Lecture8- Transportataion-SPP.ppsx
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Transportation (Cont.)
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Transcript of Lecture8- Transportataion-SPP.ppsx
Transportation(Cont.)
Transportation
1. Find a starting solution
Solving Procedure:
•North-west corner method•Least cost method•Vogel’s approximation method
2. Improve the solution by an entering
variable, and a leaving variable.
Stepping stone method
10 0 20 11
5 10
12 7 9 20
5 15 5
0 14 16 18
5
Improving the north-west corner solution
+18 -2
-5
-15 +9 +9
+0-10+0-7+20-18 = - 15
Multipliers Method (U-V method)
10 0 20 11
5 10
12 7 9 20
5 15 5
0 14 16 18
5
Table 1Improving the north-west corner solution
u1
u2
u3
v1v2 v3 v4
= 0
= 7
= 5
= 10 = 0 = 2 = 13
jiij vuc For all cells of basic variables
Multipliers Method (U-V method)
10 0 20 11
5 10
12 7 9 20
5 15 5
0 14 16 18
5
Table 1Improving the north-west corner solution
+18 -2
-5
-15 +9 +9
u1
u2
u3
v1v2 v3 v4
= 0
= 7
= 5
= 10 = 0 = 2 = 13
jiijij vucc For all cells of non- basic variables
U-V method
10 0 20 11
5 10
12 7 9 20
5 15 5
0 14 16 18
5
Table 2
Improving the north-west corner solution
• The number of variables should be kept 6.
5
0 15
0 10
• Re-evaluate u and v to calculate c at each cell. And so on
Destination1 2 3 4
Source
110 0 20 11
212 7 9 20
30 14 16 18
Demand 5 15 15 10
Table 3
Sum (Supply) = Sum (Demand)
Non-Standard transportation problems
45
/
/55
45
Add a dummy destination (D).
Supply
15
25
515
Table 3
Transportation table
55
Add a dummy destination (D).
DestinationSupply
1 2 3 4
Source
110 0 20 11
15
212 7 9 20
25
30 14 16 18
15
Demand 5 15 15 10
D
0
0
0
10
Questions on Transportation
Assignment
1 2 3
2 3 41
Workers (i) i = 1,2,…,m
Machines (j) j = 1,2,…,n
Problem Definition
c1,1
c1,2
c2,1 c3,4
Cost (ci,j )
4
c4,3
m
Given Data
1 2 3 41 1 4 6 3 12 9 7 10 9 13 4 5 11 7 14 8 7 8 5 1
1 1 1 1
Wor
kers
Machines
Solved Example
M1 M2 M3W1 5 7 9W2 14 10 12W3 15 13 16
-5
-10
-13
1- Subtract from each row the minimum value
M1 M2 M3W1 0 2 4W2 4 0 2W3 2 0 3
2- Subtract from selected column the minimum value
M1 M2 M3W1 0 2 2W2 4 0 0W3 2 0 1
-2
W1 M1W2 M3
W3 M2
Special cases for assignment problems1 2 3 4
1 1 4 6 32 9 7 10 93 4 5 11 74 8 7 8 5
-1
-7
-4
-5
1 2 3 41 0 3 5 22 2 0 3 23 0 1 7 34 3 2 3 0
-3
Special cases for assignment problems
1 2 3 41 0 3 2 22 2 0 0 23 0 1 4 34 3 2 0 0
- Connect all the zeros with the minimum number of lines;- Each zero should be crossed out only ONCE
- Subtract smallest uncrossed value from uncrossed elements- Add it at all intersections
Special cases for assignment problems
1 2 3 41 0 3 2 22 2 0 0 23 0 1 4 34 3 2 0 0
1 2 3 41234
2 1 1
0 3 23
4
00 0 2
02 0 0
Special cases for assignment problems
1 2 3 41 0 2 1 12 3 0 0 23 0 0 3 24 4 2 0 0
W1 M1W2 M3
W3 M2
W4 M4
- The procedure of connecting zeros should be repeated until a feasible solution is reached.
