unit 3 lecturer notes of transportation problem of or by dr s v prakash
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8/3/2019 Unit 3 Lecturer notes of Transportation problem of OR by Dr S V Prakash
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06IP/IM74 OPERATIONS RESEARCH
UNIT - 3: Transportation Problem
(By Dr. S.V Prakash, Asst. Prof. (Mech Dept.), MSRIT, Bangalore)
Introduction:
The objective of the transportation problem is to transport various quantities of a
single homogenous commodity, which are initially stored at various origins to various
destinations in such a way that the total transportation cost is minimum.
Definitions:
Basic Feasible solution: A feasible solution to a m-origin, n-destination problem is said to
be basic if the number of positive allocations are equal to (m+n-1).
Feasible Solution: A set of positive individual allocations which simultaneously removes
deficiencies is called a feasible solution.
Optimal Solution: A feasible solution (not basically basic) is said to be optimal if it
minimises the total transportation cost.
Mathematical Formulation of Transportation Problems
• Suppose there are ‘m’ ware houses (w1,w2,w3, _, _, wm),
where the commodity is stocked and ‘n’ markets where it is needed.
• Let the supply available in wear houses be a1, a2, a3, _,_,_ am and
• The demands at the markets (m1, m2, m3, _, _, mn) be b1, b2, b3, _, _ , _ bn.
• The unit cost of shipping from ware house i to a market j is Cij (C11,C12,_, _ Cn),
• Let X11, X12,X13,_, _, Xmn be the distances from warehouse to the markets
• we want to find an optimum shipping schedule which minimises the total cost of
transportation from all warehouses to all the markets
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The total minimum transportation cost is
Z = i=1Σm
j=1Σn
Xij * Cij i.e. Z = X11C11 + X12C12 + _ _ _ + XmnCmn
Types of Transportation Problems
1. Minimisation Balanced Transportation Problems
2. Minimisation Unbalanced Transportation Problems
3. Maximisation Balanced Transportation Problems
4. Maximisation unbalanced Transportation Problems
5. All the above models with degeneracy.
Ware
houses
Markets
m1 m2 m3 - - mn
Supplies
W1
W2
W3
-
-
wm
C11X11 C12X12 C13X13 - - - C1nX1n
C21X21 C22X22 C23X23 - - - C1nX1n
C31X31 C32X32 C33X33 - - - C1nX1n
-
-
Cn1Xm1 Cn2Xm2 Cn3Xm3 - - - CmnXmn
a1
a2
a3
-
-
am
Demand b1 b2 b3 - - - - bn i=1Σm
ai = j=1Σn
b j
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Methods of solving Transportation Problems
1. North- West Corner Rule method
2. Row-minima Method
3. Column minima method
4. Matrix Minima Method or least cost method
5. Vogel's Approximation method (VAM)
Methods for checking Optimality
1. Modified Distribution Method, UV or MODI method
PROBLEMS:
1. Solve the following transportation problem by
North-West corner rule, Row Minima, Column Minima,
Matrix Minima and VAM Method:
Solution:
This is a balanced transportation problem, since supply is equal to demand
Factories W1 W2 W3 W4 Supply
F1
F2
F3
6 4 1 5
8 9 2 7
4 3 6 2
14
16
05
Demand 6 10 15 4 35
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1. North-West corner rule Method:
Factories W1 W2 W3 W4 Supply
F1
F2
F3
6(6) 4(8) 1 5
8 9(2) 2(14) 7
4 3 6(1) 2(4)
14
16
05
Demand 06 10 15 04 35
The Total feasible transportation cost
= 6(6) + 4(8) + 9(2) + 2(14) + 6(1) + 2(4) = Rs. 128/-
2. Row Minima Method:
Factories W1 W2 W3 W4 Supply
F1
F2
F3
6 4 1(14) 5
8(6) 9(5) 2(1) 7(4)
4 3(5) 6 2
14
16
05
Demand 06 10 15 04 35
The Total feasible transportation cost
= 1(14)+8(6)+9(5)+2(1)+7(4)+3(5)
= Rs.152/-
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3. Column-Minima Method:
The Total feasible transportation cost
= 6(1)+4(10)+1(3)+2(12)+7(4)+4(5)
= Rs. 121/-
4. Matrix-Minima Method or Least Cost method:
Factories W1 W2 W3 W4 Supply
F1
F2
F3
6 4 1(14) 5
8(6) 9(9) 2(1) 7
4 3(1) 6 2(4)
14
16
05
Demand 06 10 15 04 35
The Total feasible transportation cost
= 1(14)+8(6)+9(9)+2(1)+3(1)+2(4)
= 156/-
Factories W1 W2 W3 W4 Supply
F1
F2
F3
6(1) 4(10) 1(3) 5
8 9 2(12) 7(4)
4(5) 3 6 2
14
16
05
Demand 06 10 15 04 35
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5. VAM- Vogel’s approximation method:
Step-I: Against each row and column of the matrix, denote the difference between the two
least cost in that particular row and column.
