1. Transportation and Assignment Problems
Chapter 6
Transportation and Assignment Problems

2. Introduction
Introduce two linear programming model formulations—transportation and assignment problems
Represent unique mathematical solution approaches.
Solution approaches are variations of the traditional simplex solution

3. The transportation Model
Applied to a wide variety of problems:
(1) a product is transported from certain sources to certain destinations
(2) each source is capable of supplying a fixed number of units of the product and each destination has a fixed demand for the product
The objective is to determine shipping schedule that minimizes the total shipping cost while satisfying supply and demand limits

4. The General Format . . C11:X11 a1 b1 1 1 a2 b2 2 2 . bm am Cmn:Xmn n m
Represented by the network
Any arc has two pieces of information:
Transportation cost per unit
Amount shipped
The amount of supply at source is a and demand at destination is b
Determine the unknown shipped amounts that will minimize the total transportation cost
C11:X11 a1 b1 1 1
Units of
Demand a2 b2
Units of
Supply 2 2 . . . bm am
Cmn:Xmn n m

5. Example Distribution Area 1 2 3 6 8 10 7 11 4 5 12
Oranges are grown, picked and stored in warehouse 1, warehouse 2, and warehouse 3 with daily capacities of 150, 175, and 275 truckloads
Supply oranges to three distribution areas with daily demands of 200, 100, and 300 truckload
Table gives the transportation cost ($100s) per truckload between the warehouses and the distribution areas
Determine how many truckload to transport from each of the warehouses to each distribution areas in order to minimize the total cost of transportation
Distribution Area
W 1 A R E H O U
S 3 6 8 10 7 11 4 5 12

6. Transportation Tableau
Convert problem more conveniently using transportation tableau
Each cell represents amount transported from one source to one destination
Amount placed in each cell is the value of a decision variable
Smaller box within each cell contains the unit transportation cost
Last column and last row of tableau also show the amount of supply and demand constraints
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6
Xi1 8
Xi2 10
Xi3
150 7
X21 11
X22
X23
175 4
X31 5
X32 12
Xi33
275
Demand
200
100
300
600

7. Linear Programming Formulation
LP formulation for this problem is:
First three constraints represent the supply
Last three constraints represent the demand
Note that these constraint are equations rather than ≤ inequalities
This type is referred to as a balanced transportation model
Min Z=6X11+8X12+10X13+ 7X21+11X22+11X23+4X31+5X32+12X33
Subject to:
X11+X12+X13=150
X21+X22+X23=175
X31+X32+X33=275
X11+X21+X31=200
X12+X22+X32=100
X13+X23+X33=300
Xij≥0

8. Transportation Algorithm
Balanced one will be used to demonstrate the solution of a transportation problem
Solved using TORA and Excel and follow the exact steps of the simplex method
Take advantage of special structure to organize the computation in a more convenient way
Transportation algorithm provides insight into the use of the theoretical primal-dual relationships
Manual solution methods will be presented
Then computerized solution will be demonstrated

9. Starting the Basic Solution
Special structure allows creating a starting basic solution using one of the three methods
Northwest-Corner method
Least-cost method
Vogel approximation method

10. The Northwest Corner Method
An initial allocation is made to the cell in the upper left-hand corner of the tableau
Amount is the most possible
Allocate as much as possible to cell 11
Amount is 150 truckloads
Initial allocation is:
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6
150 8 10 7 11
175 4 5 12
275
Demand
200
100
300
600

11. Second Allocation Destination 2 3 150 50
Next allocate to either cell 21 or cell 12
Cell 12 no longer represents a feasible allocation
Cell 21 represents only feasible alternative, and as much as possible is allocated to this cell.
The amount allocated to cell 21 can be either 175 truckloads or 50 truckloads
150 truckloads of the 200 truckloads have already been supplied
Because 50 truckloads is the most constrained amount, it is allocated to cell 21
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6
150 8 10 7 50 11
175 4 5 12
275
Demand
200
100
300
600

12. Third Allocation Destination 2 3
Third allocation is made in the same way as the second allocation
Only feasible cell is cell 22
Most that can be allocated is either 100 truckloads or 125 truckloads
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6
150 8 10 7 50 11
100
175 4 5 12
275
Demand
200
300
600

13. Fourth and Fifth Allocations
Fourth allocation is 25 truckloads to cell 23, and the fifth allocation is 275 truckloads to cell 33, both of which are shown as:
All of row and column allocations should add up to the constraint requirement.
Cost is computed by substituting the cell allocations in the objective function
Z=6X11+8X12+10X13+ 7X21+ 11X22+11X23+4X31+5X32+12X33
Z= 6(150)+8(0)+10(0)+7(50)+11(100)+11(25)+4(0)+5(0)+12(275)=$5,925
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6
150 8 10 7 50 11
100 25
175 4 5 12
275
Demand
200
300
600

