Smu Opim624 Homework 1

OPIM624: DECISION ANALYSIS

Q1 (a) Formulate the LP model either in word form or in algebraic form. (5m)

LP Model in Algebraic Form

Decision variables

Let X1 = Quantity of Applesauce

Let X2 = Quantity of Apple Juice

Let Y1 = Advertising Cost for Applesauce

Let Y2 = Advertising Cost for Apple Juice

Objective Function

To maximise profit

= 0.95(X1) + 1.15(X2) – Y1 – Y2

= $20,750.00

Constraints

Maximum Money to spend:                 0.8(X1) + 0.6(X2) + Y1 + Y2 < 16,000

Demand for X1:                         X1 – 3Y1 < 5000

Demand for X2:                         X2 – 5Y2 < 4000

Minimum Quantity of Applesauce:         -0.8(X1) + 0.2(X2) < 0

Maximum Quantity of Applesauce:         0.4(X1) – 0.6(X2) < 0

Non-negative constraints

X1, X2, Y1, Y2 > 0

1 (b) Set up your model in excel. (5m)

[pic 1]

1 (c) Use ASP to obtain the answer and the sensitivity reports. Interpret the sensitivity report (8m) 

According to ASP, The Andrews Apple Products Company is recommended to:

• Produce 4,200 applesauce
• Produce 16,800 apple juice at 16800 units
• Spend $2,560 on advertising for apple juice
• Not spend on advertising for applesauce

By implementing the above recommendations collectively, The Andrews Apple Products Company will achieve a net profit of $20,750.

SENSITIVITY REPORT

[pic 2]

1. As there is a zero in the allowable decrease column in the decision variable cells section, an alternative optimal solution exists.
2. Reduced cost of zeroes in the decision variables cells means any additional unit will not give any additional profit.

1. In the constraints section, the zeroes in the shadow price column suggest that there will not be any incremental profit if we increase output by 1.

1 (d) It is clear that the problem has alternative optimal solutions. Determine an alternative optimal solution. Note that you would need to reformulate the model for this purpose. (7m)

An alternative optimal solution could be achieved by having X1 meeting the demand while maintaining the profit maximization at the same amount of $20,750 as achieved previously.

Additional constraints

1. 0.95(X1) + 1.15(X2) - Y1 - Y2 = $20,750
2. Demand for X1 was changed from X1 – 3(Y1) < 5000 to Demand for X1: X1 – 3(Y1) = 5000

[pic 3]

2 (a) Develop a linear programming model (in mathematical form or in words) that will provide the maximum yield for the portfolio. (5m)