LINEAR PROGRAMMING ASSIGNMENT Harvard Case Solution & Analysis

LINEAR PROGRAMMING ASSIGNMENT Case Solution

Problem 1

Let Xijbe the amount of the space in square feet which the company leases in month ifor jmonths.

Then, the total variables which we would be having would be 6 as follows:

= x1,1; x1,2; x1,3; x2,1; x2,2; x3,1.

Then, the LP model would be as follows:

Minimize 280(x1,1 + x2,1 + x3,1) + 450(x1,2 + x2,2) + 600(x1,3)

Subject to:

x1,1 + x1,2 + x1,3 >= 25000

x1,2 + x1,3 + x2,1 + x2,2 >=10000

x1,3 + x2,2 + x3,1 >=20000

x1,1; x1,2; x1,3; x2,1; x2,2; x3,1 >=0

            The objective function shown above would minimize the total cost of leasing space. The first three constraints make sure that the required amount of space is leased for months 1, 2 & 3 respectively. The fourth constraint is for the non-negativity for all the variables.

Problem 2

The assignment for this problem would be as follows:

Row Reductions12345
142130
2431-0
345230
Column Reductions12345
100000
2010-0
303100
AssignmentTotal Cost
Contractor 1Job 1 + Job 214
Contractor 2Job 3 + Job 55
Contractor 3Job 46
Optimal Total Cost 25

Problem 3

            The LP model is as follows:

Min = 4x1− 5x2− 3x3

  1. t. x1 + x2 + x3 + x4 = 10

x3 ≥ 1 x1 + 2x3 ≤ 4 x1 ≥ 0,x2 ≤ 0

This could be transformed into standard form as follows:

Min Z = 4x1− 5x2− 3x3

Subject to:

x1 + x2 + x3 + x4 = 10

x3 – s1= 1 x1 + 2x3 + s2= 4 x1 – s3= 0,x2 +s4= 0

x1, x2, x3, x4, s1, s2, s3, s4 >= 0

We can see that the constraint system has m=4 and n = 4 variables, therefore we need to get 4-4=0 variables equal to zero to solve for phase 1 problem. For example if we substitute x2, x2, x3 and x4 to 0 then:

S1: 0

S2: 0

S3: 0

S4: 0

            This is clearly feasible and also called as the basic feasible solution.................

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