14. Steps Steps of the northwest corner method are:
Allocate as much as possible to the cell in the upper left-hand corner
2. Allocate as much as possible to the next adjacent feasible cell
3. Repeat step 2 until all supply and demand requirements have been met

15. Least Cost Method Destination 2 3
Allocate to the cells with the lowest costs
Cell 31 has the minimum cost of $4
Allocate as much as possible to this cell
Most we can allocate is 200 truckloads
All of the remaining cells in column 1 have now been eliminated
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6 8 10
150 7 11
175 4
200 5 12
275
Demand
100
300
600

16. Subsequent Allocations
Next allocation is made to cell that has the minimum cost and also is feasible
This cell is 32 which has a cost of $5
Most allocated is 75 truckloads
Next allocation of 25 truckloads is made to cell 12, which has the minimum cost of $8
Next allocations are made to cells 13 and 23
These allocations are:
Total cost is $4,550, as compared to a total cost of $5,925 for the initial northwest comer solution
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6 8 25 10
125
150 7 11
175 4
200 5 75 12
275
Demand
100
300
600

17. Steps of the Least Cost Method
Steps of the least cost method are summarized here
Allocate as much as possible to the feasible cell with the minimum transportation cost
adjust the supply and demand requirements
Repeat step 1 until all the requirements have been met

18. Vogel’s Approximation Method
Third method is the Vogel's Approximation Method
Based on the concept of penalty cost or regret cost
If a decision maker incorrectly chooses from other alternative routes, a wrong decision will result.
Penalty may be suffered for making a wrong decision
Develop a penalty cost for each source and destination
Destination 1 can be supplied by either source 1, source 2, or source 3
Best decision would be source 3 because cell 31 has the minimum cost of $4
If a wrong decision were made, a penalty of $2 would result

19. General Rule Destination 2 3
General rule is to subtract the minimum cell cost from the next higher cell cost in each row and column
Calculations of penalty costs are shown at the right and at the bottom of Table
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6 8 10
150 7 11
175 4 5 12
275
Demand
200
100
300
600 2 4 1 2 3 1

20. First Allocation Destination 2 3 150 175 275 200 100 300 600
Initial allocation is made in the row or column that has the highest penalty cost
Row 2 has the highest penalty cost
Allocate as much as possible with the minimum cost
Cell 21 has the lowest cost and the most that can be allocated to is 175 truckloads
With this allocation the greatest penalty cost has been avoided
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6 8 10
150 7
175 11 4 5 12
275
Demand
200
100
300
600 2 4 1 2 3 1

21. Second Allocation Destination 2 3 150 175 100 275 200 300 600
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6 8 10
150 7
175 11 4 5
100 12
275
Demand
200
300
600
All penalty costs must be recomputed after each allocation is made
In some cases will change; in other cases will not change
Repeat same step and allocate to row or column with highest penalty cost
Allocate as much as possible to cell 32 2 1 2 3 1

22. Third Allocation Destination 2 3
Row 3 now has the highest penalty cost of $8 with minimum cost of $4
Allocate 25
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6 8 10
150 7
175 11 4 25 5
100 12
275
Demand
200
300
600 2 8 2 1

23. Final Allocations Destination 2 3 Column 3 has a penalty cost
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6 8 10
150 7
175 11 4 25 5
100 12
275
Demand
200
300
600
Column 3 has a penalty cost
Does not exist for both rows 1 and 3
Allocations are made to cell 13 and cell 33
Total cost of this initial solution method is $5,125 2

24. Steps of Vogel's Approximation Method
Use following steps in Vogel's Approximation Method
Determine penalty costs by subtracting the minimum cell cost from next higher cell cost in each row and column
Select row or column with highest penalty cost
Ties can be broken arbitrarily
Choose lowest transportation cost in row or column with highest penalty cost and allocate as much as possible
Repeat steps 1, 2, and 3 until all requirements have been met

25. Optimal Solution Methods
Once an initial basic solution has been determined, next step is to solve for optimal solution
See whether or not a transportation route not presently being used (empty cell) would result in a lower total cost
Steps used are exact parallel of the simplex algorithm
Two optimal methods:
stepping-stone solution method
method of multipliers
Method of multipliers will be demonstrated
Use initial solution obtained by one the transportation algorithm
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6 8 25 10
125
150 7 11
175 4
200 5 75 12
275
Demand
100
300
600

