Please: a) Formulate the IP (Integer Linear Programming) b) Define decision vari

Question

Please:

a) Formulate the IP (Integer Linear Programming)

b) Define decision variables and label constraints

c) then solve using Lingo or excel or any other software.

Texbook: Operation Research, Hamdy Taha, 9th edition. Exercise 4, PS 9.1C, p. 328.

4. Three industrial sites are considered for locating manufacturing plants. The plants send their supplies to three customers. The supply at the plants, the demand at the customers, and the unit transportation cost from the plants to the customers are given in the following table. Unit transportations cost ($) Customer Supply Plant 10 17 15 1200 15 14 10 12 20 1800 1400 1300 Demand 1700 1600 In addition to the transportation costs, fixed costs are incurred at the rate of $1200 m as an ILP $11,000, and $12,000 for plants 1,2, and 3, respectively. Formulate the problem as an and find the optimum solution. are change

Explanation / Answer

TOTAL no. of supply constraints : 3
TOTAL no. of demand constraints : 3
Problem Table is

The rim values for S1=1800 and D1=1200 are compared.
The smaller of the two i.e. min(1800,1200) = 1200 is assigned to S1 D1
This meets the complete demand of D1 and leaves 1800 - 1200 = 600 units with S1
Table-1

The rim values for S1=600 and D2=1700 are compared.
The smaller of the two i.e. min(600,1700) = 600 is assigned to S1 D2
This exhausts the capacity of S1 and leaves 1700 - 600 = 1100 units with D2
Table-2

The rim values for S2=1400 and D2=1100 are compared.
The smaller of the two i.e. min(1400,1100) = 1100 is assigned to S2 D2
This meets the complete demand of D2 and leaves 1400 - 1100 = 300 units with S2
Table-3

The rim values for S2=300 and D3=1600 are compared.
The smaller of the two i.e. min(300,1600) = 300 is assigned to S2 D3
This exhausts the capacity of S2 and leaves 1600 - 300 = 1300 units with D3
Table-4

The rim values for S3=1300 and D3=1300 are compared.
The smaller of the two i.e. min(1300,1300) = 1300 is assigned to S3 D3
Table-5

Final Allocation Table is

Here, the number of allocation is equal to m + n - 1 = 3 + 3 - 1 = 5
The solution is feasible.
Total cost = 10 × 1200 + 15 × 600 + 14 × 1100 + 20 × 300 + 11 × 1300 = 56700

The minimized total cost = 56700

D1 D2 D3 Supply S1 10 15 12 1800 S2 17 14 20 1400 S3 15 10 11 1300 Demand 1200 1700 1600

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