2. A manufacturing company produces three products at two different plants. The
ID: 426934 • Letter: 2
Question
2. A manufacturing company produces three products at two different plants. The cost of producing a unit at each plant is shown in Table 1. 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 10,000 z22 10,000 2 6,000 +22 8,000 35,000 X21 X12 All variables let, xij = number of units of product j produced at plant i. Use the LINDO output to answer the questions a )What would the cost of producing product 2 at plant 1 have to be for the firm to make this choice? b)What would total cost be if plant 1 had 9,000 units of capacity? c )If it cost S9 to produce a unit of product 3 at plant 1, then what would be the new optimal solution? Tablel PlantI Product 10Explanation / Answer
a) The optimal function shows that the value of x12 as 0. This means that the cost of production of product 2 at plant 1 will be 0.
b) We can see that the row number 2 corresponds to the plant 1 constraint. Now it has a dual price of 2. This means that for every unit reduced for plant 1 constraint, the objective function will increase by 2 units. If we reduce the plant capacity to 9000 (1000 reduction) then the objective function will increase by 2*1000 = 2000. Currently the optimal solution is 128000, it will become 130000.
c) Product 3 at plant 1 is represented by x13. We can see in the section “Ranges in which the basis is unchanged”, that x13 has an allowable increase of 1. This means that if we increase x13 from 8 to 9 the basis will not change. This interprets as only the optimal solution will change. Now from objective function value we know that x13 has 4000 units. This means that if we increase the cost of x13 to 9 the optimal solution will increase in value by 4000*1 = 4000. The new optimal solution will be 128000+4000 = 132000
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