Tp for b.tech. (mechanical)
62
Method to solve Transportation Problems
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Transportation Problem
Transcript of Tp for b.tech. (mechanical)
Method to solve
Transportation Problems
Transportation Problems
and
Method to solve such problems
C11 C12 . . C1n
C21 C22 . . C2n
. . . . .
. . . . .
Cm1 . . . Cmn
Destination (Warehouse) 1 2 . . n
1
2
.
.
m
Origin
(Plant)
CasesCases
•Cij = Cost of shipping unit item from ith
Origin to jth Destination
Transportation ProblemTransportation Problem
b1 b2 . . bn iabj
a1
a2
.
am
bjia
Cij
Pij
bjia
B. Unbalanced Minimization with Cij
Categories of Transportation ProblemsCategories of Transportation Problems
bjia
bjia
bjia
bjia
A. Balanced Minimization with Cij
D. Unbalanced Maximization with Pij
C. Balanced Maximization with Pij
C11 C12 . . C1n
C21 C22 . . C2n
. . . . .
. . . . .
Cm1 . . . Cmn
Destinations(Warehouses)
1 2 . . n
1
2
.
.
m
Origins(P
lants )
• Xij = No. of items shipped from ith Origin
to jth Destination
A.A. Balanced Minimization ProblemBalanced Minimization Problem
b1 b2 . . bn bjia
a1
a2
.
am
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
Methods to get IBFS :Methods to get IBFS :
5. VAM: Vogel’s 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)
4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
2
3
3
3 4 6 7 10
5
13
12
30
102
2 6 5
2 2
2 2
511
10
NWCMNWCM
Z = 148Z = 148
4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
2
3
3
3 4 6 7 10
5
13
12
30
102
2 6 5
2
Therefore NWCM solution is :Therefore NWCM solution is :
Z = 148Z = 148
4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
2
3 3
3 4 6 7 10
5
13
12
30
10
6
21
5
3
3 2
1 3
6
RMMRMM
3
Z = 85Z = 85
4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
2
3 3
3 4 6 7 10
5
13
12
30
10
6
21
5
3
Therefore RMM solution is :Therefore RMM solution is :
Z = 85Z = 85
4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
2
3
2
3 4 6 7 10
5
13
12
30
4
6
7
3
2
2 2
11
6
CMMCMM
6
4
6
Z = 108Z = 108
4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
2
3
2
3 4 6 7 10
5
13
12
30
4
6
7
3 2
Therefore CMM solution is :Therefore CMM solution is :
6
Z = 108Z = 108
4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
2
3
3 4 6 7 10
5
13
12
30
2
3
10
2
3
6
MMMMMM
1
5
6
1
3 33
Z = 85Z = 85
4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
2
3
3 4 6 7 10
5
13
12
30
2 10
Therefore MMM solution is :Therefore MMM solution is :
5
6
1
33
Z = 85Z = 85
4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
2
3
3 4 6 7 10
5
13
12
30
2 0
1
1
VAMVAM
0
51 1
10 3
PenaltyPenaltyNumberNumber
PenaltyPenaltyNumberNumber
2 2
2 1
6 4 7
-1
2
2
7
VAM SolutionVAM Solution
4
3 6
10
1
3
3
0 -3
-1
0
4
3
6
5
4
5
5
-2
7
As (VCE)32 = -2, the solution under test is not optimal.
Z = 89
ui
vj
Filled Cells = m+n-1 = 7 , Hence O.K.
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
VAMVAM
04
1
1
1
4 3 2
5 5 2
6 4 7
1 3 4
1
2
3 3 6 7
1
3
12
164
3
2
1
1 1
3
0
4 3 2
6 4 7
1 3 4
1
3
3 6 4
1
12
133
VAMVAM
1
2 1 5
1
3 6 3 2
4 2 3 2 6
5 4 5 2 1
6 5 4 7 7
1 2 3 4 5
1
2
3
3 4 6 7 10
5
13
12
30
VAM solution is : VAM solution is :
103
3
14
63
Z = 89Z = 89
Optimality Test on IBFSOptimality Test on IBFS
Methods : 1. Stepping Stone Method 2. MODI Method
MODI Method Of Checking Optimality
Condition To Be Satisfied :
Filled Cells = m + n - 1
Where m = No. of Rows n = No. of columns
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)
2 2
2 1
6 4 7
-1
2
2
7
VAM SolutionVAM Solution
4
3 6
10
1
3
3
0 -3
-1
0
4
3
6
5
4
5
5
-2
7
As (VCE)32 = -2, the solution under test is not optimal.
Z = 89
ui
vj
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
-1
0
4
3
6
5
4
5
5
-2
7
4 1
3 3
1 4
ui
vj
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.
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, under test, is optimal, giving Z = 83.
ui
vj
6 5 4
-1
MMM SolutionMMM Solution
3
1
10
5
2
3
2
2
4
4
5
1
ui
vj
-1
-3
0
2 0
6
6 3
-1
0
4
5
2
3
5
7
6
7
Z = 85
Filled Cells = m+n-1 = 7 , Hence O.K.
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
2 3
1 4
ui
vj
-1 0
-3
1 3
0
-1
-1
0
6 5 4
3
2
2 2
2 1
6 4
Hence, improved solution will be :
1
3 6
10
4
3
3 5
Z = 85 – 2
Z = 83
This solution is to be checked by MODI Method for optimality.
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, this solution,under test, is optimal, giving Z = 83.
ui
vj
What to do when the closed figure consists of more than four lines ?
1
For cases when closed figure is not Equal or Rectangle :
7
3
5
2
4
4 5
2
3
What to do when number of
filled cells
are
less than (m + n – 1) ?
