Last week you were asked to formulate this linear program. The formulation with
ID: 456314 • Letter: L
Question
Last week you were asked to formulate this linear program. The formulation with explanation is as follows. The LINDO output is also included. Answer the six questions.
Comfort Plus Inc. (CPI) manufactures a standard dining chair used in restaurants. The demand forecasts for quarter 1 (January – March) and quarter 2 (April – June) are 3700 and 4200 chairs, respectively. CPI has a policy of satisfying demand in the quarter in which it occurs.
The chair contains an upholstered seat that can be produced by CPI or purchased from DAP, a subcontractor. DAP currently charges $12.50 per seat, but has announced a new price of $13.75 effective April 1. CPI can produced 3800 seats per quarter at a cost of $10.25 per seat.
Seats that are produced or purchased in quarter 1 and used to satisfy demand in quarter 2 cost CPI $1.50 each to hold in inventory, but the maximum inventory cannot exceed 300 seats.
CPI1 = number of seats produced by CPI in quarter 1
DAP1 = number of seats purchased from DAP in quarter 1
INV = number of seats carried in inventory from quarter 1 to quarter 2
CPI2 = number of seats produced by CPI in quarter 2
DAP2 = number of seats purchased from DAP in quarter 2
LP
min
10.25CPI1
+ 12.5DAP1
+ 1.5INV
+ 10.25CPI2
+ 13.75DAP2
s.t.
CPI1
+ DAP1
– INV
>
3700
INV
+ CPI2
+ DAP2
>
4200
CPI1
<
3800
CPI2
<
3800
INV
<
300
CPI1, CPI2, DAP1, DAP2, INV
>
0
LP OPTIMUM FOUND AT STEP 4
OBJECTIVE FUNCTION VALUE
1) 82175.00
VARIABLE VALUE REDUCED COST
CPI1 3800.000000 0.000000
DAP1 0.000000 0.250000
INV 100.000000 0.000000
CPI2 3800.000000 0.000000
DAP2 300.000000 0.000000
ROW SLACK OR SURPLUS DUAL PRICES
2) 0.000000 -12.250000
3) 0.000000 -13.750000
4) 0.000000 2.000000
5) 0.000000 3.500000
6) 200.000000 0.000000
NO. ITERATIONS= 4
RANGES IN WHICH THE BASIS IS UNCHANGED:
OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE
COEF INCREASE DECREASE
CPI1 10.250000 2.000000 INFINITY
DAP1 12.500000 INFINITY 0.250000
INV 1.500000 2.000000 0.250000
CPI2 10.250000 3.500000 INFINITY
DAP2 13.750000 0.250000 2.000000
RIGHTHAND SIDE RANGES
ROW CURRENT ALLOWABLE ALLOWABLE
RHS INCREASE DECREASE
2 3700.000000 100.000000 200.000000
3 4200.000000 INFINITY 300.000000
4 3800.000000 200.000000 100.000000
5 3800.000000 300.000000 3800.000000
6 300.000000 INFINITY 200.000000
To receive full credit You should be able to justify your answer based on the output alone without resolving the linear program.
a.What is the optimal solution including the optimal value of the objective function?
b.If the per-unit inventory cost increased from $1.50 to $2.50, would the optimal solution change? Would the optimal value of the objective function change?
c.If in quarter 2 CPI’s per-seat production cost increased by $1.25 and DAP changed its mind about the announced price increase (thus leaving it at $12.50 per seat), would the optimal solution change?
d.If DAP reduced its per seat selling price in quarter 1 from $12.50 to $12.30, should CPI purchase any seats in quarter 1?
e.How much is it worth to CPI to increase its inventory capacity from 300 to 400?
f.If CPI increased its production capacity by 100 seats in both quarters 1 and 2, what would be the savings for CPI (ignoring the capacity expansion expense)?
min
10.25CPI1
+ 12.5DAP1
+ 1.5INV
+ 10.25CPI2
+ 13.75DAP2
s.t.
CPI1
+ DAP1
– INV
>
3700
INV
+ CPI2
+ DAP2
>
4200
CPI1
<
3800
CPI2
<
3800
INV
<
300
CPI1, CPI2, DAP1, DAP2, INV
>
0
Explanation / Answer
a. OBJECTIVE FUNCTION VALUE - 82175.00
VARIABLE VALUE
CPI1 3800.000000
DAP1 0.000000
INV 100.000000
CPI2 3800.000000
DAP2 300.000000
b. By increasing the inventory holding cost from $ 1.5 to $2.5 , optimal solution will not change, as it is within the allowable range. Optimal value of objectvie function will increase by $100.
c. Optimal solution will not change
d. No. CPI should buy in qtr 1 only if the price drops below 12.25
e. It is not worth to increase the inventory capacity, as it does not lead to any cost benefit. the upper allowable range of inventory capacity is infinite, so increasing the capacity to any quantity is not worthwhile
f. this will result in total savings of $ 550. ( = 100*(13.75-10.25-1.5) + 100*(13.75-10.25)
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