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(4)Transportation Method
Special method.
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Transportation Problems
and Method to solvesuch problems
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C11 C12 . . C1n
C21 C22 . . C2n
. . . . .
. . . . .
Cm1 . . . Cmn
Destination (Warehouse)
1 2 . . n
1
2
.
.
m
Origin(Plant)
Cases
Cij = Cost of shipping unit item from ith
Origin to jth Destination
Transportation Problem
b1 b2 . . bn iabj
a1
a2
.
am
bjia
Cij
Pij
bjia
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B. Unbalanced Minimization with Cij
Categories of Transportation Problems
bjia
bjia
bjia
bjia
A. Balanced Minimization with Cij
D. Unbalanced Maximization with Pij
C. Balanced Maximization with Pij
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C11 C12 . . C1n
C21 C22 . . C2n
. . . . .
. . . . .
Cm1 . . . Cmn
Destinations(Warehouses)
1 2 . . n
1
2.
.
m
Origins(Plants)
Xij = No. of items shipped from ith Origin
to jth Destination
A.Balanced Minimization Problem
b1 b2 . . bn bjia
a1
a2.
am
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Formulation of Transportation Problem
as LPP
1 1
1
1
.
/ , ( )
( )
0 & .
m n
i j
n
j
m
i
Min Z Cij Xij
s t Xij ai for all i I
Xij bj for all j II
All Xij for all i j
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Methods to get IBFS :
5. VAM: Vogels Approximate Method
1. NWCM: North West Corner Method
2. RMM: Row Minima Method
3. CMM: Column Minima Method
4. MMM: Matrix Minima Method
(Least Cost Entry Method)
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4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
23
3
3 4 6 7 10
5
13
12
30
1022 6 5
2 2
2 2
5
11
10
NWCM
Z = 148
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4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 51
23
3
3 4 6 7 10
5
13
12
30
1022 6 5
2
Therefore NWCM solution is :
Z = 148
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4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 51
23 3
3 4 6 7 10
5
13
12
30
106
21
5
3
3 2
1
3
6
RMM
3
Z = 85
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4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 51
23 3
3 4 6 7 10
5
13
12
30
106
21
5
3
Therefore RMM solution is :
Z = 85
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4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 51
23
2
3 4 6 7 10
5
13
12
30
46
7
3
2
2 2
11
6
CMM
6
4
6
Z = 108
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4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 51
23
2
3 4 6 7 10
5
13
12
30
46
7
3 2
Therefore CMM solution is :
6
Z = 108
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4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 51
23
3 4 6 7 10
5
13
12
30
2
3
10
2
3
6
MMM
1
5
6
13 33
Z = 85
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4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 51
23
3 4 6 7 10
5
13
12
30
2 10
Therefore MMM solution is :
5
6
133
Z = 85
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4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 51
23
3 4 6 7 10
5
13
12
30
2 0
1
1
VAM
0
51 1
10
3
PenaltyNumber
PenaltyNumber
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4 2 3 2
5 4 5 2
6 5 4 7
1 2 3 4
1
2
3
3 4 6 7
5
3
12
202 0
2
1
VAM
04
1
1
1
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4 3 2
6 4 7
1 3 4
1
3
3 6 4
1
12
133
2
VAM
1
2 1 5
1
12
3 6 3 63
6 3
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4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 51
23
3 4 6 7 10
5
13
12
30
VAM solution is :
10
3
3
14
63
Z = 89
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Optimality Test on IBFS
Methods : 1. Stepping Stone Method2. MODI Method
MODI Method Of Checking Optimality
Condition To Be Satisfied :
Filled Cells = m + n - 1
Where m = No. of Rowsn = No. of columns
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MODI Method for Optimality Test
Get all ui & vj starting with any ui or vj as zero,
such that ui + vj = cij in filled cells.
Get vacant cell Evaluation of all vacant cells.
