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Transportation ModelSubject – QADM
Professor Kazi
Batch 97, Group 5• Atul Sande, Roll no. 04• Krupesh Shah, Roll no. 07• Vengadeshwaran Perumal, Roll no. 19• Pooja Chilap, Roll no. 32• Sagar Kuckian, Roll no. 36
OutlineO Transportation Modeling
O Benefits and ApplicationsO Methods Of Transportation
O Obtain the Initial Feasible Solution usingO Northwest - Corner RuleO Intuitive Least Cost Method / Minimization MethodO Vogel’s Approximation Method
O Obtain Feasible Solution for Optimality usingO Modified Distribution Method
O Conclusion
Transportation ModelingO What is Transportation Model?
A Transportation Model (TP) consists of determining how to route products in a situation where there are several destinations in order that the total cost of Transportation is minimised
O Can be used to help resolve distribution and location decisions
O Need to know:O The origin points and the capacity or supply per period
at eachO The destination points and the demand per period at
eachO The cost of shipping one unit from each origin to each
destination
Transportation Problem (TP)
A
B
C
D
E
F
SUPPLY DEMAND
Transportation Problem
FromTo
Andheri Bandra Chandivali
Dadar 5 4 3
Elphinston 8 4 3
Fort 9 7 5
Transportation Matrix
From
To Andheri Bandra Chandivali
Dadar
Elphinston
Fort
Factory capacit
y
Warehouse requirement
300
300
300 200 200
100
700
5
5
4
4
3
3
9
8
7
Cost of shipping 1 unit from factory to Bandra warehouse
Dadarcapacityconstraint
Cell representing a possible source-to-destination shipping assignment (Elphinston to Chandivali)
Total demandand total supplyChandivali
warehouse demand
DEVELOPING AN INITIAL SOLUTION— THE NORTHWEST CORNER RULE
Once the data have been arranged in tabular form, we must establish an initial feasible solution to the problem.
One systematic procedure, known as the northwest corner rule, requires that we start in the upper left hand cell (or northwest corner) of the table and allocate units to shipping routes as follows:
•Exhaust the supply ( factory supply) at each row before moving down to the next row.•Exhaust the (warehouse) requirements of each column before moving to the next column, on the right.•Check that all the supply and demands are met.
Demand Not Equal To Supply
A situation occurring quite frequently in real-world problems is the case where total demand is not equal to total supply.
These unbalanced problems can be handled easily by introducing a dummy Demand or dummy Supply.
In the event that total supply is greater than total demand, a dummy destination, with demand exactly equal to the surplus, is created.
If total demand is greater than total supply, we introduce a dummy source (factory) with a supply equal to the excess of demand over supply.
In either ease, cost coefficients of zero are assigned to each dummy location.
To (A) (B) (C)
(D)
(E)
(F)
Warehouse requirement 300 200 200
Factory capacity
300
300
250
850
5
5
4
4
3
3
9
8
7
From
50200
250
50
150
Dummy
150
0
0
0
150
Northwest Corner rule
Northwest – Corner Rule
To (A)Andheri
(B)Bandra
(C)Chandivali
(D) Dadar
(E) Elphinston
(F) Fort
Warehouse requirement 300 200 200
Factory capacity
300
300
100
700
5
5
4
4
3
3
9
8
7
From
100
100
100
200
200
Means that the firm is shipping 100 bathtubs from Fort to Bandra
Northwest Corner
Computed Shipping Cost
RouteFrom To Tubs Shipped Cost per Unit Total Cost
D A 100 5 500E A 200 8 1,600E B 100 4 400F B 100 7 700F C 200 5 1,000
Total: 4,200
This is a feasible solution but not necessarily the lowest cost
alternative
DrawbacksO The Northwest- corner rule is easy to use,
but this approach totally ignores the costsO Demand not equal to supplyO Called an unbalanced problemO Resolved by introducing dummy source or
dummy destination where in the aim of transportation model is minimization of cost so introducing a dummy source is not a good solution.
