modi method
DESCRIPTION
TRANSCRIPT
WHAT IS MODI METHOD
• It is used to save the time over stepping stone method
• It provides a new means of finding the unused route with the largest negative improvement index.
• Once largest index identified ,we are required to trace only one path, just as with the stepping stone approach, this helps to determine the maximum number of unit that can be shipped by the best unused route.
STEPS
1.Construct a transportation table with the given cost of transportation and rim requirement.
2.Determine IBFS.
3.For current basic feasible solution check degeneracy and non-degeneracy.
rim requirement=stone square(non-degeneracy)
rim requirement != stone square(degeneracy)
4.Find occupied matrix.
5.Find unoccupied matrix.
Steps (contd…)
6.Find opportunity cost of unoccupied cells using formula:
opportunity cost =actual cost-implied cost dij= cij - (ri+kj)7.Unoccupied cell evaluation: (a) if dij>0 then cost of transportation
unchanged. (b) if dij=0 then cost of transportation
unchanged. (c) if dij<0 then improved solution can be obtain
and go to next step.
STEPS(contd…)
8.Select an unoccupied cell with largest –ve opportunity cost among all unoccupied cell.
9.Construct closed path for the occupied cells determined in step 8.
10.Assign as many as units as possible to the unoccupied cell satisfying rim conditions.
11.Go to step 4 and repeat procedure until
All dij>=0 i.e reached to the optimal solution.
SPECIAL CASES
• Balanced problem• Unbalanced problem• Non -degeneracy• Degeneracy :occurs in two cases
1. Degeneracy occurs in initial basic solution.
2. Degeneracy occurs in during the test of optimality.
• Profit maximization
.
PROBLEM1:Shipping costs are Rs. 10 per kilometer.. What shipping schedule should be used. if the matrix given below the kilometers from source to destination.
destination a b c availability
x 50 30 220 1 y 90 45 170 3 z 50 200 50 4
Requirement 3 3 2 8
A B C
X
Y
Z
30 220
50
90 45 170
50 200 50
30 220
1
2
3
2
2
2 2
3
P1 P2 P3 P4
20 20 20 20
45 45 45 _
150 150 _ _
P1 40 15 120
P2 40 15 _
P3 40 15 _
IBFS BY USING VAM METHOD
3030 220
90 45 170
50 200 50
1 E
3
2 2
STONE SEQUARE =4
RIM REQUIREMENT =M+N-1=3+3-1=5
DEGENERACY
NOW STONE SEQUARE =5
RIM REQUIREMENT =5
NON –DEGENERACY
50 30
45
50 50
50 30 50
0
15
0
OCCUPIED MATRIX UNOCCUPIED MATRIX
10525
170
50
65
30
65
170
OPTIMUL SOLUTION
X A 50*1=50
X B 30*E=_
Y B 45*3=135
Z A 50*2=100
Z C 50*2=100
385 *10=3850
PROBLEM2:DETERMINE THE OPTIMUM SOLUTION FOR THE COMPANY OF TRASPOTATION PROBLEM(USING NWCM AND MODI METHOD)
8 8 15
15 10 17
3 9 10
REQUIREMENT 150 80 50
120
80
80
CAPACITY
F1
F2
F3
W1 W2 W3
WAREHOUSE
FACTORY
W1 W2 W3
F1
F2
F3
8 8 15
15
3
10 17
9 10
120
30 50
30 50
150 80 50
120
80
80
IBFS WITH NWCM
OCCUPIED MATRIX UNOCCUPIED MATRIX
8
15 10
9 10
15 10 11
-7
0
-1
15 10 11
-7
0
-1
5 11
6
-11
3 4
11
14
1588
15 10 17
3 9 10
10
120
30 50
30 50+
+_
_
8
15
3
8 15
10 17
9 10
120 E
80
30 50
STONE SEQUARE=RIM REQUIREMENTDEGENERACY OCCUAR
LOOP CONSTRUCT
OCCUPIED MATRIX UNOCCUPIED MATRIX
8 8
10
3 10
3 3 10 3 3 10
5
7
0
5
7
0
5
6
0
0
15
1710
3
OPTIMUM SOLUTIONF1 W1 8*120 =960
F1 W2 8*E = _
F2 W2 10*80 =800
F3 W1 3*30= 90
F3 W3 10*50 =500
2,350 RS
