ME30B Exam 2002/2003 Question 1.2

The question paper is available here (the link goes to the UWI website - anyone can access it!).

1. Amount to produce:

P1: 100 units

P2: 330 units

P3: 0 units

P4: 242 units

Note: Values are from the first table with headings Variable, Value, Reduced Cost.

2. Binding constraints are:

Constraint 1 – Machining hours

Constraint 4 – Raw material

Constraint 5 – P1 order

Note: We look at the table (the second) that says Constraint, Slack/Surplus, Dual Price. We are looking for the constraints that have a value of 0 in the Slack/ Surplus Column. We know which constraints they are by comparing their RHS ranges from the fourth table (with headings Constraint, Current RHS, Lower Limit and Upper Limit) with the amount of material available (i.e. stated) in the production problem.

3. Dual Prices are

Constraint 1: 9.2 – For every one hour of increase in machine time, increases the profit by \$9.2 within its allowable range

Constraint 2: 0 – Increasing or decreasing the assembly hours makes no difference to the profit within its allowable range

Constraint 3: 0 - Increasing or decreasing the packaging time makes no difference to the profit within its allowable range

Constraint 4: 2 – Increasing the raw material available by 1 unit, increases the profit by \$2 within its allowable range

Constraint 5: -5.5 – Increasing the Product 1 order by l unit decreases the profit by \$5.5 within its allowable range

Note:  For Constraint 5, this means accepting to make an order of 100 P1 units is not profitable. These values are obtained from the second table with headings constraints, slack/surplus and dual prices.

4. Change the Machine time, increase it to 1 – it increases the profit by \$9.2, most profitable amongst all three constraints. Decreasing Constraint 5 by 1 will save \$5.5, and increasing Constraint 2 by 1 will increase profit by only \$2

5. If P3 profitability changes from \$6 to \$8, the optimal solution remains the same. \$8 is within the allowable range for the coefficients (-infinity to \$8.73) for the current optimal solution. (See table 3 with headings variable, current coefficient, lower limit and upper limit). If P3 profitability changes from \$6 to \$9, the optimal solution will change since the P3 has now become profitable i.e. > \$8.73  (or 9 - 6 ≥ 2.7 : its reduced cost, from first table).

6. Raw material corresponds to constraint 4. The upper limit of constraint 4 is 1602. Beyond 1602, raw material becomes in excess. Therefore, constraint 4 can increase by 1602 – 1500 = 102 units of raw material (we do not need the other 48 units i.e. 102 + 48 = 150 units). Since 1 unit increase in raw materials realizes an increase of by \$2 (from the its dual price), then if the cost of buying the raw material is \$1.50 more than the original amount, the company can still make a profit of \$0.50 for each of the new102 material. That is new profit will be \$6,186 + (0.50) × (102) = \$6,237

7. The order of 100 P1 units corresponds to constraint 5. Constraint 5 lower limit is 0, thus canceling the order of 100 P1 units beforehand would have meant an extra profit of 5.5 ×  100 = \$550. Remember from its dual price, every one unit of P1 order decreased, increases the profit by \$5.50 (OR one unit increase of P1 decreases profit by \$5.5). This means it costs us \$550 to produce 100 P1. Therefore, a buyout of \$600 will ensure us a profit of \$600 – \$550 = \$50, from the compensation deal.

8. P5 has a profit of \$7. If we compare this with P4, which has a profit of \$7.20, we note that P5 requires more machining, assembly and packaging time than P4, same amount of material time, but a lower profit. Although P5 has a higher profit than P1, P1 must be produced to meet the order. It is more than likely that P5 should not be produced, but the only way to know for sure, is to run the LP program again with the new product.