Transportation and assignment models.ppt

© 2008 Prentice-Hall, Inc.
Chapter 10
To accompany
Quantitative Analysis for Management, Tenth Edition,
by Render, Stair, and Hanna
Power Point slides created by Jeff Heyl
Transportation and
Assignment Models
© 2009 Prentice-Hall, Inc.

© 2009 Prentice-Hall, Inc. 10 – 2
Learning Objectives
1. Structure special LP problems using the
transportation and assignment models
2. Use the northwest corner, VAM, MODI, and
stepping-stone methods
3. Solve facility location and other application
problems with transportation models
4. Solve assignment problems with the
Hungarian (matrix reduction) method
After completing this chapter, students will be able to:
After completing this chapter, students will be able to:

Chapter Outline
10.1
10.1 Introduction
10.2
10.2 Setting Up a Transportation Problem
10.3
10.3 Developing an Initial Solution: Northwest
Corner Rule
10.4
10.4 Stepping-Stone Method: Finding a
Least-Cost Solution
10.5
10.5 MODI Method
10.6
10.6 Vogel’s Approximation Method: Another
Way to Find an Initial Solution
10.7
10.7 Unbalanced Transportation Problems

Chapter Outline
10.8
10.8 Degeneracy in Transportation Problems
10.9
10.9 More Than One Optimal Solution
10.10
10.10 Maximization Transportation Problems
10.11
10.11 Unacceptable or Prohibited Routes
10.12
10.12 Facility Location Analysis
10.13
10.13 Assignment Model Approach
10.14
10.14 Unbalanced Assignment Problems
10.15
10.15 Maximization Assignment Problems

Introduction
 In this chapter we will explore two special
linear programming models
 The transportation model
 The assignment model
 Because of their structure, they can be
solved more efficiently than the simplex
method
 These problems are members of a category
of LP techniques called network flow
network flow
problems
problems

Introduction
 Transportation model
 The transportation problem
transportation problem deals with the
distribution of goods from several points of
supply (sources
sources) to a number of points of
demand (destinations
destinations)
 Usually we are given the capacity of goods at
each source and the requirements at each
destination
 Typically the objective is to minimize total
transportation and production costs

Introduction
 Example of a transportation problem in a network
format
100 Units
300 Units
300 Units 200 Units
200 Units
300 Units
Factories
(Sources)
Des Moines
Evansville
Fort Lauderdale
Warehouses
(Destinations)
Albuquerque
Boston
Cleveland
Capacities
Capacities Shipping Routes
Shipping Routes Requirements
Requirements
Figure 10.1

Introduction
 Assignment model
 The assignment problem
assignment problem refers to the class of
LP problems that involve determining the most
efficient assignment of resources to tasks
 The objective is most often to minimize total
costs or total time to perform the tasks at hand
 One important characteristic of assignment
problems is that only one job or worker can be
assigned to one machine or project

Introduction
 Special-purpose algorithms
 Although standard LP methods can be used to
solve transportation and assignment problems,
special-purpose algorithms have been
developed that are more efficient
 They still involve finding and initial solution
and developing improved solutions until an
optimal solution is reached
 They are fairly simple in terms of computation

Introduction
 Streamlined versions of the simplex method are
important for two reasons
1. Their computation times are generally 100 times faster
2. They require less computer memory (and hence can
permit larger problems to be solved)
 Two common techniques for developing initial
solutions are the northwest corner method and
Vogel’s approximation
 The initial solution is evaluated using either the
stepping-stone method or the modified
distribution (MODI) method
 We also introduce a solution procedure called the
Hungarian method
Hungarian method, Flood’s technique
Flood’s technique, or the
reduced matrix method
reduced matrix method

Setting Up a Transportation Problem
 The Executive Furniture Corporation
manufactures office desks at three locations: Des
Moines, Evansville, and Fort Lauderdale
 The firm distributes the desks through regional
warehouses located in Boston, Albuquerque, and
Cleveland
 Estimates of the monthly production capacity of
each factory and the desks needed at each
warehouse are shown in Figure 10.1

 Production costs are the same at the three
factories so the only relevant costs are shipping
from each source
source to each destination
destination
 Costs are constant no matter the quantity
shipped
 The transportation problem can be described as
how to select the shipping routes to be used and
how to select the shipping routes to be used and
the number of desks to be shipped on each route
the number of desks to be shipped on each route
so as to minimize total transportation cost
so as to minimize total transportation cost
 Restrictions regarding factory capacities and
warehouse requirements must be observed

