# Geopolitica

Páginas: 17 (4164 palabras) Publicado: 6 de marzo de 2011
PROBLEMA 1
1. Gepbab Corporation produces three products at two different plants. The cost of producing a unit at each plant is shown in Table 6. Each plant can produce a total of 10,000 units. At least 6,000 units of product 1, at least 8,000 units of product 2, and at least 5,000 units of product 3 must be produced. To minimize the cost of meeting these demands, the following LP should besolved:

MIN 5 X11 + 6 X12 + 8 X13 + 8 X21 + 7 X22 + 10 X23
SUBJECT TO
X11 + X12 + X13 <= 10000
X21 + X22 + X23 <= 10000
X11 + X21 >= 6000
X12 + X22 >= 8000
X13 + X23 >= 5000
END

Here, xij = number of units of product j produced at plant i. Use the LINDO output in Figure 9 to answer the following questions:

LP OPTIMUM FOUND AT STEP 4

OBJECTIVE FUNCTIONVALUE

1) 128000.0

VARIABLE VALUE REDUCED COST
X11 6000.000000 0.000000
X12 0.000000 1.000000
X13 4000.000000 0.000000
X21 0.000000 1.000000
X22 8000.000000 0.000000
X23 1000.000000 0.000000

ROW SLACK OR SURPLUSDUAL PRICES
2) 0.000000 2.000000
3) 1000.000000 0.000000
4) 0.000000 -7.000000
5) 0.000000 -7.000000
6) 0.000000 -10.000000

NO. ITERATIONS= 4

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES
VARIABLECURRENT ALLOWABLE ALLOWABLE
COEF INCREASE DECREASE
X11 5.000000 1.000000 7.000000
X12 6.000000 INFINITY 1.000000
X13 8.000000 1.000000 1.000000
X21 8.000000 INFINITY 1.000000
X22 7.000000 1.000000 7.000000X23 10.000000 1.000000 1.000000

RIGHTHAND SIDE RANGES
ROW CURRENT ALLOWABLE ALLOWABLE
RHS INCREASE DECREASE
2 10000.000000 1000.000000 1000.000000
3 10000.000000 INFINITY 1000.000000
4 6000.000000 1000.0000001000.000000
5 8000.000000 1000.000000 8000.000000
6 5000.000000 1000.000000 1000.000000

1. What would the cost of producing product 2 at plant 1 have to be for the firm to make this choice?

2. What would total cost be if plant 1 had 9,000 units of capacity?

MIN 5 X11 + 6 X12 + 8 X13 + 8 X21 + 7 X22 + 10 X23
SUBJECT TO
X11 + X12+ X13 <= 9000
X21 + X22 + X23 <= 10000
X11 + X21 >= 6000
X12 + X22 >= 8000
X13 + X23 >= 5000
END

LP OPTIMUM FOUND AT STEP 2

OBJECTIVE FUNCTION VALUE

1) 130000.0

VARIABLE VALUE REDUCED COST
X11 6000.000000 0.000000
X12 0.000000 1.000000
X13 3000.0000000.000000
X21 0.000000 1.000000
X22 8000.000000 0.000000
X23 2000.000000 0.000000

ROW SLACK OR SURPLUS DUAL PRICES
2) 0.000000 2.000000
3) 0.000000 0.000000
4) 0.000000 -7.000000
5) 0.000000 -7.000000
6)0.000000 -10.000000

NO. ITERATIONS= 2

RANGES IN WHICH THE BASIS IS UNCHANGED:

OBJ COEFFICIENT RANGES
VARIABLE CURRENT ALLOWABLE ALLOWABLE
COEF INCREASE DECREASE
X11 5.000000 1.000000 7.000000
X12 6.000000 INFINITY 1.000000...

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