(VCE)ij = cij(ui + vj)
VAM S l ti
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2 2
2 1
6 4 7
-1
2
2
7
VAM Solution
4
3 6
10
1
3
3
0-3 -10
4 3 6
5 4 55-2 7
As (VCE)32 = -2, the solution under test
is not optimal.
Z = 89
ui
vj
F difi i f i i l i
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2 2
2 1
6 4 7
-1
2
2
7
For modification of existing solution :
4
3 6
10
1
3
3
0-3 -10
4 3 6
5 4 5
5-2 7
4 1
3 3
1 4
ui
vj
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2 2
2 1
6 4
Hence, improved solution will be :
1
3 6
10
4
3
3 5
Z = 89(2) (3)
Z = 83
This solution is to be checked by MODI Method
for optimality.
i
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2 2
2 1
6 4
Improved solution :
1
3 6
10
4
3
3 5 0
6 5 4
- 3
5
- 3
4
4 3 6
5 4 5
7 7
As all vacant cell evaluations are positive,the solution is optimal, giving Z = 83.
ui
vj
MMM S l ti
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6 5 4
-1
MMM Solution
3
1 10
5
2
3 2
2
445
1
ui
vj
-1 -3
0 2
0
6 63
-1
0
4
5
2 3
57
6
7
Z = 85
For modification of existing solution :
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2 2
2 1
6 4 7
For modification of existing solution :
3 6
10
5
2
4 3 6
5 4 5
5 7
1
5
23
1 4
ui
vj
-10
-31
3
0-1
-1
0
6 5 4 3 2
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2 2
2 1
6 4
Hence, improved solution will be :
1
3 6
10
4
3
3 5
Z = 852
Z = 83
This solution is to be checked by MODI Method
for optimality.
I d l ti
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2 2
2 1
6 4
Improved solution :
1
3 6
10
4
3
3 5 0
6 5 4
- 3
5
- 3
4
4 3 6
5 4 5
7 7
As all vacant cell evaluations are positive,the solution is optimal, giving Z = 83.
ui
vj
F h l d fi i t E l
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1
For cases when closed figure is not Equal or
Rectangle :
7
3
5
2
4
45
2 3
Problem :
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The following table gives minimization transportation problem with
given allocation, check whether given solution is optimal. If not, get
the optimal solution.
Problem :
A
B
C
D
10 2225
24 2810
27 2515
10 1520
Demand 25 20 25
20
10
18
20
25
10
15
20
X Y Z Supply
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5
10
18
15
20
18
15
2210
10
2025
20
10
15
-1
-6
-20
Z = 1270
515 0
24 28
2527
10
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20
25 520
20
Hence, improved solution will be :
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20
5 5
20
Hence, improved solution will be :
Z 870 (3) ( 870)
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22
10
18
12
18
10
10
205 20
10
15
Z = 870(3) ( = 870
10
20
-12-2 0
22
24
27
20
28
25
15
)
Hence, this solution is Optimal.
Problem :
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12
-20
04
2012
8
28120
13530
Z = 7100
3216 4
32 12
3216
160
25
-32
Problem :
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30
120 95
5525
25
Hence, improved solution will be :
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12
12
04
2012
8
2895
-8
13555
Z = 7100(32) (25) = 6300
016 4
25
32 12
3216
160
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4
4
04
2012
8
28
40150
Z = 6300(8) (95) = 5540816 4
120
32 12
3216
65 95
Hence, this solution is Optimal.
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Home Assignments :
[ 1 ] Mr XYZ has suggested the following feasible solution
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[ 1 ] Mr XYZ has suggested the following feasible solution
for a minimizing transportation problem as given below.
Check whether the solution is optimal. If not, get optimal
solution.A B C D
1 10 2 20 11
2 12 7 9 20
3 3 14 16 18
15
1015
5
[ 2 ] For following transportation (minimization) problem
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[ 2 ] For following transportation (minimization) problem,
present total cost of transportation is Rs. 3100. Is it possible to
reduce this by proper scheduling ? What can be the saving ?
