## LINEAR PROGRAMMING ASSIGNMENT Case Solution

## 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 Reductions | 1 | 2 | 3 | 4 | 5 |

1 | 4 | 2 | 1 | 3 | 0 |

2 | 4 | 3 | 1 | - | 0 |

3 | 4 | 5 | 2 | 3 | 0 |

Column Reductions | 1 | 2 | 3 | 4 | 5 |

1 | 0 | 0 | 0 | 0 | 0 |

2 | 0 | 1 | 0 | - | 0 |

3 | 0 | 3 | 1 | 0 | 0 |

Assignment | Total Cost | ||

Contractor 1 | Job 1 + Job 2 | 14 | |

Contractor 2 | Job 3 + Job 5 | 5 | |

Contractor 3 | Job 4 | 6 | |

Optimal Total Cost | | 25 |

**Problem 3**

The LP model is as follows:

Min = 4*x*1− 5*x*2− 3*x*3

- t.
*x*1 +*x*2 +*x*3 +*x*4 = 10

*x*3 ≥ 1 *x*1 + 2*x*3 ≤ 4 *x*1 ≥ 0*,x*2 ≤ 0

This could be transformed into standard form as follows:

Min Z = 4*x*1− 5*x*2− 3*x*3

Subject to:

*x*1 + *x*2 + *x*3 + *x*4 = 10

*x*3 – s1= 1 *x*1 + 2*x*3 + s2= 4 *x*1 – s3= 0*,x*2 +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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