Transportation Problem LECTURE 18 By Dr. Arshad Zaheer.
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Transcript of Transportation Problem LECTURE 18 By Dr. Arshad Zaheer.
Transportation Problem
LECTURE 18By
Dr. Arshad Zaheer
RECAP
Transportation model Purpose Initial Feasible Solution North West Corner Method Least Cost Method Optimal Solution Stepping Stone Method Modi Method
Illustration 1
• Minimize the transportation cost
Sources D1 D2 D3 Capacity
S1 2 4 3 15
S2 5 3 7 25
S3 8 7 3 30
Demand 10 20 4070
70
Initial Solution by North West Corner RuleSources Destination Capacity
D1 D2 D3
S12
Xij
4
Xij
3
Xij15
S25
Xij
3
Xij
7
Xij25
S38
Xij
7
Xij
3
Xij30
Demand 10 20 40 70
S1D1 box is the most north column so we fill this first. See its column total which is 10 and row total which is 15. we place the less no in the box so is 10. Now the column total has been exhausted/completely filled and we mark zero to below box and the row total decrease to 5.
Sources Destination Capacity
D1 D2 D3
S12
10
4
Xij
3
Xij5
S25
0
3
Xij
7
Xij25
S38
0
7
Xij
3
Xij30
Demand 0 20 40 70
S1D2 box is the most adjacent west column so we fill this first now. See its column total which is 20 and row total which is 5. we place the less no in the box so is 5. Now the row total has been completely filled and we give zero to respective row total and the column total decreases to 15.
Sources Destination Capacity
D1 D2 D3
S12
10
4
5
3
00
S25
0
3
Xij
7
Xij25
S38
0
7
Xij
3
Xij30
Demand 0 15 40 70
S2D2 box is the most adjacent north column to the previous cell just filled. So we fill this first. See its column total which is 15 and row total which is 25. We place the less no in the box so is 15. Now the column total has been exhausted and we assign zero to relevant column total and the row total decreases to 10.
Sources Destination Capacityy
D1 D2 D3
S12
10
4
5
3
00
S25
0
3
15
7
Xij10
S38
0
7
0
3
Xij30
Demand 0 0 40 70
S2D3 box is the most adjacent west column so we fill this first. See its column total which is 40 and row total which is 10. we place the less no in the box so is 10. Now the row total has been totally filled and the column total decrease to 30.
Sources Destination Capacity
D1 D2 D3
S12
10
4
5
3
00
S25
0
3
15
7
100
S38
0
7
0
3
Xij30
Demand 0 0 30 70
S3D3 box is the last column so we fill this first. See its column and row total which is 30 . So we place 30 in the box and both column and row balance will be utilized fully.
Every time while finding the initial solution the column and row total of last box will be equal as in the given case 30.
Sources Destination Capacity
D1 D2 D3
S12
10
4
5
3
00
S25
0
3
15
7
100
S38
0
7
0
3
300
Demand 0 0 0 70
Initial solutionSources Destination Capacity
D1 D2 D3
S12
10
4
5
3
015
S25
0
3
15
7
1025
S38
0
7
0
3
3030
Demand 10 20 40 70
No of Basic Variables= m+n-1=3+3-1=5
m= No of sourcesn= No of destinations
Sources Destination Capacity
D1 D2 D3
S12
10
4
5
3
015
S25
0
3
15
7
1025
S38
0
7
0
3
3030
Demand 10 20 40 70
Basic variables Non basic variables
Total cost
Total cost = Xij * CijTotal is calculated on the basis of all the basic
variables because non basic variables are zero and their cost will also becomes zero when multiplied with zero units.
= 10 x 2 +5 x 4 +15 x 3 + 10 x 7+30 x 3=245
Criteria for Optimality
• All the shadow cost, which must be non negativeShadow cost, for basic variables is always zero so there
is no need to find their shadow cost
Shadow cost:Vij = (Ui + Vj) –CijSo we need to find the U and V.
