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Transportation Problems

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Transportation Problems - Lesson Summary

Manufacturing problems: In these problems, the number of units of different products, which are to be produced and sold by an industry or a firm when each product requires a fixed manpower, machine hours, etc., in order to make maximum profit, are determined.

Diet problems: In these problems, we determine the amount of different kinds of nutrients and constituents that need to be included in a diet, so that the cost of the desired diet is minimum, such that it contains a certain minimum amount of each constituent/nutrient.

Transportation problems: In these problems, we determine a transportation schedule in order to find the way of transporting a product from factories situated at different locations to different market places at the minimum cost.

Ex:

An aeroplane can carry a maximum of 200 passengers. A profit of Rs. 400 is made on each executive class ticket and Rs. 300 on each economy class ticket. The airline reserves at least 20 seats for executive class. However, at least 4 times as many passengers prefer to travel by economy class than by executive class. Determine how many tickets of each type must be sold in order to maximise the airline's profit. What is the maximum profit?

Solution:

Let the number of tickets of executive class be x and the number of tickets of economy class be y.

⇒ x ≥ 0 and y ≥ 0 … (1)

Maximum number of passengers = 200

⇒ x + y £ 200 ... (2)

The airline reserves at least 20 seats for executive class.

⇒ x ≥ 20 … (3)

Also given that at least 4 times as many passengers prefer to travel by economy class than by executive class.

⇒ y ³ 4x … (4)

Profit made on each executive class ticket = Rs. 400

Profit made on each economy class ticket = Rs. 300

⇒ Total profit = 400x + 300y

The objective function is Z = 400x + 300y.

Z = 400x + 300y is to be maximised using the constraints given in (1) to (4).

x ≥ 0 and y ≥ 0 … (1)

x + y £ 200 ... (2)

x ≥ 20 … (3)

y ³ 4x … (4)

x ≥ 0 and y ≥ 0

x + y £ 200

i) x + y = 200

x

0

200

y

200

0

The ordered pairs are (0, 200) and (200, 0).

y ³ 4x

ii) y = 4x

x

0

20

y

0

80

The ordered pairs are (0, 0) and (20, 80).

x ≥ 20

iii) x = 20 is a line parallel to the Y-axis intersecting the X-axis at (20, 0).

The shaded region is the feasible region, which is bounded.

The corner points of the feasible region are A (20, 80), B (40, 160) and C (20, 180).

Corner points (x, y)

Value of the objective function, Z=400x + 300y

A (20, 80)

Z=400(20)+300(80) = 32,000

B (40, 160)

Z=400(40)+300(160) = 64,000

C (20, 180)

Z= 400(20)+300(180) = 62,000



The maximum value of Z is 64,000, which occurs at the point, (40, 160).

So 40 tickets of executive class and 160 tickets of economy class should be sold to get the maximum profit of Rs. 64,000.

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