Question: 3. Consider the following linear program: Max 3x1 + 4x2 ($ Profit) s.t. x1 + 3x2
3. Consider the following linear program:
| Max | 3x1 + 4x2 ($ Profit) |
| s.t. | x1 + 3x2 <=12 |
|
| 2x1 + x2 <= 8 |
|
| x1 <= 3 |
|
| x1, x2 >= 0 |
The Management Scientist provided the following solution output:
OPTIMAL SOLUTION
Objective Function Value = 20.000
| Variable | Value | Reduced Cost |
| X1 | 2.400 | 0.000 |
| X2 | 3.200 | 0.000 |
| Constraint | Slack/Surplus | Dual Price |
| 1 | 0.000 | 1.000 |
| 2 | 0.000 | 1.000 |
| 3 | 0.600 | 0.000 |
OBJECTIVE COEFFICIENT RANGES
| Variable | Lower Limit | Current Value | Upper Limit |
| X1 | 1.333 | 3.000 | 8.000 |
| X2 | 1.500 | 4.000 | 9.000 |
RIGHT HAND SIDE RANGES
| Constraint | Lower Limit | Current Value | Upper Limit |
| 1 | 9.000 | 12.000 | 24.000 |
| 2 | 4.000 | 8.000 | 9.000 |
| 3 | 2.400 | 3.000 | No Upper Limit |
| a. | What is the optimal solution including the optimal value of the objective function? |
| b. | Suppose the profit on x1 is increased to $8. Is the above solution still optimal? What is the value of the objective function when this unit profit is increased to $8? |
| c. | If the unit profit on x2 was $9 instead of $4, would the optimal solution change? If yes, what is the optimal solution? |
| d. | If simultaneously the profit on x1 was reduced to $2.0 and the profit on x2 was raised to $5.5, what is the optimal solution? |
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