Solutions to Sensitivity Questions
- Produce 400 at 1, 200 at 2, and 400 at 3, for a cost of
- It costs $30,000 to produce one more, saves $30 to produce 1
- Production remains the same, so we same $2,000(200)=$400,000.
This type of analysis is valid for any cost of at least $8000.
- It is costing us $4,000 per car that must be produced. Value of
reducing to 200 cars is $800,000; value of reducing to 0 cars is
$1,600,000. Cost of increasing by 100 cars is $400,000. Increasing
by 200 cars is outside range, but total cost is at least $800,000.
- Material is worth $5,000 per unit. We are willing to buy 300
units at that price. After that we will be willing to spend an
unknown (but lower) amount per unit.
- Producing such a car costs us $20,000 worth of material, but
saves us the marginal $30,000 of producing with our current plants.
Therefore the cost can be no more than $10,000.
- It must reduce its cost to 0 for us to consider it.
Alternatively, it could reduce its material usage to 4.6.
- It doesn't matter how expensive it gets, we will always produce
at plant 1 (due to our extreme material shortage).
- Quantities of A, B, C, and D are 1500, 1000, 1000, 2833.33
respecitvely, for a profit of $10,166.67
- .50 for printing, 0 for cutting, .167 for folding. 1000 hours
for printing, no limit for cutting, and 7000 hours for folding.
- Either B or C. Profit improves by 100(.333) = $33.33.
- 2 minutes to print costs $1, 2 minutes to cut cost $0, 3 minutes
to fold costs $0.50, so the profit must be at least $1.50. You cannot
tell the effect of requiring 1000 without rerunning.
- Knowing the cost more accurately will not change the production
decision. Since the total effect is $0.25 (2833) = $708, it seems
unlikely that knowledge is worth $500.