26. Method of Multipliers
See whether or not an empty cell would result in a lower total cost
Associate the multipliers ui and vj with row i and column j
Computed for all cells with allocations by using:
ui+vj= cih
cij is the unit transportation cost for cell ij
u1+v2= c12 or u1+v2 = 8
Formulas for remaining cells are: u1+v3= 10 , u2+v3= 11 , u3+v1= 4 , u3+v2= 5 vj v1 v2 v3 ui To
From
Destination
Supply u1 1 6 8 25 10
125
150 u2 2 7 11
175 u3 3 4
200 5 75 12
275
Demand
100
300
600

27. ui and vj Values 25 125 175 200 75 Five equations with six unknowns
Assign only one of the unknowns a value of zero
Let u1 = 0, we can solve for all remaining ui and vj values as:
X12: u1+v2 = 8
0+v2 = 8; v1 = 8
X13: u1+v3= 10
0+v3= 10; v3= 10
X23: u2+v3= 11
u2+10= 11; u2= 1
X32: u3+v2= 5
u3+8= 5; u3= -3
X31: u3+v1= 4
-3+v1= 4; v1= 7
ui and vj values can be substituted into the tableau shown: vj
V1=8
v2=7
V3=10 ui To
From
Destination
Supply
U1=0 1 6 8 25 10
125
150
U2=1 2 7 11
175
U3=-3 3 4
200 5 75 12
275
Demand
100
300
600

28. Entering Variable Use following formula to evaluate all empty cells
ui+vj-cij=kij
kij is cost increase or decrease
Formula yields:
X11: k11=u1+v21 – c12=0+7-6=+1
X21: k21=u2+v1 – c21=1+7-7=+1
X22: k22=u2+v2 – c22=1+8-11=-2
X33: k33=u3+v3 – c33= =-5
Is equivalent to computing the row-z of simplex method
Entering variable is the most positive coefficient vj
V1=8
v2=7
V3=10 ui To
From
Destination
Supply
U1=0 1 6 8 25 10
125
150
U2=1 2 7 11
175
U3=-3 3 4
200 5 75 12
275
Demand
100
300
600

29. Leaving Variable
Suppose cell 11 is chosen as the entering variable. Having determined the entering variable, we need to determine the leaving variable in the following manner.
First, construct a closed loop that starts and ends at the entering variable cell 11.
The loop consists of connected horizontal and vertical segments only (no diagonals are allowed).
Except for the entering variable cell, each corner of the closed loop must coincide with a basic variable.
Therefore, the closed loop for the entering variable cell 11 is: vj
V1=8
v2=7
V3=10 ui To
From
Destination
Supply
U1=0 1 6 8 25 10
125
150
U2=1 2 7 11
175
U3=-3 3 4
200 5 75 12
275
Demand
100
300
600

30. Maximum Allocation Look at the path for cell 11
Maximum allocated to cell 11 is the number of truckloads that must be subtracted from cells 12 and 31 and added to cell 32
Choose 25 truckloads
25 are added to cell 11, subtracted from cell 12, added to cell 32, and subtracted from cell 31 vj
V1=8
v2=7
V3=10 ui To
From
Destination
Supply
U1=0 1 6 25 8 10
125
150
U2=1 2 7 11
175
U3=-3 3 4 5
100 12
275
Demand
200
300
600

31. Total Costs Reallocation is:
Total cost will be reduced by $1 for each gallon allocated vj
V1=8
v2=7
V3=10 ui To
From
Destination
Supply
U1=0 1 6 25 8 10
125
150
U2=1 2 7 11
175
U3=-3 3 4 5
100 12
275
Demand
200
300
600

32. The Second Iteration Repeat computation of ui and vj values
These values are:
X11: u1+v1 = 6
0+v1 = 8; v1 = 6
X13: u1+v3= 10
0+v3= 10; v3= 10
X23: u2+v3= 11
u2+10= 11; u2= 1
X31: u3+v2= 4
u3+6= 4; u3= -2
X32: u3+v2= 5
-2+v2= 4; v2= 7 vj
V1=6
v2=7
V3=10 ui To
From
Destination
Supply
U1=0 1 6 25 8 10
125
150
U2=1 2 7 11
175
U3=-2 3 4 5
100 12
275
Demand
200
300
600

