inter birth) time = 7 minutes the birth rate in the state is computed
as:
l = 24{hours} * 60{minutes} = 205.7 births per day.
7
The number of birth in the state per year =:
lt = 205.7 * 365 = 75.080 births per year
The probability of no births in any one-day is computed as:
Pn(t) = (lt)n e-lt
n!
P0(1) = (205.7 * 1)0 e-205.7 * 1 » 0.
0!

OPERATIONS RESEARCH
Suppose that we are interest in the probability of issuing 45 birth certificates by the end of a period
of 3 hours given that 35 certificates were issued in the first 2 hours.
We observe that since birth occur according to a Poisson process the required probability reduces
to having 45 –35 = 10 births in one (3 – 2) hours.
Given l = 60/7 = 8.57 births per hour.
P10(1) = (8.57 * 1)10 e-8.57 * 1
10!
P10(1) = (2137018192) * (0.000189712)
3628800
» 0.11172
Question 1:
Suppose that the clerk who enters the information from the birth certificate into a computer
normally wait at least 5 certificates to be accumulated. What it the probability that the clerk will be
waiting a new batch every hour.
Solution:
l = / = births per hour.
P5(1) = (* 1)5 e- * 1
5!
» 0.92868
PURE DEATH MODEL / PROCESS:
Pure death model is process situation where customers are withdrawals. Consider the situation
of stocking units of an item at start of the week to meet customer’s demand during the week. If
we assume that customer demand occurs at the rate m units per week and that the demand
process is completely random, the associated probability of having n items remaining in stock
after time t is given by the following Truncated Poisson distribution.
Pn(t) = (mt)N - n e-mt where n = 1, 2, 3 … N.
(N – n)!

OPERATIONS RESEARCH
Pn(t) = 1 – NΣ
n-1 * Pn(t)
Question 2:
At the beginning of each week, 15 units of an inventory item are stocked for use during the
week. Withdrawals from stock occur only during the first 6 days (business is closed on Sundays)
and follow a Poisson distribution with 3 units per day.
When the stock level reaches 5 units, a new order of 15 units is placed for delivery at the
beginning of next week. Because of the nature of the item all units left at the end of the week are
discarded.
Solution:
We can recognize that the consumption rate is m = 3 units per day.
Suppose that we are interested in computing the probability of having 5 units (reorder level) on
day t then we compute it as:
Pn(t) = (mt)N - n e-mt
(N – n)!
P5(t) = (3t)15 - 5 e-3t
(15 – 5)!
NB: P5(t) represents the probability of reordering on day t. This probability peaks at t = 3 and
then declines as we advance through the week. If we are interested in the probability of
reordering by day t we must compute the cumulative probability of having 5 units or less on day
t.
i.e. Pn = 5(t) = P0(t) + P1(t) + …..+ P5(t).
STEADY STATE MEASURES OF PERFORMANCE:
The steady state measures of performance is used to analyze the operation of the queuing situation
for the purpose of making recommendations about the design of the system. Among these
measures of performance are:
The expected number of customers waiting.
The expected waiting time per customer.
The expected utilization of the service facility.
The system comprises both the queue and the service facility.
Let:

OPERATIONS RESEARCH
Ls = expected number of customers in system.
Lq = expected number of customers in Queue.
Ws = expected waiting time in system.
Wq = expected waiting time in Queue.
Suppose that we are considering a Service Facility with c parallel servers then from the definition of
Pn we get:
Ls = ¥Σn=0 * npn
Lq = ¥Σn=c + 1 * (n – c)pn
NB: A strong relationship exists between Ls and Ws (also Lq and Wq) so that either measure is
automatically determined from the other.
Let leff be the effective average arrival rate (independent of the number in the system n) then:
Ls = leff * Ws
Lq = leff * Wq
The value of leff is determined from the state dependent ln and the probabilities Pn is given as:
leff = ¥Σn=0 * lnpn
NB: A direct relationship exists between Ws and Wq, which is equal to:
Expected waiting time in system = Expected waiting time in queue + expected service time.
Ws = Wq + 1/m.
Given that m is the Service rate per busy server, the Expected Service time = 1/m.
Ws = Wq + 1/m.
Multiplying both sides by leff we obtain:
Ls = Lq + leff/m = expected number of customers in system.
The expected utilization of a service is defined as a function of the average number of busy servers.
Since the difference between Ls and Lq must equal the Expected number of busy servers we obtain:
Expected number of busy servers = C = Ls – Lq = leff/m.