Step-II: Select the maximum value noted as per step-I, in this row or column select the cell
which has the least cost
Step-III: Allocate the maximum possible quantity
Step-IV: After fulfilling the requirements of that particular row or column, Ignore that
particular row or column and recalculate the difference by the two lowest cost for each of
the remaining rows or columns, Again select the maximum of these differences and allocate
the maximum possible quantity in the cell with the lowest cost in that particular /
corresponding row or column.
Step-V: Repeat the procedure till the initial allocation is completed
5. VAM- Vogel’s approximation method:
Factories W1 W2 W3 W4 Supply
F1
F2
F3
6 (4) 4(10) 1 5
8(1) 9 2(15) 7
4(1) 3 6 2(4)
14
16
05
Demand 06 10 15 04 35
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The Total feasible transportation cost
= 6(4) + 4(10) + 8(1) + 2(15) + 4 (1) + 2(4)
= 114/-
II- Check for degeneracy:
If (m+n-1) is not equal to the number of allocated cells, then it is called degeneracy in
transportation problems,
Where m= number of rows, n= number of columns.
This will occur if the source and destination is satisfied simultaneously.
The degeneracy can be avoided by introducing a dummy allocation cell. To equate the
number of allocated cells equal to (m+n-1)
For the above problem (m+n-1) = (3+4-1) =6 = Number of allocations = 6
Hence there is no degeneracy.
III- Checking optimality using MODI method:
• For allocated cells Cij – (Ui +V j) = 0
• For unallocated cells Cij – (Uij) +Vj) ≥ 0
The Total feasible transportation cost
= 6(4)+4(10)+8(1)+2(15)+4(1)+2(4)
= 114/-
V1=0 V2=-2 V3 =-6 V4= -2
6(4) 4(10) 1 5
8(1) 9 2(15) 7
4(1) 3 6 2(4)
U1=6
U2=8
U3= 4
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Problem.2
There are 3 Parties who supply the following quantity of coal P1= 14t, P2=12t, P3= 5t.
