Operations Management MBA Sem II
51
Operations Management MBA Sem II Module IV Transportation
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Operations Management MBA Sem II. Module IV Transportation. …the physical distribution of goods and services from several supply centers to several demand centres. Transportation. - PowerPoint PPT Presentation
Transcript of Operations Management MBA Sem II
Operations ManagementMBA Sem II
Module IV
Transportation
…the physical distribution of goods and services from several supply centers to several demand centres.
Transportation
The structure of transportation problem involves a large no. of shipping routes from several supply origins to several demand centres.
Objective : To determine the number of units of an item that should be shipped from an origin to a destination in order to satisfy the required quantity of goods or services at each destination centre.
Method
1. Formulate the problem and arrange the data in matrix form
2. Obtain an initial basic feasible solution by
- North West Corner Method
- Least Cost method
- Vogel’s Approximation Method
The solution must satisfy all the supply and demand constraints
The number of positive allocations must be m+n-1, where m is the no. of rows
and n is the no. of columns
3. Test the initial solution for optimality (MODI)
4. Update the solution
Initial basic solution by North west corner method
Q
A company has three production facilities S1,S2, S3 with production capacity of 7,9 and 18 units (in 100’s) per week of a product, respectively. These units are to be shipped to four warehouses D1, D2, D3 and D4 with requirement of 5,6,7 and 14 units (in 100’s) per week, respectively.
The transportation costs(in rupees) per unit between factories to warehouses are
given in the table in the next slide.
Minimize the total transportation cost.
D1 D2 D3 D4 CAPACITY
S1 19 30 50 10 7
S2 70 30 40 60 9
S3 40 8 70 20 18
Demand 5 8 7 14 34
• The no. of positive allocations (occupied cells) = m + n -1 = 6
D1 D2 D3 D4 CAPACITY
S1 19 5 30 2 50 10 7
S2 70 30 6 40 3 60 9
S3 40 8 70 4 20 14 18
Demand 5 8 7 14 34
Total cost = 5 *19 + 2 * 30 + 6*30 + 3 * 40 +
4 * 70 + 14* 20 = Rs. 1,015
Initial basic solution by VAM
Choose largest value 22 in columnD2
Choose cell with lowest cost
Satisfy it
D1 D2 D3 D4 capacity Row differences
S1 19 30 50 10 7 9
S2 70 30 40 60 9 10
S3 40 8 70 20 18 12
Demand 5 8 7 14 34Column differences
21 22 10 10
ghj
D1 D2 D3 D4 cap Row diff
S1 19 30 50 10 7 9
S2 70 30 40 60 9 20
S3 40 8 70 20 18 20
Demand 5 8 7 14 34Column differences
21 22 10 10
8
5
D1 D2 D3 D4 capacity
Row diff
S1 19 30 50 10 7 40
S2 70 30 40 60 9 20
S3 40 8 70 20 18 50
Demand 5 8 7 14 34
Column differences
10 10
8
5
10
D1 D2 D3 D4 capacity
Row diff
S1 19 30 50 10 7 40
S2 70 30 40 60 9 20
S3 40 8 70 20 18
Demand 5 8 7 14 34Column differences
10 50
5
8 10
2
D1 D2 D3 D4 capacity
Row diff
S1 19 30 50 10 7 40
S2 70 30 40 60 9 20
S3 40 8 70 20 18
Demand 5 8 7 14 34
Column differences
10 50
5
2
27
8 10
Total cost : 5*19 + 2*10 + 7 *40 + 2*60
+ 8 *8 + 10 *20 = Rs 779
Q
consider the following transportation problem involving 3 sources and four destinations. The cell entries represent the cost of transportation per unit. Obtain the initial basic feasible soln using
1. NWCM2. VAMFind the optimal soln (after NWCM) using
UV method
Dest
source
1 2 3 4 supply
1 3 1 7 4 300
2 2 6 5 9 400
3 8 3 3 2 500
Demand 250 350 400 200 1200
• By VAM
Dest
source
1 2 3 4 supply
1 3 1 7 4 300
2 2 6 5 9 400
3 8 3 3 2 500
Demand 250 350 400 200 1200
300
250 150
50 250 200
m+n-1 = 6
Hence this is a feasible soln
Total Cost = Rs. 2850
By NWCM
Dest
source
1 2 3 4 supply
1 3 1 7 4 300
2 2 6 5 9 400
3 8 3 3 2 500
Demand 250 350 400 200 1200
250 50
300 100
300 200
• Total cost = Rs. 4400
Optimal soln
Row 1,2,3 are assigned values U1,U2,U3
AND
Col 1, col2, col3 and col 4 are assigned variables V1,V2,V3,V4
• FOR BASIC cells Ui + Vj = cij
• Take U1 = 0
V1=3 V2=1 V3=0 V4=-1
1 2 3 4 supply
U1=0 1 3 1 7 4 300
U2=5 2 2 6 5 9 400
U3=3 3 8 3 3 2 500
250 350 400 200 1200
25050
300100
200300
If all pij<=0; optimality is reached
Compute Pij (penalties) for the non basic cells by using the formula :
Pij = Ui + Vj - cij
V1=3 V2=1 V3=0 V4=-1
1 2 3 4 supply
U1=0 1 3 1 7 4 300
U2=5 2 2 6 5 9 400
U3=3 3 8 3 3 2 500
250 350 400 200 1200
250 50
300 100
300 200
6
-ve -ve
-ve
-ve 1
If all Pij are <= 0; then optimality is reached;
Cell (2,1) with penalty 6, has the most positive penalty. So, there is scope for improving soln.
