LINEAR PROGRAMMING

INTRODUCTION TO Q.T.
QT are the set of scientific techniques intended to express the business
constraints and the risk underlying a business operation into measurable terms
and thereby reduce the decision making process to a more analytical and
objective one.

L.P.P
LPP is a mathematical technique for allotting the limited resources of a firm in an
optimize manner.
Linear equation :- any equation which is in the form :- ax+by+cz = p where a,b,c
are constants and x,y,z are variables.
The problem most commonly faced by management is to decide the manner in
which the limited resources should be used to achieve the desired objectives like
profit maximization, cost minimization etc.
 LINEAR PROGRAMMING
DECISION VARIABLES:- These are the unknowns to be determined from the
solution e.g. - x,y,z in the linear equation.

Constraints :- These represent the mathematical equations of the limitation

imposed by the problem situations.

Objective function :- This represents the mathematical equation of the major

objective of the system in terms of the decision variables

Non negativity Condition :- The decision variables must be either 0 OR +ve i.e
x,y,z>= 0.
Feasible solution :- A set of values of the decision variables which satisfies all the
constraints and the non-negativity condition is a feasible solution. A problem
can have many feasible solutions.
Optimum Solution:- This is a feasible solution which optimizes the objective
function. Normally an optimal solution is unique.
 LINEAR PROGRAMMING
Stages of L.P.P :- Following are the three stages of L.P.P
1) Problem identification through collection of data.
2) Problem formulation i.e. to formulate the mathematical problem from the given
data.
3) Problem solving.
Problem Formulation of L.P.P
Steps
Step 1 : Find the key decisions to be made from the study of the problem.
Step 2: Identify the decision variables and assign symbols like x,y,z or x1,x2,x3 etc.
Step 3 : Mention the objective function quantitatively and express it as a linear
function of the decision variables.
Step4 :- Express the constraints also as linear equalities or inequalities in terms of
the decision variables.
Step 5: Express the objective function , constraints, and the non negativity together
to form the formulation of LPP.
 Problems of L.P.P.
Problem:- Mr. Rao, the owner of a Readymade garments shop wishes
to publish advertisements in two local daily newspapers, One Marathi and
one English. The expected coverage through the advertisements is 1000
people and 1500 people per advertisement respectively. Each
advertisement in a Marathi newspaper costs Rs. 3000 and for and English
daily it is Rs. 5000. Mr. Rao has decided not to place more than 10
advertisements in the Marathi newspaper and wants to place at least 6
advertisements in the English daily. The total advertisement budget is Rs.
50000 Formulate the problem as a L.P. Model.
 LPP Formulation
Solution :- For LPP Formulation we need Objective function, Constraints and Non
Negativity condition.
Let x1 - Number of advertisement to be placed in Marathi Newspaper
x2- Number of advertisement to be placed in English Newspaper
1. Objective Function :- Coverage through Marathi Newapaper = 1000 X x1
Coverage through English Newapaper = 1500 X x2
Therefore Total Coverage is C = 1000X1 + 1500X2
Objective function is to maximize C = 1000X1+ 1500X2
2. Constraints :
(a) Advertisement expenditure for Marathi Newspaper = 3000x1
Advertisement expenditure for English Newspaper = 5000x2
Therefore Total Advertisement Expenditure = 3000x1 + 5000x2
Total Advertisement Budget = Rs. 50000
Therefore 3000x1 + 5000x2 ≤ 50000
(b) Maximum number of advertisements in Marathi newspaper is 10
i.e. x1 ≤ 10 Similarly minimum number of advertisement in english daily is 6
 LPP Formulation
i.e. x2 ≥ 6

Therefore L.P. Model of the problem is

Maxmimise C = 1000X1 + 1500X2 (Coverage Maxmisation)

Subject to 3000x1 + 5000x2 ≤ 50000 (Budget Constraints)
x1 ≤ 10 (Marathi Newspaper Constraint)
x2 ≥ 6 (English daily constraint)
x1, x2 ≥ 0 (Non negativity condition)
 Problems of L.P.P.
Q.:- A company makes three products X,Y,Z which go through three
departments- drill, lathe, and assembly. The hours of department time
required by each of the products, the hours available in each of the
departments and the profit contribution of each of the products are given in
the following table:-
Products Time Required per Unit (Hr.) Profit
Contribution (Rs.
Per unit)
Drill Lathe Assembly
X 3 3 8 9

Y 6 5 10 15

Z 7 4 12 20

Hrs 210 240 260

available

The marketing department of the company indicates that the sales

potential for products X and Y is unlimited but for Z it is not more than
30 units. Determine optimum production schedule.
 Problems of L.P.P.
Question :- A company manufacture two products A and B . The profit
per unit of A and B is Rs. 60/- and Rs. 80/- respectively. The
company is required to supply 200 units of product A to its regular
customers. Product A requires machining on machine M1 only and
per unit of A, one hour of M1 is required. Product B requires
Machine M2 only and machine hours on M2 has enough hours
available to manufacture any number of units of product B . M1 has
400 hours available. Product A and B both require one labour hour
each and company has 500 labour hours available. To determine
the number of units of A and B to be manufactured. Formulate the
LPP.
 GRAPHICAL METHOD OF SOLUTION OF LPP
(A) Basic feasible solution :- solution which satisfies the non negativity
condition as well as the consraints. A problem can have many feasible
solutions.
(B) Optimum solution :- This is a basic feasible solution which optimizes the
objective function. A problem normally has a unique optimum solution.

