How to solve the transportation problem in modi method?
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Example: MODI Method (Modified Distribution Method)
Consider the transportation problempresented in the following table.
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Distribution centre D1D2D3D4SupplyPlantP1193050127P27030406010P34010602018Requirement 58715
Determine the optimal solution of the above problem.
Solution.
An initial basic feasible solution is obtained by Matrix Minimum Method and is shown in table 1.
Table 1
Distribution centre D1D2D3D4SupplyPlantP11930507P2306010P36018Requirement 58715
Initial basic feasible solution
12 X 7 + 70 X 3 + 40 X 7 + 40 X 2 + 10 X 8 + 20 X 8 = Rs. 894.
Calculating ui and vj using ui + vj = cij
Substituting u1 = 0, we get
u1 + v4 = c14 ⇒ 0 + v4 = 12 or v4 = 12
u3 + v4 = c34 ⇒ u3 + 12 = 20 or u3 = 8
u3 + v2 = c32 ⇒ 8 + v2 = 10 or v2 = 2
u3 + v1 = c31 ⇒ 8 + v1 = 40 or v1 = 32
u2 + v1 = c21 ⇒ u2 + 32 = 70 or u2 = 38
u2 + v3 = c23 ⇒ 38 + v3 = 40 or v3 = 2
Consider the transportation problempresented in the following table.
On small screens, use horizontal scrollbar to view full table
Distribution centre D1D2D3D4SupplyPlantP1193050127P27030406010P34010602018Requirement 58715
Determine the optimal solution of the above problem.
Solution.
An initial basic feasible solution is obtained by Matrix Minimum Method and is shown in table 1.
Table 1
Distribution centre D1D2D3D4SupplyPlantP11930507P2306010P36018Requirement 58715
Initial basic feasible solution
12 X 7 + 70 X 3 + 40 X 7 + 40 X 2 + 10 X 8 + 20 X 8 = Rs. 894.
Calculating ui and vj using ui + vj = cij
Substituting u1 = 0, we get
u1 + v4 = c14 ⇒ 0 + v4 = 12 or v4 = 12
u3 + v4 = c34 ⇒ u3 + 12 = 20 or u3 = 8
u3 + v2 = c32 ⇒ 8 + v2 = 10 or v2 = 2
u3 + v1 = c31 ⇒ 8 + v1 = 40 or v1 = 32
u2 + v1 = c21 ⇒ u2 + 32 = 70 or u2 = 38
u2 + v3 = c23 ⇒ 38 + v3 = 40 or v3 = 2
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