2. WHAT IS TRANSPORTATION PROBLEM ?
• The transportation problem is a special type of linear programming
problem where the objective is to minimize the cost of distributing a
product from a number of sources or origins to a number
of destinations.
• Because of its special structure, the usual simplex method is not
suitable for solving transportation problems. These problems require a
special method of solution.
3. AIM OF TRANSPORTATION PROBLEM:
• To find out the optimum transportation schedule keeping in mind cost of
Transportation to be minimized.
• The origin of a transportation problem is the location from which shipments
are despatched.
• The destination of a transportation problem is the location to which
shipments are transported.
• The unit transportationcost is the cost of transporting one unit of the
consignment from an origin to a destination
4. • Determination of a transportation plan of a single commodity
• From a number of sources
• To a number of destinations,
• Such that total cost of transportation id minimized.
OBJECTIVES OF TRANSPORTATION
PROBLEM
5. • It is used to compute transportation routes in such a way as to minimize
transportation cost for finding out locations of warehouses.
• It is used to find out locations of transportation corporations depots where
insignificant total cost difference may not matter.
• Minimize shipping costs from factories to warehouses(or from warehouses to retail
outlets).
• Determine lowest cost location for new factory, warehouse, office ,or other
outlet facility.
• Find minimum cost production schedule that satisfies firms demand and production
limitations.
APPLICATIONS OF TRANSPORTATION
6. • The transportation problem is one of the most frequently encountered application in
real life situations and is a special type of linear programming problem.
• The transportation problem indicatesthe amount of consignment to be transported
from various origins to different destinations so that the total transportation cost is
minimized without violating the availability constraints and the requirement
constraints.
CONCLUSION
7. TWO TYPES OF TRANSPORTATION
PROBLEM:-
• BALANCED TRANSPORTATIONPROBLEM:- Where the total supply
equally total demand.
• UNBALANCED TRANSPORTATIONPROBLEM:- Where the total supply
is not equally total demand.
PHASES OF SOLUTION OF TRANSPORTATION
PROBLEM:-
▪ PHASE I:- Obtains the Initial Basic Feasible Solution.
▪ PHASE II:- Obtains the Optimal Basic Solution.
8. INITIAL BASIC FEASIBLE
SOLUTION:-
▪ NORTH-WEST CORNER RULE(NWCR)
▪ LEAST COST METHOD(LCM)
▪ VOGLE APPROXIMATION METHOD(VAM)
OPTIMUM BASIC SOLUTION:-
▪ MODIFIED DISTRIBUTION METHOD (MODI
METHOD)
▪ STEPPING STONE METHOD
9. NORTH-WEST CORNER RULE (NWCR):-
• DEFINATION:- The North-West Corner Rule is a method
adopted to compute the Initial Feasible Solution of the
transportation problem. The name North-West Corner is given to
this method because the basic variables are selected from the
extreme left corner.
• It is most systematic and easiest method for obtaining Initial
Feasible Basic Solution.
10. STEPS IN NORTH-WEST CORNER METHOD
▪ STEP 1:- Select the upper left (north-west) cell of the transportation
matrix and allocate minimum of supply and demand, i.e min(A1,B1)
value in that cell.
▪ STEP2:-
➢IfA1<B1, then allocation made is equal to the supply available at the
first source(A1 in first row), then move vertically down to the cell (2,1).
➢If A1>B1, then allocation made is equal to demand of the first
destination (B1 in first column), then move horizontally to the cell (1,2).
➢IfA1=B1, then allocate the value ofA1 or B1and then move to cell
(2,2).
▪ STEP3:- Continue the process until an allocation is made in the south-
east corner cell of the transportation table.
11. In the table, three sourcesA,B and C with the Production Capacity of 50units, 40units, 60
units of product respectively is given. Every day the demand of three retailers D,E,F is to
be furnished with at least 20units, 95units and 35units of product respectively. The
transportation costs are also given in the matrix.
Step 1:- Check whether Total Demand is equal to Total Supply. In case the demand is more
then supply, then dummy origin is added to the table. The cost associated with the dummy
origin will be zero.
Step 2:- Select the North-West or Extreme left corner of the matrix, assign as many units as
possible to cellAD, within the supply and demand constraints. Such as 20units are assigned
to the first cell, that satisfies the demand of destination D while the supply is in Surplus.
Now Assign 30 units to the cellAE. Since 30units are available with the sourceA, the
supply gets fully saturated.
SOURCE TO D E F SUPPLY
A 5 8 4 50
B 6 6 3 40
C 3 9 6 60
DEMAND 20 95 35 150/150
12. STEP 3:- Now move vertically, and assign 40 units to cell BE. The
supply of source B also gets fully saturated. Again move vertically,
assign 25units to cell CE, demand of destination E is fulfilled.