There are 3 consumers who require the coal as follows C1=6t, C2=10t, C3=15t. The cost
matrix in Rs. Per ton is as follows. Find the schedule of transportation policy which
minimises the cost:
Solution:
Factories W1 W2 W3 Supply
F1
F2
F3
6 8(5) 4(9)
4(6) 9 3(6)
1 2(5) 6
14
12
05
Demand 6 10 15 31
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Therefore the total feasible transportation cost
= 8(5)+ 4(9) + 4(6) +3 (6) + 2(5)
= Rs. 128/-
II. Check for Degeneracy:
(m+n-1) = (3+3-1) = 5 = Number of allocations
Hence there is no degeneracy
III- Checking optimality using MODI method:
•
For allocated cells Cij – (Ui +V j) = 0
• For unallocated cells Cij – (Uij) +V j) ≥ 0
Since all the marginal costs for the unallocated cells are positive, it gives an optimal
solution and the total minimum transportation cost = Rs. 128/-
V1=5 V2=8 V3=4
6 8(5) 4(9)
4(6) 9 3(6)
1 2(5) 6
U1= 0
U2=-1
U3=-6
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Factories W1 W2 W3 W4 Supply
F1
F2
F3
5 7 3 8
4 6 9 5
2 6 4 5
300
500
200
Demand 200 300 400 100 1000
Since the unit cost of production is Rs.4, 3 and 5
at the three factories, therefore Total cost = Production cost + cost
Factories W1 W2 W3 W4 Supply
F1
F2
F3
9 11 7(300) 12
7(100) 9(300) 12 8(100)
7(100) 11 9(100) 10
300
500
200
Demand 200 300 400 100 1000
Therefore the total feasible solution = 7(300) + 7(100) +9(300) + 8(100) +7(100) + 9(100)
= Rs. 7900/-
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Therefore the total feasible transportation solution= 7(300) + 7(100) + 9(300) + 8(100) + 7(100) + 9(100)
= Rs.7900/-
II. Check for Degeneracy:
(m+n-1) = (3+4-1) = 6 = Number of allocations
Hence there is no degeneracy
III- Checking optimality using MODI method:
• For allocated cells Cij – (Ui +V j) = 0
• For unallocated cells Cij – (Uij) +Vj) ≥ 0
Since all the marginal costs for the unallocated cells are positive, it gives an optimal
solution and the total minimum cost = Rs. 7900/-
Factories W1 W2 W3 W4 Supply
F1
F2
F3
9 11 7(300) 12
7(100) 9(300) 12 8(100)
7(100) 11 9(100) 10
300
500
200
Demand 200 300 400 100 1000
V1 V2 V3 V4
9 11 7(300) 12
7(100) 9(300) 12 8(100)
7(100) 11 9(100) 10
U1= -2
U2=0
U3=0
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Problem-4
A company has three plants supplying the same product to the five distribution centers.
Due to peculiarities inherent in the set of cost of manufacturing, the cost/ unit will vary
from plant to plant. Which is given below. There are restrictions in the monthly capacityof each plant, each distribution center has a specific sales requirement, capacity
requirement and the cost of transportation is given below.
The cost of manufacturing a product at the different plants is Fixed cost is Rs 7x105,
4x 105
and 5x 105.
Whereas the variable cost per unit is Rs 13/-, 15/- and 14/- respectively.
Determine the quantity to be dispatched from each plant to different distribution centers,
satisfying the requirements at minimum cost.
Solution:
Factories W1 W2 W3 W4 W5 Supply
F1
F2
F3
5 3 3 6 4
4 5 6 3 7
2 3 5 2 3
200
125
175
Demand 60 80 85 105 70 400 500
Factories W1 W2 W3 W4 W5 D Supply
F1
F2
F3
18 16 16 19 17 0