Special cases for assignment problems
NOTE:
Shortest path problems
Problem definition
- Given n nodes in a network that are denoted by
(1, 2, 3…n), it is required to find the shortest path
from the source node to the sink node.
Illustrative example
- Determine the shortest path from the source (1) to the sink (6), and the length of the path.
1
2
3
4 5
63
7
4
9
3
6
33
Source
Sink
1
2
Steps to solve
1
2
3
4 5
63
7
4
9
3
6
33
Source
Sink
- Start at node (1). Its distance is zero.- Distances to neighboring nodes are written.- Distances to non-neighboring nodes are infinity.
L(0) = [ 0 , 3 , 7 , 4 , ∞ , ∞ ]
1
2
Steps to solve
1
2
3
4 5
63
7
4
9
3
6
33
Source
Sink
L(0) = [ 0 , 3 , 7 , 4 , ∞ , ∞ ]
1
2
* .- Give the zero a permanent label (*).- Give the minimum value a temporary label (.)- Calculate the distances from the temporary label.
Steps to solve1 2 3 4 5 6
L(0) = [ 0 , 3 , 7 , 4 , ∞ , ∞ ]
L(1) = [ 0 , 3 , 5 , 4 , ∞ , 12 ]
From To Current New
2 3 7 3 + 2
2 4 4 3 + ∞
2 5 ∞ 3 + ∞
2 6 ∞ 3 + 9
1
2
3
4 5
63
7
4
9
3
6
33
1
2
Node of Temporary label
Nodes with no labels
Their current values
New values starting the temporary nodeCompare and choose the minimum
* .*
* .
- Temp. labels are permanent.- Choose new temp. label
Steps to solve1 2 3 4 5 6
L(1) = [ 0 , 3 , 5 , 4 , ∞ , 12 ]
L(2) = [ 0 , 3 , 5 , 4 , 7 , 12 ]
From To Current New
4 3 5 4+ 1
4 5 ∞ 4+ 3
4 6 12 4+ ∞1
2
3
4 5
63
7
4
9
3
6
33
1
2
Node of Temporary label
Nodes with no labels
Their current values
New values starting the temporary node
Compare and choose the minimum
* .*
* .
- Temp. labels are permanent.- Choose new temp. label
*
*
Steps to solve1 2 3 4 5 6
L(2) = [ 0 , 3 , 5 , 4 , 7 , 12 ]
L(3) = [ 0 , 3 , 5 , 4 , 7 , 11 ]
From To Current New
3 5 7 5+ 3
3 6 12 5+ 6
1
2
3
4 5
63
7
4
9
3
6
33
1
2
* .*
* .
- Temp. labels are permanent.- Choose new temp. label
*
*
*
*
Steps to solve1 2 3 4 5 6
L(3) = [ 0 , 3 , 5 , 4 , 7 , 11 ]
L(4) = [ 0 , 3 , 5 , 4 , 7 , 10 ]
From To Current New
5 6 11 7+ 3
1
2
3
4 5
63
7
4
9
3
6
33
1
2
**
* .*
*
*
*
*
* *- Shortest route distance is 10- Shortest path is ???
Solved Problem1 2 3 4 5 6
L(0) = [ 0 , 3 , 7 , 4 , ∞ , ∞ ]
L(1) = [ 0 , 3 , 5 , 4 , ∞ , 12 ]
L(2) = [ 0 , 3 , 5 , 4 , 7 , 12 ]
L(3) = [ 0 , 3 , 5 , 4 , 7 , 11 ]
L(4) = [ 0 , 3 , 5 , 4 , 7 , 10 ]* * * * * *
* * * * .
* * . *
* * .
* .
Path: 1 , 4 , 5 , 6
Illustrative example
1
2
3
4 5
63
7
4
9
3
6
33
Source
Sink
1
2
Path: 1 , 4 , 5 , 6
Another example
1
2
3
4
7
8
5
1
78
9 6
Source
Sink
2
45
63
10
93
5
- Given the roads of the city, go from (1) to (8) by the shortest route.