Move horizontally in the matrix and assign 35units to cell CF, both
the demand and supply of origin and destination get saturated. Now
Total Cost can be computed.
Total Cost can be computed by multiplying the units assigned to
each cell with the concerned transportation cost.
INTIAL TRANSPORTATION COST
TOTAL COST = (20*5)+(30*8)+(40*6)+(25*9)+(35*6)= Rs 1015.
SOURCES / TO D E F SUPPLY
A 5 (20) 8 (30) 4 50
B 6 6 (40) 3 40
C 3 9 (25) 6 (35) 60
DEMAND 20 95 35 150/150
13. LEASTCOST METHOD
USES OF LEAST COSTMETHOD
TRANPORTATION PROBLEM(BALANCED)
TRANPORTATION PROBLEM(UNBALANCED)
LEAST COST METHOD
Contents
17. 1 The minimum cost in the matrix is Rs 3, but there is a tie in the cell BF, and CD, now the question arises in which cell we
shall allocate. Generally, the cost where maximum quantity can be assigned should be chosen to obtain the better initial
solution. Therefore, 35 units shall be assigned to the cell BF.
18. 5
Again the minimum cost in the matrix is Rs 3. Therefore, 20 units shall be assigned to the cell CD. With this, the demand of
retailer D gets fulfilled. Only 40 units are left with the source C.
19. 40
The next minimum cost is 8, assign 50 units to the cellAE. The supply of sourceAgets saturated.
The next minimum cost is Rs 9; we shall assign 40 units to the cell CE. With his both the demand and supply of all the
sources and origins gets saturated.
20. Total Cost = 50*8 + 5*6 + 35*3 +20*3 +40*9 = Rs 955.
The total cost can be calculated by multiplying the assigned
quantity with the concerned cost of the cell. Therefore,
21. Example of Unbalanced Transportation Problem
Plant Warehouse Supply
W1 W2 W3
A 28 17 26 500
B 19 12 16 300
Demand 250 250 500
The total demand is 1000, whereas the total supply is 800.
Si<Dj
Total supply < total demand.
23. VOGEL APPROXIMATION METHOD
(VAM METHOD)
The Vogel Approximation Method is an improved version of
the Minimum Cell Cost Method and the Northwest Corner
Method that in general produces better initial basic feasible
solution, that report a smaller value in the objective
(minimization) function of a balanced Transportation Problem.
(sum of the supply = sum of the demand).
Applying the V
ogel Approximation Method requires the
following STEPS:
• Step 1: Determine a penalty cost for each row (column) by
subtracting the lowest unit cell cost in the row (column) from
the next lowest unit cell cost in the same row (column).
24. • Step 2: Identify the row or column with the greatest penalty cost. Break the
ties arbitrarily (if there are any). Allocate as much as possible to the variable
with the lowest unit cost in the selected row or column. Adjust the supply and
demand and cross out the row or column that is already satisfied. If a row and
column are satisfied simultaneously, only cross out one of the two and allocate
a supply or demand of zero to the one that remains.
• Step 3:
• If there is exactly one row or column left with a supply or demand of zero,
stop.
• If there is one row (column) left with a positive supply (demand), determine
the basic variables in the row (column) using the Minimum Cell Cost Method.
Stop.
• If all of the rows and columns that were not crossed out have zero supply and
demand (remaining), determine the basic zero variables using the Minimum
Cell Cost Method. Stop.
• In any other case, continue with Step 1.
25. The concept of Vogel’s Approximation Method can be well understood
through an ILLUSTRATION given below:
• First of all the difference between two least cost cells are calculated for
each row and column, which can be seen in the iteration given for each
row and column. Then the largest difference is selected, which is 4 in this
case. So, allocate 20 units to cell BD, since the minimum cost is to be
chosen for the allocation. Now, only 20 units are left with the source B.
26. • Column D is deleted, again the difference between the least cost cells
is calculated for each row and column, as seen in the iteration below.
The largest difference value comes to be 3, so allocate 35 units to cell
AF and 15 units to the cell AE. With this, the Supply and demand of
source A and origin F gets saturated, so delete both the row A and
Column F.
27. • Now, single column E is left, since no difference can be found out, so
allocate 60 units to the cell CE and 20 units to cell BE, as only 20
units are left with source B. Hence the demand and supply are
completely met.
28. • Now the total cost can be computed, by multiplying the units
assigned to each cell with the cost concerned. Therefore,
• Total Cost = 20*3 + 35*1 + 15*4 + 60*4 + 20*8 = Rs 555
• Note: Vogel’s Approximation Method is also called as Penalty
Method because the difference costs chosen are nothing but the
penalties of not choosing the least cost routes.