OPERATIONS RESEARCH
The Percent utilization of a service facility with c parallel servers is given by:
Percent utilization = C/C * 100
NB: In summary Given Pn we can compute the system’s measure of performance in the
following order:
Pn Ls = ¥Σn=npn Ws = Ls/leff Wq = Ws - 1/m Lq = leff * Wq C = Ls – Lq.
Question 3:
Job orders arriving at a production Facility is divided into three groups. Group1 will take the highest
priority for processing; Group3 will be processed only if there are no waiting orders from group 1
and 2. It is assumed that a job once admitted to the facility must be completed before any new job is
taken in.
Orders from groups 1,2 and 3 occur according to Poisson distribution with mean 4, 3 and 2 per day
respectively. The service times for the three groups are constant with rates 10, 9 and 10 per day
respectively.
Find the following:
I. The expected waiting time in the system for each of the three queues.
II. The expected waiting time in the system for any customer.
III. The expected number of waiting jobs in each of the three groups.
IV. The expected number waiting in the system.
Solution:

OPERATIONS RESEARCH
QUEUEING THEORY QUESTIONS
Q1. Arrivals at a telephone booth are considered to be Poisson, with an average time of 10
minutes between one arrival and the next. The length of a phone call assumed to be
distributed exponentially with mean 3 minutes. Then,
(a) What is the probability that a person arriving at the booth will have to wait?
(b) What is the average length of the queues that form from time to time?
(c) What is the average length of the queue?
(d) The telephone department will install a second booth when convinced that an arrival
would expect to have to wait at least 3 minutes for the phone. By how much must the
flow of arrivals be increased in order to justify a second booth?
(e) Find the probability that an arrival finds that 4 persons are waiting for their turn.
(f) Find the average number of persons waiting and making telephone calls.
-
= = =
m l l
l l
l
= =
= ⇒ = ´ = »
Ans Here,
1 1
l m
= and =
10 3
(a) P(W 0) = 1-P0 =
1 3 3
= = = 0.3
10 1 10
l
m
´
(b) The average length of non empty queues or expected number of customers in the non empty
queue.
(L | L 0) = Lq
, =
1
3 1.43 1 Person
1 1
3 10
m
m l
= = »
- -
(c) The average length of queues (empty + non-empty) or expected number of customers in the
queue.
Lq =
2
2
1
3 9
0 Customers
1 1 70 1
3
r
r
 
 
  = = »
- -
(d) W
q
'
l
m m l
( )
q
'
'
   - '

 
=
l ' l
'
1
Since W 3, , (say) for second booth
3
3 , giving 0.16.
1 1
3 3
As, 0.16 0.16 60 9.6 10customers / hour
Thus, as soon as the arrival rate increases to 10 customers/hour, another booth may be installed.

OPERATIONS RESEARCH
(e) With 4 people waiting for their turn in the queue implies a total of 5 people in the system. Thus,
the probability that an arrival finds 4 persons in the queue is,
( ) n
l m ´ =
The average waiting time of a customer in the queue is given by:
= = = ´
- - ´ ´
= = r - r
P n 5 (1 )
= 5
-
=
0.3 (1 0.3)
0.0017
(f) The average number of people waiting and making calls is given by the expected length of the
system. Thus,
s L
1
r
r
=
-
0.3
1 0.3
=
-
= 0.43
Q2. Customers arrive at a sales counter manned by a single person according to a Poisson
process with a mean rate of 20 per hour. The time required to serve a customer has an
exponential distribution with a mean of 100 seconds. Find the average waiting time of a
customer.
Ans Here we are given:
60 60
= 20 per hour and = 36 per hour
100
q
20 5 5 3600
l
W hours or , i.e.,125seconds.
m m l
( ) 36(36 20) 36 4 36 4
The average waiting time of a customer in the system is given by:
s
1 1 1
= =
W or hour i.e., 225seconds.
m - l (36 -
20) 16
Q3. Customers arrive at a one-window drive according to a Poisson distribution with mean
of 10 minutes and service time per customer is exponential with mean of 6 minutes. The
space in front of the window can accommodate only three vehicles including the serviced
one. Other vehicles have to wait outside this space. Calculate:
(i) Probability that an arriving customer can drive directly to the space in front of the
window.
(ii) Probability that an arriving customer will have to wait outside the directed space.
(iii) How long an arriving customer is expected to wait before getting the service?
Ans From the given information, we find that:
Mean arrival rate, l = 6 customers per hour
And mean service rate, m = 10 customers per hour
(i) Probability that an arriving customer can drive directly to the space in front of the window is
given by:

