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What is the maximum profit? $ Steinwelt Piano manufactures uprights and consoles

ID: 3122075 • Letter: W

Question

What is the maximum profit? $ Steinwelt Piano manufactures uprights and consoles in two plants, Plant I and Plant II. The output of Plant I is at most 300/month, whereas the output of Plant II is at most 250/month. These pianos are shipped to three warehouses that serve as distribution centers for the company. To fill current and projected orders, Warehouse A requires a minimum of 200 pianos/month, Warehouse B requires at least 150 pianos/month, and Warehouse C requires at least 200 pianos/month. The shipping cost of each piano from Plant I to Warehouse A, Warehouse B, and Warehouse C is $45, $45, and $65, respectively, and the shipping cost of each piano from Plant II to Warehouse A, Warehouse B, and Warehouse C is $65, $55, and $35, respectively. Use the method of this section to determine the shipping schedule that nutrition will enable Steinwelt to meet the warehouses' requirements while keeping the shipping costs to a minimum. Plant I to Warehouse A pianos. Plant I to Warehouse B pianos. Plant I to Warehouse C pianos. Plant II to Warehouse A pianos. Plant II to Warehouse B pianos. Plant II to Warehouse C pianos. What is the minimum cost? $

Explanation / Answer

WAREHOUSES

A

B

C

SUPPLY

PLANT I

45

45

65

300

PLANTII

65

55

35

250

DEMAND

200

150

200

Initial solution using Vogel’s method is shown below  

WAREHOUSES

          Row differences

A

B

C

SUPPLY

PLANT I

45 [200]

45 [100]

65

300

20

0

PLANTII

65

55 [50]

35   [200]

250

10

10

DEMAND

200

150

200

Column differences

20

10

30

20

10

-----

Checking the optimality of solution using Modified Distribution method

Step 1: Make equations of the form Ui + Vj = Cij for allocated cells. Where Ui is the row number, Vj – column number, Cij – cost of (I,j)th cell.

From the above table, equations are

U1 +V1 = 45

U1+V2 = 45

U2 +V2 = 55

U2 +V3 = 35

Solving the equations by putting U1 =0, we have V1 = 45, V2 =45, U2 =10, V3 = 25

Step 2: construct Ui+Vj table for unoccupied cells

WAREHOUSES

V1= 45

V2 = 45

V3 = 25

U1= 0

X

X

25

U2=10

55

X

X

Step 3: construct Cij table for unoccupied cells

WAREHOUSES

A

B

C

PLANT I

X

X

65

PLANT II

65

X

X

Step 4 : construct D ij = Cij- (Ui+Vj )table for unoccupied cells

WAREHOUSES

A

B

C

PLANT I

X

X

40

PLANT II

10

X

X

Since all values in Dij 0, the solution obtained in the initial stage is optimum. The solution is

                        

WAREHOUSES

A

B

C

SUPPLY

PLANT I

45 [200]

45 [100]

65

300

PLANTII

65

55 [50]

35   [200]

250

DEMAND

200

150

200

20

10

-----

Therefore,

Plant I to Warehouse A = 200 pianos

Plant I to Warehouse B = 100 pianos

Plant I to Warehouse C = 0 pianos

Plant II to Warehouse A = 0 pianos

Plant II to Warehouse B = 50 pianos

Plant II to Warehouse C = 200 pianos

Minimum cost is = 45*200 + 45 * 100 + 55* 50 + 35 * 200 = 23250 $

WAREHOUSES

A

B

C

SUPPLY

PLANT I

45

45

65

300

PLANTII

65

55

35

250

DEMAND

200

150

200

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