Q1 Suppose a media specialist has to decide how to allocate advertising in three different media. The
unit costs of a message in the three media are Rs 1000, Rs 750 and Rs 500. The total budget available for
the campaign is Rs 200,000 for the period. The first media is a monthly magazine and it is desired to
advertise not more than one insertion in one issue. At least six messages should appear in the second
media. The number of messages in the third media should be between 4 and 8. The effective audience
for the unit message in the media is shown below.
Media Expected effective audience
1 80,000
2 60,000
3 45,000
Formulate this as an LPP model to determine the optimum allocation that would maximize total
effective audience.
Q2 The board of directors of a company has given approval for the construction of a new plant. The
plant will require an investment of Rs 50 lakh. The required funds will come from the sale of a proposed
bond issue and by taking loans from two financial corporations. For the company, it will not be possible
to sell more than Rs 20 lakh worth of bonds at the proposed rate of 12%. Financial corporation A will
give loan up to Rs 30 Lakh at an interest of 16% but insists that the amount of bond debt plus the
amount owned to financial corporation B be no more than twice the amount owned to financial
corporation A. Financial corporation B will loan the same amount as that loaned by financial corporation
A but it would do so at an interest rate 18%. Formulate this as an LPP model to determine the amount of
funds to be obtained from each source in a manner that minimizes the total annual charges.
Q3 Use Graphical method to solve the following LPP
Maximize Z = x + 0.5 y
Subject to:
3x + 2 y ≤ 12
5 x = 10
x + y ≥ 8
x + y ≥ 4
x ≥ 0, y ≥ 0
Q4 Use Graphical method to solve the following LPP
Minimize Z = 3x + 2y
Subject to:
5x + y ≥ 10
x + y ≥ 6
x + 4y ≥ 12
≥ 0, y ≥ 0
Answers
Given : 5x + y ≥ 10 , x + y ≥ 6 , x + 4y ≥ 12 , x≥ 0, y ≥ 0
To find : Minimize Z = 3x + 2y
Solution:
Z = x + 0.5 y
Subject to:
3x + 2 y ≤ 12
5 x = 10 => x = 2
x + y ≥ 8 => 2 + y ≥ 8 => y ≥ 6
x + y ≥ 4 => 2 + y ≥4 => y ≥ 2
x ≥ 0, y ≥ 0
3x + 2 y ≤ 12
=> 3(2) + 2y ≤ 12
=> 2y ≤ 6
=> y ≤ 3
y ≥ 6 also y ≤ 3
Hence Data is wrong
Minimize Z = 3x + 2y
Subject to:
5x + y ≥ 10
x + y ≥ 6
x + 4y ≥ 12
x ≥ 0, y ≥ 0
From graph 4 points
( 0 , 10) , ( 1 , 5) , ( 4 , 2) , ( 12 , 0)
Z = 3x + 2y
Z = 20 for ( 0 , 10)
Z = 13 for ( 1 ,5)
Z = 14 for ( 4 , 2)
Z = 36 for ( 12 , 0)
Z is minimum at ( 1 . 5) = 13
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