Prob. A feasible solution of Minimization Transportation Problem is given below.
( i ) Is it Optimal Solution ? If not, get it.
( ii ) Is the Optimal Solution unique ? Justify.
7
58
D1 D2 D3
O1
2 4 10
O2
5 6 3
7
58
D1 D2 D3
O1
2 4 10
O2
5 6 3
Z = 83. Here filled cells = 3 instead of 4 (m+n1).
Hence, is to be put up in one independent vacant cell and consider it as filled cell.
2 4 10
5 6 35
7
8
0
2 4
3
0
1
8 7
7
1
Hence, improved solution will be :
7
3
0
4
3
2 7
5
Z = 83 – (7) (1) = 76
1
6 5 3
10
5 6
Hence, this solution is Optimal.
7
Typical Questions
12
-20
0 4
20 12
8
28 120
135
30
Z = 7100
32 16 4
32
12 32
16
160 25
-32
[ 1 ] Filled Cells = m+n-1 = 6 , Hence O.K.
30
120 95
55 25
25
Hence, improved solution will be :
12
12
0 4
20 12
8
28 95
-8
135
55
Z = 7100 – (32) (25) = 6300
0 16 4
25
32
12 32
16
160
55
95 25
135
160
120
95
150
65
40
Hence, improved solution will be :
4
4
0 4
20 12
8
28
40 150
Z = 6300 – (8) (95) = 5540
8 16 4
120
32
12 32
16
65 95
Hence, this solution is Optimal.
X Y Z
A4 2 4 15
B12 8 4 15
10 10 10
10 5
10 5
5
5
MMM :
Z = 5*4 + 10*2 + 5*12 + 10*4 = 140
X Y Z
A
B12
10
10
5
5
Optimality Test :
0
4 2
8
4
4
8 2
105
5
10 5
5
2 4
4
Filled Cells = m+n-1 = 4 , Hence O.K.
X Y Z
A4 2
B12 4
10
10
5
5
0
4 2
6
2
4
8
Hence, improved solution will be
Hence, this is Optimal Solution.
Z = 140 2*5 = 130
X Y Z
A4 2
B10 4
10
10
5
5
0
4 2
6
2
4
8
If C21 is changed to 10, then
Hence, this will be one of the Multiple Optimal Solutions.
0
Find IBFS of following “Minimization Transportation Problem” using “Least Cost Entry Method” and apply “MODI Method” to check optimality of this IBFS.
P Q R S T
A
B
C
D
E
73 40 09 79 20 8
62 93 96 08 13 7
96 65 80 50 65 10
57 58 29 12 87 3
56 23 87 18 12
5
6 8 10 5 4
[ 3 ]
P Q R S T
A
B
C
D
E
73 40 09 79 208
62 93 96 08 137
96 65 80 50 6510
57 58 29 12 873
56 23 87 18 125
6 8 10 5 4 33
5
2
8
2
4 1 1
7
2 11
5
2
3
7 3 3
P Q R S T
A
B
C
D
E
0937
62 0862
96 6596
57 2957
23 12+54
0 –31 –28 –54 –42
5
8
41
21
2
73 44
7
The solution under the test is not Optimal, as (VCE)BT = 7.
[ 4 ] What is [ Cij (ui + vj) ] in a vacant cell ?
Ans. : Vacant Cell Evaluation for ij vacant cell.
[ 5 ] What are the methods to convert “Profit Matrix” into “Relative Cost Matrix” ?
Ans. : (1) (Pij) (-1) = RCij
(3) (Pij)max - Pij = RCij(2) 1/Pij = RCij
(5) VAM: Vogel’s Approximate Method
Ans. : (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)
[ 6 ] State methods of getting IBFS of a Transportation Problem.
[ 7 ] What is indicated by “Penalty Number” in VAM ?
Ans. : Penalty for not selecting the proper cell for allocation.
[ 8 ] Name IBFS of Transportation Problem when number of filled cells is more than (m+n1).
Ans. : Redundant Solution.
[ 9 ] Give full form of following abbreviations :
NWCM, VCE, IBFS, MODI
MODI – Modified Distribution
IBFS – Initial Basic Feasible Solution
Ans. : NWCM – North West Corner Method
VCE – Vacant Cell Evaluation
Home Assignments :
[ 1 ] 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 ?
CentresA B C Supply
X 25 35 10 150
Y 20 5 80 100
Demand 50 50 150
Factories
[ 2 ] 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
[ 3 ] Get optimal solution of following “Minimization Transportation Problem”.
D1 D2 D3
O120 15 25 40
O215 10 20 55
30 25 50
[ 4 ] Find Optimal Solution of following Minimization Transportation Problem, Is it unique Optimal Solution ? If not, get alternative Optimal Solution : How many Optimal Solutions are possible ?
P Q R
A 2 5 6 10
B 4 3 4 6
4 7 5
[ 5 ] Following is one of the feasible solution of “Maximization Transportation Problem”. Is it optimal solution ? If not, get optimal solution. Is it a case of unique optimal solution ?
A B C
X 414
4 911
Y 34
516
8
7
58
[ 6 ] A feasible solution of Minimization Transportation Problem is given below.
( i ) Is it Optimal Solution ? If not, get it.
( ii ) Is the Optimal Solution unique ? Justify.
365O2
1042O1
D3D2D1
6
Thank youThank youFor any Query or suggestion :
Contact :Dr. D. B. Naik Professor & Head, Training & Placement (T&P)S. V. National Institute of Technology (SVNIT), Ichchhanath, Surat – 395 007 (Gujarat) INDIA.
Email ID : [email protected]@[email protected]
Phone No. : 0261-2201540 (D), 2255225 (O)