Centres
A B C Supply
X 25 35 10 150
Y 20 5 80 100
Demand 50 50 150
Factories
[ 3 ] A m f t t t hi 8 l d f hi d t
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[ 3 ] A manufacturer wants to ship 8 loads of his product as
shown in table below. The matrix gives km from origin
O to destination D. The shipping costs are Rs. 10 per load per
km. Find the optimal solution.
D1 D2 D3
O1 50 30 220 1
O2 90 45 170 3
O3 250 200 50 4
4 2 2
[ 4 ] Solve the following minimization Transportation
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[ 4 ] Solve the following minimization Transportation
problem.
D1 D2 D3 D4 Supply
O1 2 3 11 7 6
O2 1 1 6 1 1
O3 5 8 15 10 10
Demand 7 5 3 2
[ 5 ] In following transportation problem X, Y, Z are
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[ ] g p p , ,
warehouses and P, Q, R, S are customers. The capacity at
the warehouses and the demand from customers are
shown around the perimeter. Per unit transportation costsare shown in the cells.
P Q R S
X 7 8 11 10 30
Y 10 12 5 4 45
Z 6 10 11 9 35
20 28 17 33
( i ) Find VAM solution. Is this solution optimal ? If not,
get optimal solution.
( ii ) By how much should YP cost need to be reduced in
order to make shipments along this route worth while ?
[ 6 ] Using NWC rule method get initial feasible solution for
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[ 6 ] Using NWC rule method, get initial feasible solution for
following minimization transportation problem. Is this
solution optimal ? If not, get optimal solution and
corresponding optimal value of objective function.
A B C Plant
capacity
W 4 8 8 55
X 16 24 16 25
Y 8 16 24 35
Project Requirement 35 45 35
[ 7 ] A company has three plants at P1 P2 P3 which supply to
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[ 7 ] A company has three plants at P1, P2, P3 which supply to
warehouses W1, W2, W3, W4 and W5. Monthly plant capacities
are 800, 500 and 900 units respectively. Monthly warehouse
requirements are 100, 400, 500, 400 and 800 units respectively.The unit transportation costs in rupees are given in table below.
W1 W2 W3 W4 W5
P1 5 8 6 5 3P2 4 7 7 3 6
P3 8 4 6 4 2
( i ) Determine the optimal allocation for the company.
( ii ) If the warehouse W1 is closed because of reduced
demand and the corresponding units are shipped to warehouse
W3 by raising its capacity find the total cost of shipment. Is it
optimal schedule ? If not, get optimal schedule.
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Transshipment Problems :
Transportation Problem :
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7 3
9 3
3 3
5
3 8
6
5
O1
O2
D1 D2
Z = 45
Transportation Problem :
Transshipment Problem :
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Transshipment Problem :
0 2 7 3
2 0 9 3
7 8 0 4
2 2 2 0
O1
O1
O2
O2
D1
D1
D2
D2
11 11 + 6
11 + 5
11
11 11
11
11 + 3 11 + 8
11
11
3
6
5
8
Z = 39
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O1
D2
6
D13
O2
5
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Home Assignments :
[ 1 ]
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0 1 3 4
1 0 2 4
3 2 0 1
4 4 1 0
O1
O1
O2
O2
D1
D1
D2
D2
5
25
20 10
[ 2 ]
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0 2 2 1
1 0 2 3
2 2 0 2
1 3 2 0
O1
O1
O2
O2
D1
D1
D2
D2
8
3
7 4
[ 3 ]
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0 4 20 5 25 12
10 0 6 10 5 20
15 10 0 8 45 7
20 25 10 0 30 6
20 18 60 15 0 10
10 25 30 23 4 0
O1
O1
O2
O2
D1
D1
D2
D2
O3
O4
O3 O4
Th k
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Thank you
For any Query or suggestion :Contact :Dr. D. B. Naik
Professor, Training & Placement SectionSardar Vallabhbhai National Institute ofTechnology, Surat
Ichchhanath, Surat395007, Gujarat.
Email: [email protected]