Steps for obtaining Graphical Solution :-

Setp 1. Formulate the given problem in L.P. format.
Setp 2 Plot the graph constraints:
(a) Make the constant term on the right hand side of the constraints as
positive.
(b) Replace the inequality sign in a constraint by an equality.
(c) Plot the constraint line
(d) Repeat the procedure to draw all the constraint lines
 GRAPHICAL METHOD OF SOLUTION OF LPP
Steps for obtaining Graphical Solution :-
Step 3 :- Identify the Feasible Region or the solution space
(a) Each constraint line divides the XY plane into two parts :- one
which includes the origin and the other away from the origin.
(b) For the constraints of the type ( ≥ ) mark the region away
from the origin. For the constraints of the type (≤) mark the region
towards the origin about the constraint line. For the type (= ) the region
is the line itself. Use arrow (→ ) to indicate the respective regions.
(c) Mark the region which is common to all these regions. This is
the Feasible Region.
Step 4 :- To find Optimum Solution :-
Corner Point Method :-
(a) Identify all the corner points of the feasible region. To obtain their
coordinates solve the equations of the corresponding intersecting lines.
(b) Find the value of the objective function at all these corner points.
(c ) For a maximization problem, the co-ordinates of the corner
point where Z is maximum give the solution. Similarly , for a
minimization problem the coordinates of the corner point
corresponding to minimum value of Z give the solution.
 Problem (LPP SOLUTION)
(1) Anita electric company produces two products P1 and P2 that are
produced and sold on weekly basis. The weekly production can not
exceed 25 for product P1 and 35 for product P2 because of limited
facilities. The company employs total of 60 workers. Product P1
requires 2 man weeks of labour, where as product P2 requires
only 1. Profit margin on P1 is Rs. 60 and on P2 is Rs. 40. Formulate
it as a LPP and solve for maximum profit.

(2) Maximize Z = 10 x1 + 15 x2
subject to x1≤ 3
x2≤5
3x1 + 4x2 = 29
x1, x2 ≥ 0
 Measures of Central Tendency
Measure of Central Tendency is the techniques of analysis and interpretation of
data. There is a tendency of the collected data to concentrate about a particular
value, called as ‘central tendency’ and that particular value is called as ‘the
measure of central tendency’ or average.

This is very useful for :

1. Describing the characteristic of entire distribution.
2. Comparison of various distribution.
3. Computing other statistical measures.
The Requisites of a Good Measure of Central Tendency are :
1. It should be rigidly defined.
2. It should be easy to understand and simple to compute.
3. It should be based on all observations.
4. It should be suitable for further mathematical treatment.
The commonly used measures of central tendency are - Arithmatic Mean, Median
and Mode.
 Arithmatic Mean
1. For Individual Observations : Mean is given by their sum divided by the total
number of observations. Thus , if x1,x2, ……xn are N given observation than
Arithmatic Mean of x i.e x̄ = x1+x2+ ………..+xn
_______________ = Σx / N
N
Example : If there are 6 observations with values 12, 18, 28, 16, 16, 15 then by
direct method arithmatic mean x̄ = 12+18+28+16+16+15

__________________
6
= 105/6 = 17.5
2. For Ungrouped Frequency Distribution :
Values (x) : x1 x2 x3 ……………. Xn
Frequency (f) : f1 f2 f3 ……………. Fn
Then Arithmatic Mean x̄ = f1x1 + f2x2+ …………+fnxn
__________________ = Σfx / Σf = Σfx/ N
f1 + f2 +f3 + ……….. +fn
 Arithmatic Mean
If values of x and f are large it is convenient to use the Deviation method as :-

Arithmatic mean of x = A + Σfd/N where A = Assumed mean and

d = x-A (Deviation from Assumed mean)
Example : Following table gives the students performance in class test are as follows-

Let Assumed Mean A = 3

We have by direct method, Arithmatic Mean x̄ = Σfx / Σf = 74/25 = 2.96
OR by Deviation method, Arithmatic Mean = A + Σfd / N = 3 + (-1)/25 = 3 – 0.04
= 2.96
 Arithmatic Mean
3. For Grouped Frequency Distribution : Here we have
Arithmatic Mean x̄ = Σfx / Σf = Σfx / N
Where x = mid value of each class and
f = frequency of the class
and Σf = N = Total frequency = Total number of observations
Also, it is more convenient to use Deviation Method :
x̄ = A + Σfd / N where A = Assumed mean , d = x –A

