A company must meet the following demands for a product: January. 30 units; Febr
ID: 458191 • Letter: A
Question
A company must meet the following demands for a product: January. 30 units; February, 30 units; Match, 20 units. Demand may be backlogged at a cost of $5/unit/month. All demand must be met by the end of March. Thus, if 1 unit of January demand is met during March, a backlogging cost of 5(2) = $10 is incurred. Monthly production capacity and unit production cost during each month are given in Table 66. A holding cost of $20/unit is assessed on the inventory at the end of each month. a Formulate a balanced transportation problem that could be used to determine how to minimize the total cost (including backlogging, holding, and production costs) of meeting demand. b Use Vogel's method to find a basic feasible solution, c Use the transportation simplex to determine how to meet each month's demand. Make sure to give an interpretation of your optimal solution (for example, 20 units of month 2 demand is met from month 1 production).Explanation / Answer
a) Please find the matrix for transportation problem
January
February
March
Dummy
Supply
January
400
420
440
0
35
February
425
420
440
0
30
March
420
415
410
0
35
Demand
30
30
20
20
We have added a dummy of 20 to balance the supply and demand.
b) Vogel's method is same as northwest corner method.
January
February
March
Dummy
Supply
January
400
30
420
5
440
0
35
February
425
420
25
440
5
0
30
March
420
415
410
15
0
20
35
Demand
30
30
20
20
c) TPP
January
V1
February
V2
March
V3
Dummy
V4
Supply
January
U1
400
30
420
5
440
0
35
February
U2
425
420
25
440
5
0
30
March
U3
420
415
410
15
0
20
35
Demand
30
30
20
20
u1 = 0
u1 + v1 = 400, v1 = 400 [c13] = u1 + v3 –c13 = 0+440 – 440 = 0
[c14] = u1 + v4 – c14 = 0+ 30 -0 = 30
u1 + v2 = 420, v2 = 420, [c21] = u2 + v1 –c21 = 0 + 400 -425 = -25
u2 + v2 = 420, u2 = 0, [c24] = u2 + v4 – c24 = 0 + 30 -0 = 30
u2 + v3 = 440, v3 = 440, [c31] = u3 + v1 – c31 = -30 +400 – 420 = -50
u3 + v3 = 410, u3 = -30 [c32] = u3 + v2 – c32 = -30 + 420 – 415 = -25
u3 + v4 = 0, v4 = 30
c24 will enter the bases and c23 will leave the bases.
January
V1
February
V2
March
V3
Dummy
V4
Supply
January
U1
400
30
420
5
440
0
35
February
U2
425
420
25
440
0
5
30
March
U3
420
415
410
20
0
15
35
Demand
30
30
20
20
u1 = 0 [c13] = u1 + v3 – c13 = 0 + 410 -440 = -30
u1 + v1 = 400, v1 = 400 [c14] = u1 + v4 – c14 = 0 + 0 – 0 = 0
u1 + v2 = 420, v2 = 420 [c21] = u2 + v1 – c21 = 0 + 400 -425 = -25
u2 + v2 = 420, u2 = 0 [c23] = u2 + v3 - c23 = 0 + 410 – 440 = -30
u2 + v4 = 0, v4 = 0 [c31] = u3 + v1 – c31 = 0 + 400 – 420 = -20
u3 + v4 = 0, u3 = 0 [c32] = u3 + v2 – c32 = 0+420 – 415 = 5
u3 + v3 = 410, v3 = 410
c32 will enter the bases and c34 should leave the bases.
January
V1
February
V2
March
V3
Dummy
V4
Supply
January
U1
400
30
420
5
440
0
35
February
U2
425
420
10
440
0
20
30
March
U3
420
415
15
410
20
0
35
Demand
30
30
20
20
u1 = 0 [c13] = u1 + v3 – c13 = 0 + 415 – 440 = -25
u1 + v1 = 400, v1 = 400 [c14] = u1 + v4 – c14= 0 + 0 – 0 = 0
u1 + v2= 420, v2 = 420 [c21] = u2 + v1 – c21 = 0 + 400 – 425 = -25
u2 + v2 = 420, u2 = 0 [c23] = u2 + v3 – c23 = 0 + 415 – 440 = -25
u2 + v4 = 0, v4 = 0 [c31] = u3 + v1 – c31 = -5 + 400 – 420 = -25
u3 + v2 = 415, u3 = -5 [c34] = u3 + v4 – c34 = 0 + 0 – 0 = 0
u3 + v3 = 410, v3 = 415
This means that the tableau is optimal, [c14] =0 and [c34] =0 means that the tableau has alternative solutions.
It shows that the company should produce 35 units in January. It should sell 30 units of them in January and keep remaining 5 units to sell in February. The company also should produce 10 units in February and sell it in this month. In March it should produce 15 units for backlogged demand of February and 20 units for March demand.
Min z= 30*400+5*420+10*420+20*0+15*415+20*410 = 32725
January
February
March
Dummy
Supply
January
400
420
440
0
35
February
425
420
440
0
30
March
420
415
410
0
35
Demand
30
30
20
20
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