OPERATIONS RESEARCH
2
l l l l l
m m m m m
l l l
m m m
       
+ + =  -  +  -  +    - 
P P P 1 1 1
0 1 2
       
     2

=  1 -   1
+ +   
     
=  - 6   + 6 +  6  2
 =
98
           
1 1 or 0.784
10 10 10 1225
(ii) Probability that an arriving customer will have to wait outside the directed space is given by:
0 1 2 1- (P + P + P ) = 1- 0.784 = 0.216 or 21.6%
(iii) Expected waiting time of a customer being getting the service is given by:
l
= = =
- -
m m l
l = 12 trucks/hr.
m = 20 trucks/hr.
= = =
- -
-
m m l
q
6 3
W hour or 9 min utes.
( ) 10(10 6) 20
Q4. A company distributes its products by trucks loaded at its only loading station. Both,
company’s trucks and contractor’s trucks, are used for this purpose. It was found out that
on an average every 5 minutes, one truck arrived and the average loading time was three
minutes. 50% of the trucks belong to the contractor. Find out:
(i) The probability that a truck has to wait,
(ii) The waiting time of truck that waits, and
(iii) The expected waiting time of contractor’s trucks per day, assuming a 24-hours shift.
Ans Here we are given:
60
Average arrival rate of trucks,
=
5
Average service rate of trucks,
60
=
3
(i) The probability that a truck has to wait is given by:
12
0.6
20
r l
m
(ii) The waiting time of a truck that waits is given by:
s
1 1 1
= =
W = hour i.e., 7.5 min utes.
m l (20 12) 8
(iii) The expected waiting time of contractor’s trucks per day, assuming a 24-hours shift is given
by:
(No. of trucks per day) x (Contractor’s percentage) x (Expected waiting time of a truck)
50
= 12 ´ 24
´ ´
l
100 ( )
1 12 54
= ´ ´ =
288 or 10.8 hrs.
´
2 20 8 5
Q5. Customers arrive at the First Class Ticket counter of a Theatre at a rate of 12 per hour.
There is one clerk serving the customers at a rate of 30 per hour.

OPERATIONS RESEARCH
(i) What is the probability that there is no customer in the counter (i.e. the system is
= =
1 P P P 1 1 1 1
 
 - 
 
idle)?
(ii) What is the probability that there are more than 2 customers in the counter?
(iii) What is the probability that there is no customer waiting to be served?
(iv) What is the probability that a customer is being served and nobody is waiting?
12
= 0.4
30
r l
m
(i) P (system is idle) = P0 = 1- r = 1- 0.4 = 0.6 .
(ii) P(n2) = 1–P(n£2)
[ ]
( ) ( )
( ) ( )
2
0 1 2
2
2
2
n 1
1 1 1
1 1 1
1 1 0.4 1 0.4 0.4
1 0.6 1.56
1 0.936
0.064
Also,P( n)
P( 2)
l l l l l
m m m m m
l l l
m m m
r r r
r +
       
= - + + = -  -  +  -  +    - 
       
     
= -  -   + +   
     
= -  -  + +    
= -  -  + +    
= - ´
= -
=
=
= 0.42+1 = 0.43 = 0.064
(iii) P(no customer waiting to be served) = P0 + P1
= 0.6 + 0.24 = 0.84
(iv) P (a customer is being served and none is waiting) = P1 = 1 l l
m m
=r (1- r )= 0.4 x
0.6 = 0.24.
Q6. A TV repairman finds that the time spent on his job has an exponential distribution
with mean 30 minutes. If he repairs sets in the order in which they come and if the arrival of
sets is approximately Poisson with an average rate of 10 per 8-hour day, what is his expected
idle time each day? How many jobs are ahead of the set just brought in?
Mean arrival rate, l = 10 sets per day.
And mean service rate, m = 16 sets per day.
10 5
=
= =
16 8
r l
m

OPERATIONS RESEARCH
- r = - = .
´ = 3 hours.
5
8 5 2 L 1 sets
= =
= =
- -
 
 
  = = =
-r = - = .
P (repairman is idle) = P0 =
5 3
1 1
8 8
Expected idle time per day =
3
8
8
To determine the expected number of sets ahead of the set just brought in, we shall find the
expected number of sets in the system. Thus,
s
1 5 3 3 1
8
r
r
= = = =
- -
.
Q7. Assume that at a bank teller window the customers arrive in their cars at the average
rate of 20 per hour according to a Poisson distribution. Assume also that the bank teller
spends an average of 2 minutes per customer to complete a service, and the service time is
exponentially distributed. Customers, who arrive from an infinite population, are served on
a FCFS basis, and there is no limit to possible queue length.
(i) What is the expected waiting time in the system per customer?
(ii) What is the mean number of customers waiting in the system?
(iii) What is the probability of zero customers in the system?
(iv) What value is the utilization factor?
20 2
=
30 3
r l
m
(i) Expected waiting time in the system per customer is given by:
s
1 1 1
W = hour i.e., 6 min utes.
m l (30 20) 10
(ii) Mean number of customers waiting in the system is given by:
2
2
q
2
3 4
L
1 2 3 1
3
r
r
- -
(iii) Probability of zero customers in the system is given by P0 =
2 1
1 1
3 3
(iv) Utilization factor,
2
3
r = .

OPERATIONS RESEARCH

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OPERATIONS RESEARCH