A factory produces cakes, pies, rolls and danishes. The resources required are flour, sugar, meat and fruit. The following table shows
the per-unit resource quantities required for each product,
the price we sell each product for,
the per-unit resource prices, and
the resource quantities available to us for purchase:
Products | Resources | Selling price | |||
---|---|---|---|---|---|
flour | sugar | meat | fruit | ||
cakes | 0.500 | 0.900 | 0.000 | 0.100 | $8.40 |
pies | 0.430 | 0.090 | 0.500 | 0.000 | $20.21 |
rolls | 0.250 | 0.015 | 0.000 | 0.000 | $1.45 |
danishes | 0.180 | 0.290 | 0.000 | 0.050 | $3.99 |
Price per unit | $2.50 | $1.50 | $28.00 | $18.00 | |
Amount available | 1700.000 | 700.000 | 35.000 | 2200.000 |
After calculating the per-unit profits, we can formulate the linear program to maximise profit as follows:
Max P = 4X1 + 5X2 + 0.8X3 +
2.2X4
s.t.
0.5X1 + 0.43X2 + 0.25X3 +
0.18X4 ≤ 1700.00
0.9X1 + 0.09X2 + 0.015X3 +
0.29X4 ≤ 700.00
0.5X2 ≤ 35.00
0.1X1 + 0.05X4 ≤ 2200.00
Xi ≥ 0.
When this linear program is solved in Excel, the following Sensitivity Report is produced:
Variable Cells
Cell | Name |
Final Value |
Reduced Cost |
Objective Coefficient |
Allowable Increase |
Allowable Decrease |
---|---|---|---|---|---|---|
$B$5 | Cakes | 0 | -2.660458453 | 4 | 2.660458453 | 1E+30 |
$B$6 | Pies | 70 | 0 | 5 | 1E+30 | 3.250573066 |
$B$7 | Rolls | 5149.068768 | 0 | 0.8 | 2.091152074 | 0.686206897 |
$B$8 | Danishes | 2125.737822 | 0 | 2.2 | 13.26666667 | 0.853793103 |
Constraints
Cell | Name |
Final Value |
Shadow Price |
Constraint R.H. Side |
Allowable Increase |
Allowable Decrease |
---|---|---|---|---|---|---|
$B$14 | Flour | 1700 | 2.851002865 | 1700 | 9891.766667 | 1239.327586 |
$B$15 | Sugar | 700 | 5.816618911 | 700 | 1996.694444 | 593.506 |
$B$16 | Meat | 35 | 6.501146132 | 35 | 1656.24424 | 35 |
$B$17 | Fruit | 106.2868911 | 0 | 2200 | 1E+30 | 2093.713109 |
Now suppose that we decrease the price of danishes to $2.90 per unit. From the Sensitivity Report, what is the most you can say about the effect on the optimal production plan and profit?
Select one:
a. The optimal production plan will change. The optimal profit will be between $6828.82 and $7330.94.
b. The optimal production plan will not change. The optimal profit will be $6313.89.
c. The optimal production plan will not change. The optimal profit will be $8064.57.
d. The optimal production plan will not change. The optimal profit will be $9145.88.
e. The optimal production plan will change. The optimal profit will be between $5808.47 and $7330.94.
f. The optimal production plan will not change. The optimal profit will be $8993.98.
Answer:
a. The optimal production plan will change. The optimal profit will be between $6828.82 and $7330.94.
Explanation:
The reduced cost of danishes is 0 and the allowable increase is 13.26666667 and allowable decrease is 0.853793103.
This means that optimal values wont change if the price is not increased by 13.26666667 or decreased 0.853793103.
But, we are decreasing the selling price by = 3.99 - 2.90 = $1.09
That means unit profit will change by $1.09. Therefore the optimal solution will also change.
New profit for danishes = 2.90 -1.09= 1.11
Now, the minimum value of objective function with the optimal solution will be :
= 70*5 + 5149.068768*0.8 + 2125.737822*1.11 = 6828.82.
Therefore the asnwer is a)
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