Homework Answers
T1 and T2 be the number of Type1 and Type2 hats produced respectively.
Let L = required labor hours for producing one unit of a Type2
hat.
So, 2L = required labor hours for producing one unit of a Type1
hat.
So, total labor hours required = 2L.T1 + L.T2
Again, we know that if all the available labor hours were dedicated to Type 2 alone, the company can produce a total of 400 Type 2 hats a day.
So, available hours = 400.L
So, the labor hour constraint will be 2L.T1 + L.T2 ≤ 400L or, 2T1 + T2 ≤ 400 (canceling out L from both the side)
Objective function: Maximize.Z = Toal Revenue = 8T1 + 5T2
Subject to,
2T1 + T2 ≤ 400 (labor hours)
T1 ≤ 150 (market limit of Type1)
T1 ≤ 200 (market limit of Type2)
T1, T2 ≥ 0
-----------------------------------
(a)
| Type1 | Type2 | ||
| Qty. | 100 | 200 | Total |
| Revenue | $8 | $5 | 1800 |
(b)
| Final | Reduced | Objective | Allowable | Allowable | ||
| Cell | Name | Value | Cost | Coefficient | Increase | Decrease |
| $C$3 | Qty. Type1 | 100 | 0 | 8 | 2 | 8 |
| $D$3 | Qty. Type2 | 200 | 0 | 5 | 1E+30 | 1 |
| Final | Dual | Constraint | Allowable | Allowable | ||
| Cell | Name | Value | Price | R.H. Side | Increase | Decrease |
| $E$6 | Labor hours Total | 400 | 4 | 400 | 100 | 200 |
| $E$7 | Limit of T1 Total | 100 | 0 | 150 | 1E+30 | 50 |
| $E$8 | Limit of T2 Total | 200 | 1 | 200 | 200 | 100 |
(c)
The maximum allowable decrease for Type1 demand is 50 i.e. up to 100. The given decease is 120 i.e. within the range of feasibility. So, the dual price of 0 will be the same.
So, revenue will be 1800 - 0 x (150 - 120) = 1800 (same as before)
(d)
Dual price = 1
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