 The first step is setting up the transportation
table
 Its purpose is to summarize all the relevant data
and keep track of algorithm computations
Transportation costs per desk for Executive Furniture
TO
FROM ALBUQUERQUE BOSTON CLEVELAND
DES MOINES $5 $4 $3
EVANSVILLE $8 $4 $3
FORT LAUDERDALE $9 $7 $5
Table 10.1

 Geographical locations of Executive Furniture’s
factories and warehouses
Albuquerque
Cleveland
Boston
Des Moines
Fort Lauderdale
Evanston
Factory
Warehouse
Figure 10.2

 Transportation table for Executive Furniture
TO
FROM
WAREHOUSE
AT
ALBUQUERQUE
WAREHOUSE
AT
BOSTON
WAREHOUSE
AT
CLEVELAND
FACTORY
CAPACITY
DES MOINES
FACTORY
$5 $4 $3
100
EVANSVILLE
FACTORY
$8 $4 $3
300
FORT LAUDERDALE
FACTORY
$9 $7 $5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700
Table 10.2
Des Moines
capacity
constraint
Cell representing a
source-to-destination
(Evansville to Cleveland)
shipping assignment
that could be made
Total supply
and demand
Cleveland
warehouse
demand
Cost of shipping 1 unit from
Fort Lauderdale factory to
Boston warehouse

 In this table, total factory supply exactly
equals total warehouse demand
 When equal demand and supply occur, a
balanced problem
balanced problem is said to exist
 This is uncommon in the real world and
we have techniques to deal with
unbalanced problems

Developing an Initial Solution:
Northwest Corner Rule
 Once we have arranged the data in a table, we
must establish an initial feasible solution
 One systematic approach is known as the
northwest corner rule
northwest corner rule
 Start in the upper left-hand cell and allocate units
to shipping routes as follows
1. Exhaust the supply (factory capacity) of each row before
moving down to the next row
2. Exhaust the demand (warehouse) requirements of each
column before moving to the right to the next column
3. Check that all supply and demand requirements are met.
 In this problem it takes five steps to make the
initial shipping assignments

1. Beginning in the upper left hand corner, we
assign 100 units from Des Moines to
Albuquerque. This exhaust the supply from Des
Moines but leaves Albuquerque 200 desks short.
We move to the second row in the same column.
TO
FROM
ALBUQUERQUE
(A)
BOSTON
(B)
CLEVELAND
(C)
FACTORY
CAPACITY
DES MOINES
(D) 100
100
$5 $4 $3
100
EVANSVILLE
(E)
$8 $4 $3
300
FORT LAUDERDALE
(F)
$9 $7 $5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700

2. Assign 200 units from Evansville to Albuquerque.
This meets Albuquerque’s demand. Evansville
has 100 units remaining so we move to the right
to the next column of the second row.
TO
FROM
ALBUQUERQUE
(A)
BOSTON
(B)
CLEVELAND
(C)
FACTORY
CAPACITY
DES MOINES
(D)
100
$5 $4 $3
100
EVANSVILLE
(E) 200
200
$8 $4 $3
300
FORT LAUDERDALE
(F)
$9 $7 $5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700

3. Assign 100 units from Evansville to Boston. The
Evansville supply has now been exhausted but
Boston is still 100 units short. We move down
vertically to the next row in the Boston column.
TO
FROM
ALBUQUERQUE
(A)
BOSTON
(B)
CLEVELAND
(C)
FACTORY
CAPACITY
DES MOINES
(D)
100
$5 $4 $3
100
EVANSVILLE
(E)
200
$8
100
100
$4 $3
300
FORT LAUDERDALE
(F)
$9 $7 $5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700

4. Assign 100 units from Fort Lauderdale to Boston.
This fulfills Boston’s demand and Fort
Lauderdale still has 200 units available.
TO
FROM
ALBUQUERQUE
(A)
BOSTON
(B)
CLEVELAND
(C)
FACTORY
CAPACITY
DES MOINES
(D)
100
$5 $4 $3
100
EVANSVILLE
(E)
200
$8
100
$4 $3
300
FORT LAUDERDALE
(F)
$9
100
100
$7 $5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700