19 20 21 18 22 0
16 17 19 16 17 0
200
125
175
Demand 60 80 85 105 70 100 500
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Therefore the total feasible transportation cost =
= 16(55) + 16(85) + 17(60) + 20(25) + 0(100) +16(60) + 16(105) +17(10) = Rs6570/-
II. Check for Degeneracy:
(m+n-1) = (6+3-1) = 8 = Number of allocations
Hence there is no degeneracy
Factories W1 W2 W3 W4 W5 D Supply
F1
F2
F3
18 16(55) 16(85) 19 17(60) 0
19 20(25) 21 18 22 0(100)
16 17 19 6(105) 17(10) 0
200
125
175
Demand 60 80 85 105 70 100 500
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Therefore total feasible solution
= 1(15) +4(10)+4(35)+2(5)+0+3(22)+2(20)+4(3)= Rs. 323/-
Therefore total feasible solution
= 1(15) +4(10)+4(35)+2(5)+0+3(22)+2(20)+4(3)= Rs. 323/-
II. Check for degeneracy:
(m+n-1) = (3+6-1) = 8 = Number of allocations
Hence there is no degeneracy
III. Check for optimality:
V1=3 V2=1 V3=2 V4=4 V5=4 V6=2
4 1(15) 3 4(10) 4(35) 0(-2)
2 3 2 2(5) 3 0(30)
3(22) 5 2(20) 4(3) 4 0(-2)
U1=0
U2=-2
U3=0
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V1=3 V2=1 V3=2 V4=4 V5=4 V6=2
4 1(15) 3 4(10-x) 4(35) 0(x)
2 3 2 2(5+x) 3 0(30-x)
3(22) 5 2(20) 4(3) 4 0(-2)
U1=0
U2=-2
U3=0
10-x=0 30-x=0 Min. value of x=10
V1=1 V2=1 V3=0 V4=2 V5=4 V6=0
4 1(15) 3 4 4(35) 0(10)
2 3 2 2(15) 3(-1) 0(20)
3(22) 5 2(20) 4(3) 4 0(-2)
U1=0
U2=0
U3=2
20-X = 0: 3-x=0: therefore x=3
V1=3 V2=1 V3=2 V4=3 V5=4 V6=0
4 1(15) 3 4 4(35) 0(10)
2 3 2 2(18) 3(-1) 0(17)
3(22) 5 2(20) 4 4 0(3)
U1=0
U2=-1
U3=0
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35-x=0 and 17-x = 0, therefore x= 17
V1=3 V2=1 V3=2 V4=3 V5=4 V6=0
4 1(15) 3 4 4(18) 0(27)
2 3 2 2(18) 3(17) 0
3(22) 5 2(20) 4 4 0(3)
U1=0
U2=-1
U3=0
27-x=0 : 22-X= 0: 17-X=0: therefore X=17
V1=3 V2=1 V3=2 V4=3 V5=4 V6=0
4 1(15) 3 4 4(35) 0(10)
2(17) 3 2 2(18) 3 0
3(5) 5 2(20) 4 4 0(20)
U1=0
U2=-1
U3=0
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Factories A B C Supply
M1
M2
M3
139140 137
209207 210
254 255 255
160
120
140
Demand 90 210 120 420
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Factories A B C Supply
M1
M2
M3
116115(70) 118(90)
4(90) 48 45(30)
1 0(140) 0
160
120
140
Demand 90 210 120 420
Factories A B C Supply
M1
M2
M3
116115 118
46 48 45
1 0 0
160
120
140
Demand 90 210 120 420
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V1=4 V2=42 V3=45
116 115(70+x) 18(90-x)
4(90) 48 45(30)
1 0(140-x) 0(x)
U1=73
U2=0
U3=-42
90-X=0 and 140-X=0, therefore X=90
V1=4 V2=45 V3=45
116 115(160) 18
4(90) 48 45(30)
1 0(50) 0(90)
U1=70
U2=0
U3=-45
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Factories A B C Supply
M1
M2
M3
139 140 137
209 207 210
254 255 255
90
210
120
Demand 160 120 140 420
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Factories A B C Supply
M1
M2
M3
116 115 118
46 48 45
1 0 0
90
210
120
Demand 160 120 140 420
Factories A B C Supply
M1
M2
M3
116(90) 115 118
46(70) 48 45(140)
1 0(120) 0
90
210
120
Demand 160 120 140 420
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Factories A B C Supply
M1
M2
M3
116(90) 115 118
46(70) 48 45(140)
1(θ) 0(120) 0
90
210
120
Demand 160 120 140 420
V1=0 V2=-1 V3= -1
116(90) 115 118
46(70) 48 45(140)