O Missing out the best cost effective path
Intuitive Lowest-Cost Method
O Identify the cell with the lowest costO Allocate as many units as possible to
that cell without exceeding supply or demand; then cross out the row or column (or both) that is exhausted by this assignment
O Find the cell with the lowest cost from the remaining cells
O Repeat steps 2 and 3 until all units have been allocated
Lowest - Cost MethodTo (A)
Andheri(B)
Bandra(C)
Chandivali
(D) Dadar
(E) Elphinston
(F) Fort
Warehouse requirement 300 200 200
Factory capacity
300
300
100
700
5
5
4
4
3
3
9
8
7
From
First, 3 is the lowest cost cell so ship 100 units from Dadar to Chandivali and cross off the first row as Dadar is satisfied
Lowest Cost MethodTo (A)
Andheri(B)
Bandra(C)
Chandivali
(D) Dadar
(E) Elphinston
(F) Fort
Warehouse requirement 300 200 200
Factory capacity
300
300
100
700
5
5
4
4
3
3
9
8
7
From
100
100
Second, 3 is again the lowest cost cell so ship 100 units from Elphinston to Chandivali and cross off column C as Chandivali is satisfied
Lowest Cost MethodTo (A)
Andheri(B)
Bandra(C)
Chandivali
(D) Dadar
(E) Elphinston
(F) Fort
Warehouse requirement 300 200 200
Factory capacity
300
300
100
700
5
5
4
4
3
3
9
8
7
From
100
100
200
Third, 4 is the lowest cost cell so ship 200 units from Elphinston to Bandra and cross off column B and row E as Elphinston and Bandra are satisfied
Lowest Cost MethodTo (A)
Andheri(B)
Bandra(C)
Chandivali
(D) Dadar
(E) Elphinston
(F) Fort
Warehouse requirement 300 200 200
Factory capacity
300
300
100
700
5
5
4
4
3
3
9
8
7
From
100
100
200
300
Finally, ship 300 units from Andheri to Fort as this is the only remaining cell to complete the allocations
Lowest Cost MethodTo (A)
Andheri(B)
Bandra(C)
Chandivali
(D) Dadar
(E) Elphinston
(F) Fort
Warehouse requirement 300 200 200
Factory capacity
300
300
100
700
5
5
4
4
3
3
9
8
7
From
100
100
200
300
Total Cost = 3(100) + 3(100) + 4(200) + 9(300)= 4,100
Lowest Cost Method
To (A)Andheri
(B)Bandra
(C)Chandivali
(D) Dadar
(E) Elphinston
(F) Fort
Warehouse requirement 300 200 200
Factory capacity
300
300
100
700
5
5
4
4
3
3
9
8
7
From
100
100
200
300
Total Cost = 3(100) + 3(100) + 4(200) + 9(300)= 4,100
This is a feasible solution, and an improvement over the previous solution, but not necessarily the
lowest cost alternative
Vogel’s Approximation Method
O Calculate penalty for each row and column by taking the difference between the two smallest unit costs. This penalty or extra cost has to be paid if one fails to allocate the minimum unit transportation cost.
O Select the row or column with the highest penalty and select the minimum unit cost of that row or column. Then, allocate the minimum of supply or demand values in that cell. If there is a tie, then select the cell where maximum allocation could be made.
O Adjust the supply and demand and eliminate the satisfied row or column. If a row and column are satisfied simultaneously, only of them is eliminated and the other one is assigned a zero value.Any row or column having zero supply or demand, can not be used in calculating future penalties.
O Repeat the process until all the supply sources and demand destinations are satisfied
Vogel’s Approximation MethodSupply Row Diff.
19 30 50 10
70 30 40 60
40 8 70 20
8Demand 34
Col.Diff.
9
10
12
21 22 10 10
D1 D2 D3
S1
S2
S3
5 8 7
D4
14
7
9
18
Supply Row Diff.
19 50 10
5
70 40 60
40 70 20
Demand 34
Col.Diff.
D1 D3 D4
S1 7
S2 9
S3 10
5 7 14
21 10 10
9
20
20
Vogel’s Approximation MethodSupply Row Diff.