29. MODI METHOD(Modified Distribution Method)
• 1. Determine an initial basicfeasiblesolution using any one of the three methods given below:
• North West CornerMethod
• Matrix MinimumMethod
• VogelApproximationMethod
• 2. Determine the values of dual variables,ui and vj, using ui + vj = cij
• 3. Computethe opportunitycost using cij – ( ui + vj ).
• 4. Check the sign of each opportunitycost. If the opportunitycosts of all the unoccupied cells are either
positiveor zero, the given solutionis the optimal solution.On the other hand, if one or more unoccupied cell
has negativeopportunitycost, the given solutionis not an optimal solutionand further savingsin
transportation cost are possible.
• 5. Select the unoccupied cellwith the smallest negativeopportunitycost as the cell to be includedin the
next solution.
• 6. Draw a closed path or loop for the unoccupied cell selected in the previous step. Please note that the right
angle turn in this path is permittedonly at occupied cells and at the original unoccupied cell.
• 7. Assign alternateplus and minus signs at the unoccupied cells on the corner points of the closed path with
a plus sign at the cell being evaluated.
• 8. Determine the maximum number of units that shouldbe shipped to this unoccupied cell. The smallest
valuewith a negativepositionon the closed path indicatesthe number of units that can be shipped to the
entering cell. Now, add this quantityto all the cells on the corner pointsof the closed path marked with plus
signs, and subtract it from those cells marked with minus signs. In this way, an unoccupied cell becomes an
occupied cell.
30. Distribution centre
D1 D2 D3 D4 Supply
Plant
P1 19 30 50 12 7
P2 70 30 40 60 10
P3 40 10 60 20 18
Requireme
nt
5 8 7 15
onsider the transportation problem presented in the
following table.
31. Distribution centre
D1 D2 D3 D4 Supply
Plant P1 19 30 50 7
P2 30 60 10
P3 60 18
Require
= 2ment
5 8 7 15
An initial basic feasible solution is obtained by Matrix
Minimum Method and is shown in table 1.
Table 1
Initial basic feasible solution
12 X 7 + 70 X 3 + 40 X 7 + 40 X 2 + 10 X 8 + 20 X 8 = Rs.
894.
32. Calculating ui and vj using ui + vj = cij
Substituting u1 = 0, we get
u1 + v4 = c14
c34
c32
c31
⇒ 0 + v4 = 12 or v4 = 12
⇒ u3 + 12 = 20 or u3 = 8
⇒ 8 + v2 = 10 or v2 = 2
⇒ 8 + v1 = 40 or v1 = 32
u3 + v4 =
u3 + v2 =
u3 + v1 =
u2 + v1 = c21 ⇒ u2 + 32 = 70 or u2 = 38
u2 + v3 = c23 ⇒ 38 + v3 = 40 or v3 = 2
Distribution centre
D1 D2 D3 D4 Supply ui
Plant P1 19 30 50 7 0
P2 30 60 10 38
P3 60 18 8
Requi
remen
t
5 8 7 15
vj 32 2 2 12
34. Distribution centre
D1 D2 D3 D4 Supply ui
Plant P1 7 0
P2 10 38
P3 18 8
Requiremen
t
5 8 7 15
vj 32 2 2 12
Now choose the smallest (most) negative value from
opportunity cost (i.e., –13) and draw a closed path
from P1D1. The following table shows the closed path.
Table 4
35. Choose the smallest value with a negative position on
the closed path(i.e., 2), it indicates the number of units
that can be shipped to the entering cell. Now add this
quantity to all the cells on the corner points of the
closed path marked with plus signs and subtract it from
those cells marked with minus signs. In this way, an
unoccupied cell becomes an occupied cell.
Now again calculate the values for ui & vj and
opportunity cost. The resulting matrix is shown below.
Table 5
36. Distribution centre
D1 D2 D3 D4 Supply ui
Plant P1 7 0
P2 10 51
P3 18 8
Requir
ement
5 8 7 15
vj 19 2 –11 12
Choose the smallest (most) negative value from
opportunity cost (i.e., –23). Now draw a closed path
from P2D2 .
38. Distribution centre
D1 D2 D3 D4 Supply ui
Plant P1 7 0
P2 10 28
P3 18 8
Requi
remen
t
5 8 7 15
vj 19 2 12 12
Since all the current opportunity costs are non–negative,
this is the optimal solution. The minimum transportation
cost is: 19 X 5 + 12 X 2 + 30 X 3 + 40 X 7 + 10 X 5 + 20 X
13 = Rs. 799