5. Assign 200 units from Fort Lauderdale to
Cleveland. This exhausts Fort Lauderdale’s
supply and Cleveland’s demand. The initial
shipment schedule is now complete.
TO
FROM
ALBUQUERQUE
(A)
BOSTON
(B)
CLEVELAND
(C)
FACTORY
CAPACITY
DES MOINES
(D)
100
$5 $4 $3
100
EVANSVILLE
(E)
200
$8
100
$4 $3
300
FORT LAUDERDALE
(F)
$9
100
$7
200
200
$5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700
Table 10.3

 We can easily compute the cost of this shipping
assignment
ROUTE
UNITS
SHIPPED x
PER UNIT
COST ($) =
TOTAL
COST ($)
FROM TO
D A 100 5 500
E A 200 8 1,600
E B 100 4 400
F B 100 7 700
F C 200 5 1,000
4,200
 This solution is feasible but we need to check to
see if it is optimal

Stepping-Stone Method:
Finding a Least Cost Solution
 The stepping-stone method
stepping-stone method is an iterative
technique for moving from an initial
feasible solution to an optimal feasible
solution
 There are two distinct parts to the process
 Testing the current solution to determine if
improvement is possible
 Making changes to the current solution to
obtain an improved solution
 This process continues until the optimal
solution is reached

Stepping-Stone Method:
Finding a Least Cost Solution
 There is one very important rule
 The number of occupied routes (or squares) must
The number of occupied routes (or squares) must
always be equal to one less than the sum of the
always be equal to one less than the sum of the
number of rows plus the number of columns
number of rows plus the number of columns
 In the Executive Furniture problem this means
the initial solution must have 3 + 3 – 1 = 5
squares used
Occupied shipping
routes (squares)
Number
of rows
Number of
columns
= + – 1
 When the number of occupied rows is less than
this, the solution is called degenerate
degenerate

Testing the Solution for
Possible Improvement
 The stepping-stone method works by
testing each unused square in the
transportation table to see what would
happen to total shipping costs if one unit
of the product were tentatively shipped on
an unused route
 There are five steps in the process

Five Steps to Test Unused Squares
with the Stepping-Stone Method
1. Select an unused square to evaluate
2. Beginning at this square, trace a closed path
back to the original square via squares that are
currently being used with only horizontal or
vertical moves allowed
3. Beginning with a plus (+) sign at the unused
square, place alternate minus (–) signs and plus
signs on each corner square of the closed path
just traced

4. Calculate an improvement index
improvement index by adding
together the unit cost figures found in each
square containing a plus sign and then
subtracting the unit costs in each square
containing a minus sign
5. Repeat steps 1 to 4 until an improvement index
has been calculated for all unused squares. If all
indices computed are greater than or equal to
zero, an optimal solution has been reached. If
not, it is possible to improve the current solution
and decrease total shipping costs.

 For the Executive Furniture Corporation data
Steps 1 and 2
Steps 1 and 2. Beginning with Des Moines–Boston
route we trace a closed path using only currently
occupied squares, alternately placing plus and
minus signs in the corners of the path
 In a closed path
closed path, only squares currently used for
shipping can be used in turning corners
 Only one
Only one closed route is possible for each square
we wish to test

Step 3
Step 3. We want to test the cost-effectiveness of the
Des Moines–Boston shipping route so we pretend
we are shipping one desk from Des Moines to
Boston and put a plus in that box
 But if we ship one more
more unit out of Des Moines
we will be sending out 101 units
 Since the Des Moines factory capacity is only
100, we must ship fewer
fewer desks from Des Moines
to Albuquerque so we place a minus sign in that
box
 But that leaves Albuquerque one unit short so we
must increase the shipment from Evansville to
Albuquerque by one unit and so on until we
complete the entire closed path

 Evaluating the unused
Des Moines–Boston
shipping route
TO
FROM
ALBUQUERQUE BOSTON CLEVELAND FACTORY
CAPACITY
DES MOINES 100
$5 $4 $3
100
EVANSVILLE 200
$8
100
$4 $3
300
FORT LAUDERDALE
$9
100
$7
200
$5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700 Table 10.4
Warehouse B
$4
Factory
D
Warehouse A
$5
100
Factory
E
$8
200
$4
100
+
– +
–