50 10
40 60
70 20
10Demand 34
Col.Diff. 10 10
40
20
50
D3 D4
S1 2
S2 9
S3 10
7 14
Supply Row Diff.
50 10
2
40 60
Demand 34
Col.Diff.
20
10 50
40
7 4
S1 2
S2 9
D3 D4 Supply Row Diff.
40 60
7 2Demand 34
Col.Diff.
20
7 2
S2 9
D3 D4
Vogel’s Approximation Method
O The total transportation cost obtained by this method= 8*8+19*5+20*10+10*2+40*7+60*2= Rs.4055
O Here, we can see that Vogel’s Approximation Method involves the lowest cost than North-West Corner Method and Least Cost Method and hence is the most preferred method of finding initial basic feasible solution.
Benefits of Vogel’s Approximation Method
O VAM is an improved version of the least-cost method that generally, but not always, produces better starting solutions.
O This method is preferred over the other methods because it generally yields, an optimum, or close to optimum, starting solutions.
Obtain Feasible solution for Optimality using MODI Method
The modified distribution method, MODI for short , is an improvement over the stepping stone method for testing and finding optimal solutions
Steps Involved in MODI Method
O Find a basic solution by any standard method. If supply and demand are equal then it is a balanced transportation problem.
O Test for optimality. The number of occupied cells should equal to m + n -1. If the initial basic feasible solution does not satisfy this rule, then optimal solution cannot be obtained. Such solution is a degenerate solution
O Set up a cost matrix for allocated cells only.
Steps Involved in MODI Method
O Determine a set of number Ui for each row and a set of number Vj on the bottom of the matrix.
O Compute the value of Ui and Vj with the formula Ui + Vj = Cij to all basic(occupied) cells.
O Calculate the water value of of non-basic ( unoccupied) cells using the relation Ui+Vj=Cij.
O Compute the penalties for each unoccupied cell by using the formula Dij=Pij=Ui+Vj-Cij.
Examine whether all Pij ≤ 0. If all Pij < 0, then the solution is optimal and unique. If all Pij ≤ 0, then the solution is optimal and an alternative
solution exists. If at least one Pij > 0, then the solution is not optimal.
At the end, prepare the optimum solution table and calculate the optimum/minimum transportation cost.
Optimum Solution Using Modi Method
8 8 15
15 10 17
3 9 10
REQUIREMENT 150 80 50
120
80
80
CAPACITY
D
E
F
A B C
W1 W2 W3
F1
F2
F3
8 8 15
15
3
10 17
9 10
120
30 50
30 50
150 80 50
120
80
80
IBFS= 120(8)+30(15)+50(10)+30(9)+50(10)
IBFS=960+450+500+270+500=2680
OCCUPIED MATRIX UNOCCUPIED MATRIX
8
15 10
9 10
Vj V1=8 V2=3 V3=4
Ui
U1=0
U2=7
U3=6
Vj 8 3 4
Ui Ui
0
7
6
-5 -11
-6
11
3 4
11
14
-8 -15
-3
-17
Pij=(Ui+Vj)-CijUi + Vj = Cij
1588
15 10 17
3 9 10
10
120
30 50
30 50+
+_
_
8
15
3
8 15
10 17
9 10
120 E
80
3050
STONE SEQUARE=RIM REQUIREMENTm+n-1=5
DEGENERACY OCCUAR
Value of O is equal to the minimum of the existing allocation among the signed cells on the loop.
LOOP CONSTRUCT
120
80
80
Dj 150 80 50 280
Si
OPTIMUM SOLUTION TABLE
8 8 15
15 10 17
3 9 10
OPTIMUM COST
• F1 W1 8*120 =960
• F1 W2 8*E = _
• F2 W2 10*80 =800
• F3 W1 3*30= 90
• F3 W3 10*50 =500
____________
2350
120
30
E
80
50
Why MODI Method?
O It Is the simplex method O It is a minimum cost solution to the
transportation problem.O All the drawbacks which were in all
the three methods is covered in this modi method.
Thank You!!
