Page 1 of 7
Applications in Linear Programming
• Production Planning Problem
A factory manufactures a product each unit of which consists of 5 units of part A and 4 units of
part B. the two parts A and B require two different row materials I and II of which 120 units and
240 units respectively are available. These parts can be manufactured by three different
methods. Raw material requirements per production run and the number of units for each part
produced are given below;
Method Input per run (units) Output per run (units)
Row material I Row material II Part A Part B
1 7 5 6 4
2 4 7 5 8
3 2 9 7 3
Determine the number of production runs for each method so as to maximize the total number
of complete units of the final products.
Formulation
Let a, b and c represent the number of production runs for method 1, 2, 3 respectively.
Let z be the number of units produced.
6𝑎+5𝑏+7𝑐
Since for a unit of product needs 5 parts from A and 4 parts from B, z ≤ 5
and
4𝑎+8𝑏+3𝑐
z≤ . These two inequalities imply 6a+5b+7c-5z ≥ 0 and 4a+8b+3c-4z ≥ 0 respectively. Thus, the
4
mathematical model can be developed as follows;
Maximize z (Number of units produced)
Subject to
7a+4b+2c ≤ 120 (Row material I)
5a+7b+9c ≤ 240 (Row material II)
6a+5b+7c-5z ≥ 0
4a+8b+3c-4z ≥ 0
a≥0, b≥0, c≥0
, Page 2 of 7
• Advertising Media Selection Problem
An advertising company wishes to plan its advertising strategy in three different media –
television, radio and magazines. The purpose of advertising is to reach a large number of
potential customers as possible. Following data has been obtained from market survey;
Television Radio Magazine I Magazine
II
Cost of an advertising unit Rs.30 000 Rs.20 000 Rs.15 000 Rs.10 000
No. of potential customers 200 000 600 000 150 000 100 000
reached per unit
No. of female customers 150 000 400 000 70 000 50 000
reached per unit
The company wants to spend not more than Rs.450 000 on advertising. Following are the
further requirements that must be met;
i. At least 1 million exposures take place among female customers,
ii. Advertising on magazines be limited to Rs.150 000,
iii. At least 3 advertising units be bought on magazine I and 2 units on magazine II, and
iv. The number of advertising units on television and radio should each be between 5 and
10.
Formulation
Let a, b, c, d represent the number of these advertising units in television, radio, magazine I and
magazine II respectively,
Objective is to maximize the total number of potential customers. Therefore, the model can be
represented as follows;
Maximize z = 105 (2a + 6b + 1.5c + d) (total number of potential customers)
Subject to
30a + 20b + 15c +10d ≤ 450 (budget)
15a + 40b 7c +5d ≥ 100 (female customers)
15c + 10d ≤ 150 (expense on magazine)
C ≥ 3 (No. of units on magazine I)
d ≥ 2 (No. of units on magazine II)
5 ≤ a ≤ 10 (No, of units on television)
Applications in Linear Programming
• Production Planning Problem
A factory manufactures a product each unit of which consists of 5 units of part A and 4 units of
part B. the two parts A and B require two different row materials I and II of which 120 units and
240 units respectively are available. These parts can be manufactured by three different
methods. Raw material requirements per production run and the number of units for each part
produced are given below;
Method Input per run (units) Output per run (units)
Row material I Row material II Part A Part B
1 7 5 6 4
2 4 7 5 8
3 2 9 7 3
Determine the number of production runs for each method so as to maximize the total number
of complete units of the final products.
Formulation
Let a, b and c represent the number of production runs for method 1, 2, 3 respectively.
Let z be the number of units produced.
6𝑎+5𝑏+7𝑐
Since for a unit of product needs 5 parts from A and 4 parts from B, z ≤ 5
and
4𝑎+8𝑏+3𝑐
z≤ . These two inequalities imply 6a+5b+7c-5z ≥ 0 and 4a+8b+3c-4z ≥ 0 respectively. Thus, the
4
mathematical model can be developed as follows;
Maximize z (Number of units produced)
Subject to
7a+4b+2c ≤ 120 (Row material I)
5a+7b+9c ≤ 240 (Row material II)
6a+5b+7c-5z ≥ 0
4a+8b+3c-4z ≥ 0
a≥0, b≥0, c≥0
, Page 2 of 7
• Advertising Media Selection Problem
An advertising company wishes to plan its advertising strategy in three different media –
television, radio and magazines. The purpose of advertising is to reach a large number of
potential customers as possible. Following data has been obtained from market survey;
Television Radio Magazine I Magazine
II
Cost of an advertising unit Rs.30 000 Rs.20 000 Rs.15 000 Rs.10 000
No. of potential customers 200 000 600 000 150 000 100 000
reached per unit
No. of female customers 150 000 400 000 70 000 50 000
reached per unit
The company wants to spend not more than Rs.450 000 on advertising. Following are the
further requirements that must be met;
i. At least 1 million exposures take place among female customers,
ii. Advertising on magazines be limited to Rs.150 000,
iii. At least 3 advertising units be bought on magazine I and 2 units on magazine II, and
iv. The number of advertising units on television and radio should each be between 5 and
10.
Formulation
Let a, b, c, d represent the number of these advertising units in television, radio, magazine I and
magazine II respectively,
Objective is to maximize the total number of potential customers. Therefore, the model can be
represented as follows;
Maximize z = 105 (2a + 6b + 1.5c + d) (total number of potential customers)
Subject to
30a + 20b + 15c +10d ≤ 450 (budget)
15a + 40b 7c +5d ≥ 100 (female customers)
15c + 10d ≤ 150 (expense on magazine)
C ≥ 3 (No. of units on magazine I)
d ≥ 2 (No. of units on magazine II)
5 ≤ a ≤ 10 (No, of units on television)
