Question

1. A company has two warehouses A and B, and three retail outlets 1, 2 and...

1. A company has two warehouses A and B, and three retail outlets 1, 2 and 3. The warehouse capacities, retail outlet demands, and per-unit shipping costs (\$) are shown in the table below. Formulate a linear programming (LP) model of this transportation problem with the objective of minimizing total shipping cost.

 Retail Outlets Warehouses 1 2 3 Total Supply A \$5 \$8 \$3 500 B \$7 \$4 \$6 250 Total Demand 300 400 250

2. A department head has four managers, and three projects (A, B, C) to be executed. The managers differ in efficiency, and the projects differ in their intrinsic complexity. The estimate of the times each manager would take to complete each project is given in the matrix below. Formulate a linear programming (LP) model for determining how the projects should be assigned, to a manager, so as to minimize the total project completion time. No manager should be assigned more than one project.

Manager

Project

 1 2 3 4 A 8 14 12 11 B 6 9 10 8 C 7 10 9 11

Problem 1:

Let, xij = units shipped from warehouse ‘i’ to retail outlet j

Where, i = a, b for two warehouses

j = 1, 2, 3 for three retail outlets

Thus, there will be 2 x 3 = 6 decision variables in the LPP.

Objective Function:

Objective is to minimize the total shipping cost:

Let Cij = per unit-shipping cost time from warehouse i to outlet j

Min. Z = (Cij x Xij)

Min. z = \$5xa1 + \$8xa2 + \$3xa3 + \$7xb1 + \$4xb2 + \$6xb3

Subject To:

The constraint of problem are:

Warehouse cannot ship more than supply capacity

Demand of outlet has to be satisfied

Supply constraint of warehouses: Since the total supply capacity is less than total demand, all the capacity will be utilized.

Warehouse A: xa1 + xa2 + xa3 = 500

Warehouse B: xb1 + xb2 + xb3 = 250

Demand constraint of retail: Since demand is more than supply capacity, not all demand will be satisfied

Retail 1: xa1 + xb1 <= 300

Retail 2: xa2 + xb2 <= 400

Retail 1: xa3 + xb3 <= 250

Nonnegative Constraint: all Xij >= 0

Problem 2:

Let, xij = 1 if the manager ‘i’ is assigned to project j otherwise Xij = 0

Where, i = 1, 2,…4 for four managers

j = a, b, c for three projects

Thus, there will be 4 x 3 = 12 decision variables in the LPP.

Objective Function:

Objective is to minimize the total project completion time:

Let Cij = time required by manager i to complete project j

Min. Z = (Cij x Xij)

Min. z = 8x1a + 6x1b + 7x1c + 14x2a + 9x2b + 10x2c + 12x3a + 10x3b + 9x3c + 11x4a + 8x4b + 11x4c

Subject To:

The constraint of problem is to assign one manager to one project.

Assignment of Managers: Since the number of manager is more than number of projects, one of the manager will remained unassigned.

Manager 1: x1a + x1b + x1c <= 1

Manager 2: x2a + x2b + x2c <= 1

Manager 3: x3a + x3b + x3c <= 1

Manager 4: x4a + x4b + x4C <= 1

Assignment of Projects

Project A: x1a + x2a + x3a + x4a = 1

Project B: x1b + x2b + x3b + x4b = 1

Project C: x1c + x2c + x3c + x4c = 1