If all the class intervals are equal, then use a further simplified formula (Step
deviation method) :-

x̄ = A + hΣfd’ / N where h = class interval, d’ = x –A / h

 Arithmatic Mean
Example :- Consider the following data about the number of workers employed
in the companies in an industrial area.
Let A = Assumed Mean = 25 and Class Interval = 10

Mean Mean x̄ = A + hΣfd’ / N = 25 + 10(28/100) = 25 + 2.8 = 27.8

Merits of Mean
1. It is rigidly defined.
2. It is easy to understand and calculate.
3. It is based on all observations. And useful for further mathematical
treatment
 Median
Median refers to the value of the middle item of the distribution when the data is
arranged in an ascending or descending order. Following are the method for
calculating Median :-
1. For Individual Observations : - Arrange the data in the ascending order of the
value and then median, Md = Value of (N+1/2)th item if N is Odd. But if N is
even, we take the arithmetic mean of the two middle values.
Example:- For the data : 2, 11, 10, 9, 6 Calculate Median.
Sol. we have N = 5 and ascending order = 2, 6, 9 , 10 , 11
Median = value of (N+1)th i.e (5+1/2 = 3rd observation
Therefore Median = 9
Example :- For the data 2, 11 , 10, 9, 6, 7 calculate Median.
Sol. We have N=6 and ascending order = 2, 6, 7, 9, 10, 11
Median = value of (N+1/2)th i.e (6+1/2 = 3.5 th observation

Therefore Median = arithmetic mean of 3rd and 4th observation = 7+9/2 = 8

 Median
2. For Discrete Frequency distribution :-
(a) Arrange the data in ascending order of the values of the variable.
(b) Write down less than cumulative frequencies (c.f.)
(c ) Find (N+1/2)
(d) Identify the C.F. which is equal to (N+1/2) or just greater than it.
(e) The corresponding value of the observation is the median.
Example :- Consider the data
No. of accidents (x) : 0 2 1 3 4
No. of Shifts (f) : 2 8 9 4 1

Here N = 24
Therefore N+1/2 = 24+1/2 = 12.5
C.F. just greater than (12.5) is 19
Therefore Median = corresponding value of x = 2
 Median
3. For Grouped Frequency distribution :-
(a) Write the frequency distribution by making the classes continuous (if not
already given)
(b) Arrange the classes in ascending order and write their less than cumulative
frequencies.
(c) Find N/2
(d) Identify the c.f. which is equal to (N/2) OR just greater than it.
(e) Note the corresponding class which is the ‘median class’
(f) Use the formula

Median = l +h/f (N/2 –c)

Where l = lower limit of the median class
h = class interval
f = frequency of the median class
c = c.f. of the class preceding the median class.
 Median
Example :- Classes : 10-14 15-19 20-24 25-29 30-34
f : 4 5 6 8 7
Sol. Here we have to make the classes continuous as follows :

Here N = Σf = 30
And median = value of N/2th i.e. 15th item. This will belong to the class having c.f. equal or greater than 15 i.e. 3 rd class hence the median
class is 19.5 – 24.5
Therefore Median Md = l +h/f (N/2 – c)
Here l = 19.5 h = 5 f = 6 N = 30 and c = c.f. of preceding class = 9
Therefore Median = 19.5 + 5/6(30/2 – 9) = 19.5 + 5/6(6) = 24.5
 Mode
It is the value which occurs most frequently in the given data. Thus, it is the value
which occurs with highest frequency.
Following are the method for calculating Mode :-
1. For Individual observation :- it can be easily found as below :
Example :- For the data 2, 4, 1, 6, 6, 8, 9, 4, 6, 8 mode is 6 as it appears with
maximum frequency i.e. 3.

2. For Discrete Frequency Distribution :- It is given by the value corresponding to

the highest frequency.
Example :- x : 1 2 3 4 5
f : 6 9 16 12 10
Here mode = value (x) corresponding to highest frequency of 16.
=3
 Mode
3. For Grouped Frequency Distribution :- Here we make the classes continuous
first if not given. The class interval must also be equal and then we identify
the class corresponding to the maximum frequency as the modal class then we
use the formula mode = (f1 – f0)
l + h -------------
2f1-f0-f2
Where l = lower limit of modal class h = class interval of modal class
F1 = frequency of the modal class f0 = frequency of the class preceding the
modal class
F2 = frequency of the class succeeding the modal class

Example :- Consider the distribution

Classes : 0-10 10-20 20-30 30-40 40-50
Frequency : 12 13 19 16 13
Calculate the Mode
 Mode
Solution :- Here the distribution is continuous and the class intervals are also
equal. The model class ( class corresponding to highest frequency of 19 is
(20-30) .
Thus l = 20, h = 10 f1= 19 f0= 13 f2 = 16 and

Therefore Mode = 20 + 10 (19-13)

---------------
2(19)-13-16 = 20 + 10(6)/9 = 26.667

Merits of Mode :- 1. It is easy to calculate and understand.

2. Useful for qualitative data
3. Not affected by extreme values at all.