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Home > Science > Mathematics   »   Linear Programming Problm

 
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Old Jun 17, 2008, 11:39 AM
parag_ca
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Linear Programming Problm

Transport Ltd. provides tourist vehicles of 3 types- 20-seater vans, 8-seater big cars & 5-seater small cars. These seating capacities are excluding the drivers. The company has 4 vehicles of the 20-seater van type, 10 vehicles of the eight-seater big car types 20 vehicles of the 5-seater small car types. These vehicles have to be used to transport employees of their client company from their residences to their offices and back. All the residences are in the same housing colony. The offices are at two different places, one is the Head Office and the other is the Branch. Each vehicle plies only one round trip per day, if residence to office in the morning and office to residence in the evening. Each day, 180 officials need to be transported in Route I (from residence to Head Office & back) and 40 officials need to be transported in Route II (from Residence to Branch office & back). The cost per round trip for each type of vehicle along each route is given below.
Figs-Rs./round trip
20-seater vans 8-seater big cars 5-seater small cars
Route I-
Residence-Head Office & Back 600 400 300
Route II-
Residence-Branch Office & Back 500 300 200
You are required to formulate the information as LPP with objective of minimizing the total cost of hiring vehicles for the client company, subject to the constraints mentioned above.

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Old Jun 17, 2008, 02:26 PM   #2  
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I hope this is a start.

There are 34 vehicles altogether.

Let x = # vans, y = # large cars, z = # small cars.

....number of each used.

...total number of vehicles used.

........total number of people transported.

Minimize total cost:

The new restrictions can be found by subbing into the 2 inequalities.
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Old Jun 29, 2008, 11:31 AM   #3  
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Sorry friend
Your answer is not correct
Correct answer is :
Z=600x1 + 400x2 + 300X3 + 500y1 + 300y2 + 200y3
Subject to constraints,
x1 + y1 ≤ 4
x2 + y2 ≤ 10
x3 + y3 ≤ 20
20x1 + 8x2 + 5x3 = 180
20y1 + 8y2 + 5y3 = 40
x1, x2, x3, y1, y2, y3 ≥ 0
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