Des Moines–Boston
shipping route
TO
FROM
CAPACITY
DES MOINES 100
$5 $4 $3
100
EVANSVILLE 200
$8
100
$4 $3
300
FORT LAUDERDALE
$9
100
$7
200
$5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700 Table 10.4
Warehouse A
Factory
D
$5
Warehouse B
$4
Factory
E
$8 $4
100
200 100
201
99
1
+
– +
–
99

Des Moines–Boston
shipping route
TO
FROM
CAPACITY
DES MOINES 100
$5 $4 $3
100
EVANSVILLE 200
$8
100
$4 $3
300
FORT LAUDERDALE
$9
100
$7
200
$5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700 Table 10.4
Warehouse A
Factory
D
$5
Warehouse B
$4
Factory
E
$8 $4
100
99
1
201
200 100
99
+
– +
–
Result of Proposed
Shift in Allocation
= 1 x $4
– 1 x $5
+ 1 x $8
– 1 x $4 = +$3

Step 4
Step 4. We can now compute an improvement index
improvement index
(I
Iij
ij) for the Des Moines–Boston route
 We add the costs in the squares with plus signs
and subtract the costs in the squares with minus
signs
Des Moines–
Boston index = IDB = +$4 – $5 + $5 – $4 = + $3
 This means for every desk shipped via the Des
Moines–Boston route, total transportation cost
will increase
increase by $3 over their current level

Step 5
Step 5. We can now examine the Des Moines–
Cleveland unused route which is slightly more
difficult to draw
 Again we can only turn corners at squares that
represent existing routes
 We must pass through the Evansville–Cleveland
square but we can not turn there or put a + or –
sign
 The closed path we will use is
+ DC – DA + EA – EB + FB – FC

 Evaluating the Des Moines–Cleveland shipping
route
TO
FROM
CAPACITY
DES MOINES 100
$5 $4 $3
100
EVANSVILLE 200
$8
100
$4 $3
300
FORT LAUDERDALE
$9
100
$7
200
$5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700
Table 10.5
Start
+
+ –
–
+ –
Des Moines–Cleveland
improvement index = IDC = + $3 – $5 + $8 – $4 + $7 – $5 = + $4

 Opening the Des Moines–Cleveland route will not
lower our total shipping costs
 Evaluating the other two routes we find
 The closed path is
+ EC – EB + FB – FC
 The closed path is
+ FA – FB + EB – EA
 So opening the Fort Lauderdale-Albuquerque
route will
will lower our total transportation costs
Evansville-
Cleveland index = IEC = + $3 – $4 + $7 – $5 = + $1
Fort Lauderdale–
Albuquerque index = IFA = + $9 – $7 + $4 – $8 = – $2

Obtaining an Improved Solution
 In the Executive Furniture problem there is only
one unused route with a negative index (Fort
Lauderdale-Albuquerque)
 If there was more than one route with a negative
index, we would choose the one with the largest
improvement
 We now want to ship the maximum allowable
number of units on the new route
 The quantity to ship is found by referring to the
closed path of plus and minus signs for the new
route and selecting the smallest number
smallest number found in
those squares containing minus signs

 To obtain a new solution, that number is added to
all squares on the closed path with plus signs
and subtracted from all squares the closed path
with minus signs
 All other squares are unchanged
 In this case, the maximum number that can be
shipped is 100 desks as this is the smallest value
in a box with a negative sign (FB route)
 We add 100 units to the FA and EB routes and
subtract 100 from FB and EA routes
 This leaves balanced rows and columns and an
improved solution

 Stepping-stone path used to evaluate route FA
TO
FROM
A B C FACTORY
CAPACITY
D 100
$5 $4 $3
100
E 200
$8
100
$4 $3
300
F
$9
100
$7
200
$5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700
Table 10.6
+
+ –
–

 Second solution to the Executive Furniture problem
TO
FROM
A B C FACTORY
CAPACITY
D 100
$5 $4 $3
100
E 100
$8
200
$4 $3
300
F 100
$9 $7
200
$5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700
Table 10.7
 Total shipping costs have been reduced by (100
units) x ($2 saved per unit) and now equals $4,000

 This second solution may or may not be optimal
 To determine whether further improvement is
possible, we return to the first five steps to test
each square that is now
now unused
 The four new improvement indices are
D to B = IDB = + $4 – $5 + $8 – $4 = + $3
(closed path: + DB – DA + EA – EB)
D to C = IDC = + $3 – $5 + $9 – $5 = + $2
(closed path: + DC – DA + FA – FC)
E to C = IEC = + $3 – $8 + $9 – $5 = – $1
(closed path: + EC – EA + FA – FC)
F to B = IFB = + $7 – $4 + $8 – $9 = + $2
(closed path: + FB – EB + EA – FA)