33. The Optimal Solution Cost changes for the empty cells are now computed
X12: k12=u1+v2 – c12=0+7-8=-1
X21: k21=u2+v1 – c21=1+6-7=0
X22: k21=u2+v2 – c22=1+7-11=-3
X33: k33=u3+v3 – c33= =-4
None of is positive, the solution is optimal
Cell 21 indicates a multiple optimal solution vj
V1=6
v2=7
V3=10 ui To
From
Destination
Supply
U1=0 1 6 25 8 10
125
150
U2=1 2 7 11
175
U3=-2 3 4 5
100 12
275
Demand
200
300
600

34. The Unbalanced Transportation Model
Determination of initial solution and optimal solution were demonstrated within the context of a balanced transportation model
Face unbalanced problems more often
Demand for destination 3 is changed from 300 to 350
Create a situation in which total demand is 650 and total supply is 600
A "dummy“ row is added to balance the model
Transportation costs in the dummy row are zero
Because amounts allocated to these cells are not really transported
do not have any impact on total cos.
Dummy cells are slack variables
A dummy column is added for cases where total supply is more than total demand
Addition of a dummy row or a dummy column has no effect on determination of an initial solution or optimal solution
Because they are treated in the same way
From To
Destination
Supply
S 1 U P
P 2 L
Y 3 6 8 10
150 7 11
175 4 5 12
275
Dummy 50
Demand
200
100
350
600

35. Degeneracy
In all transportation solutions, there is a condition that was met
m rows + n columns -1 = cells with allocations
When this not met, tableau is said to be degenerate
A solution that none of the optimal solution methods will work
One of the empty cells must be artificially assigned with allocation of zero
Allocation of zero to a cell does not guarantee the determination of optimal solution
Zero must be reallocated to another cell and repeat the process until an optimal solution is obtained

36. Prohibited Routes Might see one or more of routes are prohibited
Units cannot be transported from a particular source to a particular destination
Make sure that no units in the cell representing this cell
Assign a large cost so that it will keep it from being selected as an entering variable

37. The Assignment Problems
Special form of the transportation model
The supply at each source and the demand at each destination are each limited to one unit
We need to assign four jobs to 4 workers
Because of varying skills, the cost of performing jobs vary
Determine job assignments that minimizes the total costs
Jobs
Worker 1 21 9 18 16 2 10 7 13 20 3
17.5
10.5 14 17 4 8
6.5 12

38. The Hungarian Method 1 2 3 4 Supply and demand are always equal to one
Not necessary to include supply and demand rows in the tableau
Solved as a regular transportation model
Use a simple solution algorithm call Hungarian method
Jobs
Worker 1 21 9 18 16 2 10 7 13 20 3
17.5
10.5 14 17 4 8
6.5 12

39. The Row Reductions
First step is to subtract the minimum value in each row from every value in the row
These computations are referred to as row reductions
This is the same concept in Vogle’s Approximation Method in determination of penalty costs
Row reductions are shown:
Jobs
Worker 1 12 9 7 2 3 6 13
3.5
6.5 4
1.5
5.5

40. The Column Reductions
Subtract minimum value in each column from all column values
These computations are called column reductions
Whenever a zero is present, the table reflects the assignments of jobs to workers
Job 1 can be assigned to worker 1
This Indicates that there is not a unique optimal assignment for worker 2
Have not reached an optimal solution yet
Jobs
Worker 1
10.5
5.5
1.5 2
2.5
7.5 3 4
0.5

41. The Optimality Test
A test to determine if four unique assignments exist is to draw the minimum number of horizontal or vertical lines necessary to cross out all zeros
Optimal solution results when each of the four jobs can be uniquely assigned to a different worker.
Indicate that there are only three unique assignments
Subtract the minimum value that is not crossed out, which is 1.5, from all other values not crossed out
Add this minimum value to those cells where two lines intersect.
Jobs
Worker 1
10.5
5.5
1.5 2
2.5
7.5 3 4
0.5

42. The Optimal Solution Four lines required to cross out all the zeros
Job 1 can be assigned to either worker 2 or worker 4
Job 1 is assigned to worker 2
Means row 1 and the worker 2 column can be eliminated
Job 2 is assigned to worker 1
Assign job 3 to worker 3
Leaves job 4 for worker 4
Total minimum costs is $45
Jobs
Worker 1 9 4 2 6 3
5.5
1.5
2.5
0.5

43. An Unbalanced Assignment Problem
Might face an unbalanced assignment problem when supply exceeds demand or demand exceeds supply
A dummy column is added to balance the model

44. Solving Transportation Problems by Excel and TORA
No "transportation" module in Excel
Must be solved in Excel as a linear programming model
TORA provides a module to solve transportation problems
Allows for any of the three initial solution methods to be selected
Does not have to be balanced if it is an unbalanced problem