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

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OPERATIONS RESEARCH