 An improvement can be made by shipping the
maximum allowable number of units from E to C
TO
FROM
A B C FACTORY
CAPACITY
D 100
$5 $4 $3
100
E 100
$8
200
$4 $3
300
F 100
$9 $7
200
$5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700
Table 10.8
 Path to evaluate for the EC route
Start
+
+ –
–

 Total cost of third solution
ROUTE
DESKS
SHIPPED x
PER UNIT
COST ($) =
TOTAL
COST ($)
FROM TO
D A 100 5 500
E B 200 4 800
E C 100 3 300
F A 200 9 1,800
F C 100 5 500
3,900

TO
FROM
A B C FACTORY
CAPACITY
D 100
$5 $4 $3
100
E
$8
200
$4
100
$3
300
F 200
$9 $7
100
$5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700
Table 10.9
 Third and optimal solution

 This solution is optimal as the improvement
indices that can be computed are all greater than
or equal to zero
D to B = IDB = + $4 – $5 + $9 – $5 + $3 – $4 = + $2
+ $2
(closed path: + DB – DA + FA – FC + EC – EB)
D to C = IDC = + $3 – $5 + $9 – $5 = + $2
+ $2
(closed path: + DC – DA + FA – FC)
E to A = IEA = + $8 – $9 + $5 – $3 = + $1
+ $1
(closed path: + EA – FA + FC – EC)
F to B = IFB = + $7 – $5 + $3 – $4 = + $1
+ $1
(closed path: + FB – FC + EC – EB)

Summary of Steps in Transportation
Algorithm (Minimization)
1. Set up a balanced transportation table
2. Develop initial solution using either the northwest
corner method or Vogel’s approximation method
3. Calculate an improvement index for each empty
cell using either the stepping-stone method or the
MODI method. If improvement indices are all
nonnegative, stop as the optimal solution has been
found. If any index is negative, continue to step 4.
4. Select the cell with the improvement index
indicating the greatest decrease in cost. Fill this
cell using the stepping-stone path and go to step
3.

Using Excel QM to Solve
Transportation Problems
 Excel QM input screen and formulas
Program 10.1A

Using Excel QM to Solve
Transportation Problems
 Output from Excel QM with optimal solution
Program 10.1B

MODI Method
 The MODI (modified distribution
modified distribution) method allows
us to compute improvement indices quickly for
each unused square without drawing all of the
closed paths
 Because of this, it can often provide considerable
time savings over the stepping-stone method for
solving transportation problems
 If there is a negative improvement index, then
only one stepping-stone path must be found
 This is used in the same manner as before to
obtain an improved solution

How to Use the MODI Approach
 In applying the MODI method, we begin with an
initial solution obtained by using the northwest
corner rule
 We now compute a value for each row (call the
values R1, R2, R3 if there are three rows) and for
each column (K1, K2, K3) in the transportation table
 In general we let
Ri =value for assigned row i
Kj =value for assigned column j
Cij =cost in square ij (cost of shipping
from source i to destination j)

Five Steps in the MODI Method to
Test Unused Squares
1. Compute the values for each row and column, set
Ri + Kj = Cij
but only for those squares that are currently used
but only for those squares that are currently used
or occupied
or occupied
2. After all equations have been written, set R1 = 0
3. Solve the system of equations for R and K values
4. Compute the improvement index for each unused
square by the formula
Improvement Index (Iij) = Cij – Ri – Kj
5. Select the best negative index and proceed to
solve the problem as you did using the stepping-
stone method

Solving the Executive Furniture
Corporation Problem with MODI
 The initial northwest corner solution is repeated
in Table 10.10
 Note that to use the MODI method we have added
the Ris (rows) and Kjs (columns)
Kj K1 K2 K3
Ri
TO
FROM
A B C FACTORY
CAPACITY
R1
D 100
$5 $4 $3
100
R2
E 200
$8
100
$4 $3
300
R3
F
$9
100
$7
200
$5
300
WAREHOUSE
REQUIREMENTS
300 200 200 700
Table 10.10