1(θ) 0(120) 0
U1=116
U2=46
U3=1
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Factories A B C D Supply
I
II
III
D
21 26 20 21
22 24 20 19
18 20 19 20
00 00 00 00
450
300
150
020
Demand 200 300 150 270 920
A B C D Supply
4(150) 0(300) 6 5
5(50) 2 6(130) 7(120)
8 6 7 6(150)
0 0 0(20) 0
450
300
150
020
200 300 150 270 920
Factories A B C D Supply
I
II
III
22 26 20 21
21 24 20 19
18 20 19 20
450
300
150
Demand 200 300 150 270 920 900
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V1= 5 V2=1 V3= 6 V4=7
4(150) 0(300) 6 5(-1)
5(50) 2 6(130) 7(120)
8 6 7 6(150)
0 0 0(20) 0
U1=-1
U2=0
U3=-1
U4=-6
V1= 5 V2=1 V3= 6 V4=7
4(150-X) 0(300) 6 5(X)
5(50+X) 2 6(130) 7(120-X)
8 6 7 6(150)
0 0 0(20) 0
U1=-1
U2=0
U3=-1
U4=-6
150-X= 0, 120-X= 0, therefore X=120
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V1= 5 V2=1 V3= 6 V4=7
4(150-X) 0(300) 6 5(X)
5(50+X) 2 6(130) 7(120-X)
8 6 7 6(150)
0 0 0(20) 0
U1=-1
U2=0
U3=-1
U4=-6
V1= 4 V2=0 V3=5 V4=5
4(30) 0(300) 6 5(120)
5(170) 2 6(130) 7
8 6 7 6(150)
0 0 0(20) 0
U1=0
U2=1
U3=1
U4=-5
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III. CHECK FOR OPTIMALITY
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Factories D E F G Supply
A
B
C
OT.A
OT.B
11 13 17 14
16 18 14 10
21 24 13 10
15 17 21 18
21 23 19 15
250
350
400
50
75
Demand 200 225 275 300 1125
Factories D E F G D1 Supply
A
B
C
OT.A
OT.B
11(200) 13(50) 17 14 0(θ)
16 18(175) 14 10(175) 0
21 24 13(275) 10(125) 0
15 17 21 18 0(50)
21 23 19 15 0(75)
250
350
400
50
75
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Factories D E F G D1 Supply
A
B
C
OT.A
OT.B
11(200) 13(50) 17 14 0
16 18(175) 14 10(175) 0
21 24 13(275) 10(125) 0
15 17 21 18 0(50)
21 23 19 15 0(75)
250
350
400
50
75
Demand 200 225 275 300 125 1125
8/3/2019 Unit 3 Lecturer notes of Transportation problem of OR by Dr S V Prakash
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V1=16 V2=18 V3=13 V4=10 V5=0
11(200) 13(50+θ) 17 14 0
16 18(175-θ) 14 10(175) 0(θ)
21 24 13(275) 10(125) 0
15 17(-1) 21 18 0(50)
21 23 19 15 0(75)
U1=-5
U2=5
U3=5
U4=0
U5=0
8/3/2019 Unit 3 Lecturer notes of Transportation problem of OR by Dr S V Prakash
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11(200) 13(50+θ) 17 14 0
16 18(175-θ-x) 14 10(175) 0(θ+x)
21 24 13(275) 10(125) 0
15 17(x) 21 18 0(50-x)
21 23 19 15 0(75)
8/3/2019 Unit 3 Lecturer notes of Transportation problem of OR by Dr S V Prakash
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8/3/2019 Unit 3 Lecturer notes of Transportation problem of OR by Dr S V Prakash
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8/3/2019 Unit 3 Lecturer notes of Transportation problem of OR by Dr S V Prakash
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8/3/2019 Unit 3 Lecturer notes of Transportation problem of OR by Dr S V Prakash
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Reference Books:1. Taha H A, Operation Research - An Introduction, Prentice Hall of India, 7th edition, 2003
2. Ravindran, Phillips and Solberg, Operations Research : Principles and Practice, John
Wiely & Sons, 2nd Edition
3. D.S.Hira, Operation Research, S.Chand & Company Ltd., New Delhi, 2004
4.