optimal distribution scheme to minimize costs using the Northwest Corner Rule an
ID: 413349 • Letter: O
Question
optimal distribution scheme to minimize costs using the Northwest Corner Rule and Stepping Stone Method. 5. Consider a distribution network with the below-stated costs, capacities, and requirements. Solve for the Dealer Dealer Deaer SUpPLY 46 36 400 44450 43 37 40 37 650 Plant 1 Plant 2 Plant 3 39 600 DEMAND 400 Dealer Dealer Dealer SUPPLY 43 46 Plant 1 Plant 2 Plant 3 37 40 37 650 400 450 39 400 500 600 Dealer Dealer Dealer SUPPLY 2 43 46 Plant 1 Plant 2 Plant 3 37 40 37 650 36 400 450 39 400 500 600Explanation / Answer
NWCM method does not take into account the shipping cost. The steps followed for this method can be summarized as:
Step 1: For first allocation, start with the cell at the upper-left corner of the table, that is, O1D1 cell. The allocation unit will be equal to the minimum of supply or demand, that is, . Then, update the balanced units of supply and demand in the table. If the supply is exhausted, eliminate that source and if the demand is satisfied, then eliminate that destination by striking out that source or destination from transportation table.
Step 2: For the next allocation, if the source is eliminated, then move vertically down to the next cell. However, if the destination is eliminated, move horizontally to the right of the allocated cell.
Dealer
Supply
Plant
D1
D2
D3
P1
37
43
46
650
400
250
P2
40
39
36
400
0
400
P3
37
44
44
450
250
200
Demand
400
500
600
Optimal Solution by Stepping stone method
In the Stepping-Stone method, increase or decrease in the total cost is identified by assigning one unit for each unallocated cell one-by-one. The steps to evaluate the unoccupied cells by this method are:
Step 1: Consider the IBF solution, select any unoccupied cell and assign +1 unit for that cell. Form a closed loop starting from the same cell, with alternative horizontal and vertical lines.
Step 2: Alternatively mark 1 and +1 at the corners of the loop. If +1 is marked at one end of the line, then 1 should be marked at the other end.
Step 3: Calculate change in allocation cost for the cells in loop as:
Repeat steps 1 to 3 for each unallocated cell.
Step 4: If dij > 0 for every unallocated cell, then the solution is optimal, further improvement is not required. However, if dij < 0 for at least one of the unoccupied cells, then improve the solution by following further steps.
Step 5: Cell with most negative dij value is selected for new allocation. In the loop of this cell, identify the lowest quantity () from the cells in the loop marked with 1 and allocate the same unit () to the cell with the most negative dij value.
Step 6: Perform addition or subtraction of units at each corner of loop to satisfy the rim condition.
Again, check for optimality for the new allocation by repeating steps 1 to 6.
Unallocated Cell
Closed Loop
Opportunity Cost
(dij)
P1D3
P1D3-P2D3-P2D2-P1D2-P1D3
d13 = + 46 – 36 + 39 – 43 = 6
P2D1
P2D1-P1D1-P1D2-P2D2-P2D1
d21 = + 40 – 37 + 43 – 39 = 7
P3D1
P3D1-P1D1-P1D2-P2D2-P2D3-P3D3-P3D1
d31 = + 37 – 37 + 43 – 39 + 36 – 44 = - 4
P3D2
P3D2-P2D2-P2D3-P3D3-P3D2
d32 = + 44 – 39 + 36 – 44 = -3
Maximum cost reduction is -4 for cell P3D1, if one unit is reallocated in this cell the cost reduce by $4. Thus, given solution is not optimal.
Here, improvement is done in allocation by considering P3D1 as an incoming variable. Now, let + be the maximum allocation unit for incoming cell P3D1. Now, the value is obtained as a minimum allocation value among the cells which are marked with –1.
Here, = minimum (400, 250, 450) = 250 units.
Thus, new allocation for the cells in loop will be:
Rechecking of optimality
Unallocated Cell
Closed Loop
Opportunity Cost
(dij)
P1D3
P1D3-P3D3-P3D1-P1D1-P1D3
d13 = + 46 – 44 + 37 – 37 = 2
P2D1
P2D1-P2D3-P3D3-P1D1-P2D1
d21 = + 40 – 36 + 44 – 37 = 11
P2D2
P2D2-P2D3-P3D3-P3D1-P1D1-P1D2-P2D2
d22 = + 39 – 36 + 44 – 37 + 37 – 43 = 15
P3D2
P3D2-P3D1-P1D1-P1D2-P3D3
d32 = + 44 – 37 + 37 – 43 = 1
The change in allocation cost for all the unallocated cells are greater than or equal to zero, thus optimal solution is obtained.
Optimal Allocations and cost of allocation is as follows:
Plant
Dealer
Allocation unit
Cost
1
1
150
150*37 = 5,550
1
2
500
500*43 = 21,500
2
3
400
400*36 = 14,400
3
1
250
250*37 = 9,250
3
3
200
200*44 = 8,800
Total Cost
$59,500
Dealer
Supply
Plant
D1
D2
D3
P1
37
43
46
650
400
250
P2
40
39
36
400
0
400
P3
37
44
44
450
250
200
Demand
400
500
600
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