Shadow costSources Destination Capacity
D1 D2 D3
S12
10
4
5
3
0
15
U1
S25
0
3
15
7
10
25
U2
S38
0
7
0
3
30
30
U3
Demand 10
V1
20
V2
40
V3
70
• If we find the values of U1,U2,U3, and V1,V2 and V3 than we can easily find the shadow cost
• Write the equations for the basic variables
• U1+V1=2• U1+V2=4• U2+V2=3• U2+V3=7• U3+V3=3
• There are 6 basic variables and 5 equations so we will give one variable arbitrary value which is equal to zero
Let U2=0By putting this U2 zero we can easily find the value
of all other variables which will be as follows• U1=1 V1=1• U2=0 V2=3• U3=-4 V3=7
Sources Destination Capacity
D1 D2 D3
S10 2
10
4
5
5 3
0
15
U1=1
S25
0
3
15
7
10
25
U2=0
S38
0
7
0
3
30
30
U3=-4
Demand 10
V1=1
20
V2=3
40
V3=7
70
Shadow cost of S1D1Vij = (Ui + Vj) –CijV11 = (U1 + V1) –C11
=(1+1) -2=0
So save time and do not calculate the shadow cost for basic variables. Write the shadow cost on opposite side of cost in the column
Shadow cost of S1D3Vij = (Ui + Vj) –CijV13 = (U1 + V3) –C13
=(1+7) -3=5
Sources Destination Capacity
D1 D2 D3
S10 2
10
4
5
5 3
0
15
U1=1
S2-4 5
0
3
15
7
10
25
U2=0
S3-11 8
0
-8 7
0
3
30
30
U3=-4
Demand 10
V1=1
20
V2=3
40
V3=7
70
Our criteria of optimality is non positivity of shadow cost which is not satisfied as we have S1D3 box with positive shadow cost so we use θ
Sources Destination Capacity
D1 D2 D3
S10 2
10
4
5-θ
5 3
0+θ
15
U1=1
S2-4 5
0
3
15+θ
7
10-θ
25
U2=0
S3-11 8
0
-8 7
0
3
30
30
U3=-4
Demand 10
V1=1
20
V2=3
40
V3=7
70
Add θ in cell having largest positive shadow cost this will disturb the row and column balances for which you need to add and subtract θ from some other cells
• Max θ = Min of (10,5) =5
The boxes from which the θ has been subtractedReplace the θ with 5 and make new tableau.‘
Sources Destination Capacity
D1 D2 D3
S1 2
10
4
0
3
5
15
U1
S2 5
0
3
20
7
5
25
U2
S3 8
0
7
0
3
30
30
U3
Demand 10
V1
20
V2
40
V3
70
Again find the shadow cost by finding the same rules as discussed above
Total Cost:=10*2+5*3+20*3+5*7+30*3=220 (cost decreased as compared to previous tableau)
Equations:U1+V1=2U1+V3=3U2+V2=3U2+V3=7U3+V3=3
• Let U2=0By putting this U2 zero we can easily find the
value of all other variables which will be as follows
• U1=-4 V1=6• U2=0 V2=3• U3=-4 V3=7
Sources Destination Capacity
D1 D2 D3
S1 2
10
-5 4
0
3
5
15
U1= - 4
S21 5
0
3
20
7
5
25
U2=0
S3-6 8
0
-8 7
0
3
30
30
U3= - 4
Demand 10
V1= 6
20
V2=3
40
V3=7
70
Our criteria of optimality is non positivity of shadow cost which is not satisfied as we have S2D1 box with positive shadow cost so we use θ
Sources Destination Capacity
D1 D2 D3
S1 2
10-θ
-5 4
0
3
5+θ
15
U1= - 4
S21 5
0+θ
3
20
7
5-θ
25
U2=0
S3-6 8
0
-8 7
0
3
30
30
U3= - 4
Demand 10
V1= 6
20
V2=3
40
V3=7
70
We can no add θ in S3D2 because its above and below boxes have zero units and we can not balance in zero .
• Max θ = Min of (10,5) =5
The boxes from which the θ has been subtractedReplace the θ with 5 and make new tableau.
Sources Destination Capacity
D1 D2 D3
S1 2
5
4
0
3
10
15
U1=
S25
5
3
20
7
0
25
U2=
S38
0
7
0
3
30
30
U3=
Demand 10
V1=
20
V2=
40
V3=
70
Total Cost:=5*2+10*3+5*5+20*3+30*3=215 (The cost further reduced)
Shadow cost Equations:
U1+V1=2U1+V3=3U2+V1=5U2+V2=3U3+V3=3
• Let U2=0By putting this U2 zero we can easily find the
value of all other variables which will be as follows
• U1=-3 V1=5• U2=0 V2=3• U3=3 V3=6
Sources Destination Capacity
D1 D2 D3
S1 2
5
-4 4
0
3
10
15
U1= -3
S25
5
3
20
-1 7
0
25
U2=0
S30 8
0
-7 7
0
3
30
30
U3= 3
Demand 10
V1=5
20
V2=3
40
V3=6
70All the shadow costs are non positive (negative) which is the indication of optimal solution
Optimal Distribution
• S1 ─ ─ ─ ─ > D1 = 5• S1 ─ ─ ─ ─ > D3 = 10• S2 ─ ─ ─ ─ > D1 = 5• S2 ─ ─ ─ ─ > D2 = 20• S3 ─ ─ ─ ─ > D3 = 30
Total = 70
Total Cost= 215
Thank You