Construct a loop using one basic cell and other non basic cells.
Assign + and – signs alternatively to the basic cells.
Choose the minimum among the –vely assigned basic cells = 250
Add to all +vely cells.
Subtract from all –vely assigned cells.
V1=3 V2=1 V3=0 V4=-1
1 2 3 4 supply
U1=0 1 -
3
+ 1 7 4 300
U2=5 2 2 + -
6
5 9 400
U3=3 3 8 3 3 2 500
250 350 400 200 1200
250
6
-ve
50
300
1
-ve
100
300
-ve
-ve
200
V1=-3 V2=1 V3=0 V4=-1
1 2 3 4 supply
U1=0 1 3 1 7 4 300
U2=5 2 2 -
6
5
+
9 400
U3=3 3 8 + 3 X 3 - 2 500
250 350 400 200 1200
-ve
300
-ve -ve
250 50 100
-ve
-ve 1300 200
V1=-2 V2=1 V3=1 V4=0
1 2 3 4 supply
U1=0 1 3 1 7 4 300
U2=4 2 2 6 5 9 400
U3=2 3 8 3 3 2 500
250 350 400 200 1200
-ve300
-ve -ve
250150
-ve
-ve 50
-ve
250 200
All penalties <=0;
Hence optimum solution is reached
Total = 2850
Q
Consider the transportation problem as shown in the table. Find the initial basic solution using
1. NWCM
2. VAM
APPLY UV method to find the optimal solution.
1 2 3 4 5 supply
1 10 2 16 14 10 300
2 6 18 12 13 16 500
3 8 4 14 12 10 825
4 14 22 20 8 18 375
Demand
350 400 250 150 400
The given problem is unbalanced because the total demand(1550) < total supply (2000)
Convert to a balanced transportation problem.
1 2 3 4 5 6 Supply
1 10 2 16 14 10 0 300
2 6 18 12 13 16 0 500
3 8 4 14 12 10 0 825
4 14 22 20 8 18 0 375
Demand
350 400 250 150 400 450 2000
Initial basic soln using NWCM
1 2 3 4 5 6 supply
1 10 2 16 14 10 0 300
2 6 18 12 13 16 0 500
3 8 4 14 12 10 0 825
4 14 22 20 8 18 0 375
Demand
350 400 250 150 400 450
300
50 400 50
200 150 400 75
375
Total cost = Rs. 19,700
Initial basic solution using VAM
1 2 3 4 5 6 Supply
1 10 2 16 14 10 0 300
2 6 18 12 13 16 0 500
3 8 4 14 12 10 0 825
4 14 22 20 8 18 0 375
Demand 350 400 250 150 400 450 2000
300
350 75 75
100175 150 400
375
Total cost = Rs. 12250
Asssign Ui & Vj
V1=6 V2=2 V3=12 V4=10
V5=8
V6=0
1 2 3 4 5 6 Supply
U1=0 1 10 2 16 14 10 0 300
U2=0 2 6 18 12+ 13 16 0 - 500
U3=2 3 8 4 14 - 12 10 0 + 825
U4=0 4 14 22 20 8 18 0 375
D 350 400 250 150 400 450 2000
300
350 75 75
100 175 150 400
2 375
- ve -ve-ve -ve
-ve -ve -ve
-ve
-ve -ve -ve -ve
0
2
V1=6 V2=2 V3=12 V4=10 V5=8 V6=-2
1 2 3 4 5 6 S
U1=0 1 10 2 16 14 10 0 300
U2=0 2 6 18 12 13 16 0 500
U3=2 3 8 4 14 -
12
10 + 0 825
U4=2 4 14 22 20 8
+
18 0
-
375
D
350 400 250 150 400 450 2000
-ve -ve -ve -ve -ve
-ve -ve -ve -ve
-ve -ve -ve -ve
300
350 150
0100 100 150 400 75
4 375
V1=6 V2=2 V3=12 V4=6
V5=8 V6=-2
1 2 3 4 5 6 S
U1=0 1 10 2 16 14 10 0 300
U2=0 2 6 18 12 13 16 0 500
U3=2 3 8 4 14 -
12
10 0 825
U4=2 4 14 22 20 8 18 0
-
375
D
350 400 250 150 400 450 2000
-ve -ve -ve -ve -ve
-ve -ve -ve -ve
-ve
-ve -ve -ve -ve
300
350 150
0 100 100 400 225
150 225
• Total cost = Rs. 11,500
