CHAPTER VI
Transportation Models
MODI-STEPPING STONE METHODS
Instructor : Lect.Volkan ÇETİNKAYA
1.
Determine a starting basic feasible solution.
2.
m: The number of row and n: the number of
column.When we make an allocation into
m+n-1 pieces cell, we have a basic feasible
solution.
3.
Use Modi or Stepping Stone Methods to find the
optimal solution.
 SunRay Transport Company ships truckloads of grain three silos
to four mills. The supply (in truckloads) and the demand (also in
truckloads) together with the unit transportation costs per
truckload on the different routes are summerized in the
transportation model on the table below.The unit
transportation costs,Cij,(shown in northeast corner of each box)
are hundreds of dollars. The model seeks the minimum cost of
shipping schedule Xij between silo i and mill j.
1.
Northwest-Corner Method
2.
Least Cost Method
3.
Vogel Approximation Method(VAM)
 VAM yields the best starting solution, Northwest Corner Method
yields the worst.
 On the contrary, Northwest Corner involves the least amount of
computation.
1.
Start at the northwest corner cell. Allocate as much as
possible to the selected cell and adjust the associated
amounts of supply and demand by subtracting the allocated
amount.
2.
Cross out the row or column with zero supply or demand to
indicate that no further assignments can be made in that
row or column.
3.
If exactly one row or column is left uncrossed out, STOP!
1.
Start at the cell which has the least unit transportation
cost. Allocate as much as possible to the selected cell
and adjust the associated amounts of supply and
demand by subtracting the allocated amount.(in a case
of equal cost, we can select the cell randomly.)
2.
The quality of this method is better than that of the
northwest-corner method.
 VAM is an improved version of the least-
cost method that generally , but not
always, produces better starting
solutions.
1.
For each row(column), determine a penalty measure by
subtracting the smallest unit cost element in the
row(column) from the next smallest unit cost element in the
same row(column)
2.
Identify the row or cloumn with the largest penalty.Allocate
as much as possible to the variable with the least unit cost in
the selected row or column.Adjust the supply and demand,
and cross out the satisfied row or column. If a row or column
are satisfied simultaneously, only one of the two is crossed
out, ant the remaining is assigned zero supply.
3.
(a)If exactly one row or column with zero supply or
demand remains uncrossed out, STOP!
(b)If one row(column) with positive supply(demand)
remains uncrossed out, determine the basic variables in the
row(column)by the least cost method and STOP!
(c)If all the uncrossed out rows and columns
have(remaining) zero supply and demand, determine the zero
basic varibles by the least cost method.STOP!
(d) Otherwise, go to step-1 again.
1.
Check the problem is balanced or not.If not, balance
the demand and supply.
2.
Find the starting solution by North East Method
3.
Check if there are (m+n-1) pieces variables(coeff.)
4.
Compute ui and vj values belonging Xij existing in
basic feasible solution.
5.
For each Xij which are not basic variable, calculate
Cij= Ui + Vj indirect costs.
6.
If
Cij – Cij ≤ 0
STOP! OPTIMAL SOLUTION IS FOUND.
As long as there is one result proving
GO ON ...
Cij – Cij > 0
1.
Find starting solution by any method.
2.
Determine the empty cells and find differences by creating a
loop.
3.
Select the minimum difference and make its cell entering
valuation.
4.
Determine the quantity of entering value and construct the
new solution.
5.
Determine if it is optimal by applying 2nd step.
6.
If all differences > 0 STOP! Optimal Solution found...
 Find the starting solution.
 Find the optimal solution by means of Stepping
Stone Method and MODI.
F1
F2
F3
D1
4
16
8
72
D2
8
24
16
102
D3
8
16
24
41
56
82
77
 Find the starting solution.
 Find the optimal solution by means of MODI
Method.
F1
F2
F3
D1
1
4
0
40
D2
2
3
2
60
D3
3
2
2
80
D4
4
0
1
60
60
80
100
 Find the starting solution.
 Find the optimal solution by means of MODI
Method
F1
F2
F3
D1 D2 D3 D4
2
5
7
8
3
4
10
3
1
9
8
12
100 130 180 120
150
178
202
 The transshipment models recognizes
that in real life it may be cheaper to ship
through intermediate nodes before
reaching the final destination.
Sources
Distribution Center
Destinations
 Two automotive plants, P1 and P2, are linked to five dealers, D1, D2,
D3, D4 and D5, by way of three transit center, T1, T2 and T3.
 The supply amounts at plants P1 and P2 are 125 and 175 cars, and
the demand amount at dealers, D1, D2, D3, D4 and D5 are 45, 50, 85,
50 and 70 cars.
 The capacity of the transit centers, T1, T2 and T3 are 100, 110 and 90.
 The shipping cost per car between pairs of nodes are shown on the
table.
 The way between T3 and D4 is under construction.
 Find the minimum transshipment cost.
S
T1 T2 T3 D1
P1 70 80 65
İ
D2
İ
D3
İ
D4
İ
D5
İ
125
P2 72
78
73
İ
İ
İ
İ
İ
175
T1
0
İ
İ
12
5
7
10
9
100
T2
İ
0
İ
10
6
5
9
7
110
T3
İ
İ
0
8
7
4
İ
9
90
100 110 90
45
50
85
50
70
600
600
D1
P1
P2
35
90
T1
D2
50
T2
65
50
110
T3
40
D3
45
45
70
D4
D5
T1
T2
T3
D1
D2
D3
D4
D5
P1
70
80
65
İ
İ
İ
İ
İ
125
P2
72
78
73
İ
İ
İ
İ
İ
175
T1
0
İ
T2
İ
T3
İ
0
İ
İ
12
5
7
10
9
100
İ
10
6
5
9
7
110
8
7
4
9
90
0
100 110 90
45
50
I
85
50
70