Homework SourseWivco produces Product 1 and Product 2 by processing raw mat
Wivco produces Product 1 and Product 2 by processing raw material. Up to 90 lbs of raw material may be purchased at a cost of $10/lb. One pound of raw material can be used to produce either 1 lb of product 1 or 0.33 lb of product 2. Using a pound of raw material to produce product 1 requires 2 hours of labor or 3 hours to produce 0.33 lb of product 2. A total of 200 hours of labor is available. At most 40 lbs of Product 2 can be sold. Product 1 sells for $13/lb and Product 2 for $40/lb.
a) Formulate a Linear Program (LP) that maximizes Wivco’s profit.
b) Solve the LP using LINDO.
Solution
. Let
RM = pounds of raw material processed
P1 = pounds of raw material used to produce product 1
P2 = pounds of raw material used to produce product 2
To maximize profit, Wivco should solve the following LP:
Max z =13P1 + 40(0.33)P2 – 10RM
Such that,
RMP1 + P2
2P1 + 3P2200
RM90
0.33P240
P1, P2, RM0
b) Solve the LP using LINDO.
Using LINDO to solve the LP.
LINDO Output for Wivco
MAX
13 P1 + 13.2 P2 - 10 RM
SUBJECT TO
2) - P1 - P2 + RM >= 0
3)
2 P1 + 3 P2 <=
200
4)
RM <=
90
5)
0.33 P2 <=
40
END
LP OPTIMUM FOUND
AT STEP
3
OBJECTIVE FUNCTION VALUE
1)
274.000000
VARIABLE
VALUE
REDUCED COST
P1
70.000000
0.000000
P2
20.000000
0.000000
RM
90.000000
0.000000
ROW
SLACK OR SURPLUS
DUAL PRICES
2)
0.000000
-12.600000
3)
0.000000
0.200000
4)
0.000000
2.600000
5)
33.400002
0.000000
NO. ITERATIONS=
3
RANGES IN WHICH THE BASIS IS UNCHANGED
OBJ COEFFICIENT RANGES
VARIABLE
CURRENT
ALLOWABLE
ALLOWABLE
COEF
INCREASE
DECREASE
P1
13.000000
0.200000
0.866667
P2
13.200000
1.300000
0.200000
RM
-10.000000
INFINITY
2.600000
RIGHTHAND SIDE RANGES
ROW
CURRENT
ALLOWABLE
ALLOWABLE
RHS
INCREASE
DECREASE
2
0.000000
23.333334
10.000000
3
200.000000
70.000000
20.000000
4
90.000000
10.000000
23.333334
5
40.000000
INFINITY
33.400002
| MAX | 13 P1 + 13.2 P2 - 10 RM | ||
| SUBJECT TO | |||
| 2) - P1 - P2 + RM >= 0 | |||
| 3) | 2 P1 + 3 P2 <= | 200 | |
| 4) | RM <= | 90 | |
| 5) | 0.33 P2 <= | 40 | |
| END | |||
| LP OPTIMUM FOUND | AT STEP | 3 | |
| OBJECTIVE FUNCTION VALUE | |||
| 1) | 274.000000 | ||
| VARIABLE | VALUE | REDUCED COST | |
| P1 | 70.000000 | 0.000000 | |
| P2 | 20.000000 | 0.000000 | |
| RM | 90.000000 | 0.000000 | |
| ROW | SLACK OR SURPLUS | DUAL PRICES | |
| 2) | 0.000000 | -12.600000 | |
| 3) | 0.000000 | 0.200000 | |
| 4) | 0.000000 | 2.600000 | |
| 5) | 33.400002 | 0.000000 | |
| NO. ITERATIONS= | 3 | ||
| RANGES IN WHICH THE BASIS IS UNCHANGED | |||
| OBJ COEFFICIENT RANGES | |||
| VARIABLE | CURRENT | ALLOWABLE | ALLOWABLE |
| COEF | INCREASE | DECREASE | |
| P1 | 13.000000 | 0.200000 | 0.866667 |
| P2 | 13.200000 | 1.300000 | 0.200000 |
| RM | -10.000000 | INFINITY | 2.600000 |
| RIGHTHAND SIDE RANGES | |||
| ROW | CURRENT | ALLOWABLE | ALLOWABLE |
| RHS | INCREASE | DECREASE | |
| 2 | 0.000000 | 23.333334 | 10.000000 |
| 3 | 200.000000 | 70.000000 | 20.000000 |
| 4 | 90.000000 | 10.000000 | 23.333334 |
| 5 | 40.000000 | INFINITY | 33.400002 |
