1.
Transportation Model
Subject – 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
2.
Outline
O Transportation Modeling
O Benefits and Applications
O Methods Of Transportation
O Obtain the Initial Feasible Solution using
O Northwest - Corner Rule
O Intuitive Least Cost Method / Minimization Method
O Vogel’s Approximation Method
O Obtain Feasible Solution for Optimality using
O Modified Distribution Method
O Conclusion
3.
Transportation Modeling
O 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 each
O The destination points and the demand per period at
each
O The cost of shipping one unit from each origin to each
destination
4.
Transportation Problem (TP)
SUPPLY
DEMAND
A
D
B
E
C
F
5.
Transportation Problem
To
From
Andheri
Bandra
Chandivali
Dadar
5
4
3
Elphinston
8
4
3
Fort
9
7
5
6.
Transportation Matrix
To
Andheri
Bandra
From
5
Dadar
8
Elphinston
9
Fort
Warehouse
requirement
300
Cost of shipping 1 unit from factory
to Bandra warehouse
4
Factory
Chandivali capacit
y
3
100
Dadar
capacity
constraint
Cell
representi
4
3
ng a
300
possible
source-todestinatio
7
5
n shipping
300
assignme
nt
(Elphinsto
200
200
700
n to
Chandivali
)
Total demand
Chandivali
and total supply
warehouse demand
7.
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.
8.
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.
10.
Northwest – Corner Rule
To
From
(D) Dadar
(E) Elphinston
(A)
Andheri
100
200
(C)
Chandivali
5
4
3
8
4
3
9
(F) Fort
Warehouse
requirement
(B)
Bandra
300
100
100
200
7
200
200
5
Factory
capacity
100
300
300
700
Means that the firm is shipping 100 bathtubs
from Fort to Bandra
11.
Northwest Corner
Computed Shipping Cost
Route
From
To
D
E
E
F
F
A
A
B
B
C
Tubs Shipped
Cost per Unit
100
200
100
100
200
5
8
4
7
5
Total Cost
500
1,600
400
700
1,000
Total: 4,200
This is a feasible solution but not
necessarily the lowest cost alternative
12.
Drawbacks
O The Northwest- corner rule is easy to use, but
O
O
O
O
this approach totally ignores the costs
Demand not equal to supply
Called an unbalanced problem
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.
Missing out the best cost effective path
13.
Intuitive Lowest-Cost Method
O Identify the cell with the lowest cost
O 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
14.
Lowest - Cost Method
To
From
(A)
Andheri
Warehouse
requirement
300
3
4
3
9
(F) Fort
4
8
(E) Elphinston
(C)
Chandivali
5
(D) Dadar
(B)
Bandra
7
5
200
200
Factory
capacity
100
300
300
700
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
15.
Lowest Cost Method
To
From
(A)
Andheri
5
(F) Fort
300
4
9
(E) Elphinston
4
8
(D) Dadar
Warehouse
requirement
(B)
Bandra
(C)
Chandivali
7
200
100
100
3
3
5
200
Factory
capacity
100
300
300
700
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
16.
Lowest Cost Method
To
From
(A)
Andheri
3
4
200
9
(F) Fort
Warehouse
requirement
4
8
(E) Elphinston
(C)
Chandivali
5
(D) Dadar
(B)
Bandra
300
100
100
7
200
3
5
200
Factory
capacity
100
300
300
700
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
17.
Lowest Cost Method
To
From
(A)
Andheri
5
(E) Elphinston
Warehouse
requirement
300
300
4
8
(D) Dadar
(F) Fort
(B)
Bandra
4
200
9
(C)
Chandivali
100
100
7
200
3
3
5
200
Factory
capacity
100
300
300
700
Finally, ship 300 units from Andheri to Fort as this is the only
remaining cell to complete the allocations
19.
Lowest Cost Method
To
From
(A)
Andheri
(B)
Bandra
This is a feasible solution, and 5
an
(D) Dadar
improvement over the previous
solution, but not necessarily the lowest
8
cost alternative
(E) Elphinston
200
(F) Fort
Warehouse
requirement
Total Cost
300
300
9
4
4
(C)
Chandivali
100
100
7
200
3
3
5
200
= 3(100) + 3(100) + 4(200) + 9(300)
= 4,100
Factory
capacity
100
300
300
700
20.
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
23.
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.
24.
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.
25.
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
26.
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.
27.
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.
28.
Optimum Solution Using
Modi Method
CAPACITY
A
D
B
8
C
8
15
120
E
15
10
17
80
3
9
10
80
F
REQUIREMENT
150
80
50
31.
LOOP CONSTRUCT
8
8
15
8
120
15
_
10
+
9
17
_
15
Si
120
E
10
15
17
80
80
3
10
30
+
120
50
30
3
8
50
1
30
0
9
10
50
80
DEGENERACY OCCUAR
Dj 150
80
50
280
Value of O is equal to the minimum of the
existing allocation among the signed cells on STONE SEQUARE=RIM REQUIREMENT
m+n-1=5
the loop.
33.
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.
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