 The first step is to set up an equation for each
occupied square
 By setting R1 = 0 we can easily solve for K1, R2, K2,
R3, and K3
(1) R1 + K1 = 5 0 + K1 = 5 K1 = 5
(2) R2 + K1 = 8 R2 + 5 = 8 R2 = 3
(3) R2 + K2 = 4 3 + K2 = 4 K2 = 1
(4) R3 + K2 = 7 R3 + 1 = 7 R3 = 6
(5) R3 + K3 = 5 6 + K3 = 5 K3 = –1

 The next step is to compute the improvement
index for each unused cell using the formula
Improvement index (Iij) = Cij – Ri – Kj
 We have
Des Moines-
Boston index
Des Moines-
Cleveland index
Evansville-
Cleveland index
Fort Lauderdale-
Albuquerque index
IDB = C12 – R1 – K2 = 4 – 0 – 1
= +$3
IDC = C13 – R1 – K3 = 3 – 0 – (–1)
= +$4
IEC = C23 – R2 – K3 = 3 – 3 – (–1)
= +$1
IFA = C31 – R3 – K1 = 9 – 6 – 5
= –$2

 The steps we follow to develop an improved
solution after the improvement indices have been
computed are
1. Beginning at the square with the best
improvement index, trace a closed path back
to the original square via squares that are
currently being used
2. Beginning with a plus sign at the unused
square, place alternate minus signs and plus
signs on each corner square of the closed
path just traced

3. Select the smallest quantity found in those
squares containing the minus signs and add
that number to all squares on the closed path
with plus signs; subtract the number from
squares with minus signs
4. Compute new improvement indices for this
new solution using the MODI method
 Note that new Ri and Kj values must be
calculated
 Follow this procedure for the second and third
solutions

Vogel’s Approximation Method:
Another Way To Find An Initial Solution
 Vogel’s Approximation Method
Vogel’s Approximation Method (VAM
VAM) is not as
simple as the northwest corner method, but it
provides a very good initial solution, often one
that is the optimal
optimal solution
 VAM tackles the problem of finding a good initial
solution by taking into account the costs
associated with each route alternative
 This is something that the northwest corner rule
does not do
 To apply VAM, we first compute for each row and
column the penalty faced if we should ship over
the second-best
second-best route instead of the least-cost
least-cost
route

Vogel’s Approximation Method
 The six steps involved in determining an initial
VAM solution are illustrated below beginning with
the same layout originally shown in Table 10.2
VAM Step 1
VAM Step 1. For each row and column of the
transportation table, find the difference between
the distribution cost on the best
best route in the row
or column and the second best
second best route in the row or
column
 This is the opportunity cost
opportunity cost of not using the
best route
 Step 1 has been done in Table 10.11

3 0 0
OPPORTUNITY
COSTS
TO
FROM
A B C TOTAL
AVAILABLE
D 100
$5 $4 $3
100 1
E 200
$8
100
$4 $3
300 1
F
$9
100
$7
200
$5
300 2
TOTAL REQUIRED 300 200 200 700
Table 10.11
 Transportation table with VAM row and column
differences shown

VAM Step 2
VAM Step 2. identify the row or column with the
greatest opportunity cost, or difference (column A in
this example)
VAM Step 3
VAM Step 3.Assign as many units as possible to the
lowest-cost square in the row or column selected
VAM Step 4
VAM Step 4. Eliminate any row or column that has
been completely satisfied by the assignment just
made by placing Xs in each appropriate square
VAM Step 5
VAM Step 5. Recompute the cost differences for the
transportation table, omitting rows or columns
eliminated in the previous step

3 1 0 3 0 2
OPPORTUNITY
COSTS
TO
FROM
A B C TOTAL
AVAILABLE
D 100
$5
X
$4
X
$3
100 1
E
$8 $4 $3
300 1
F
$9 $7 $5
300 2
Table 10.12
 VAM assignment with D’s requirements satisfied

VAM Step 6
VAM Step 6. Return to step 2 for the rows and
columns remaining and repeat the steps until an
initial feasible solution has been obtained
 In this case column B now has the greatest
difference, 3
 We assign 200 units to the lowest-cost square in
the column, EB
 We recompute the differences and find the
greatest difference is now in row E
 We assign 100 units to the lowest-cost square in
the column, EC

 Second VAM assignment with B’s requirements
satisfied
3 1 0 3 0 2
OPPORTUNITY
COSTS
TO
FROM
A B C TOTAL
AVAILABLE
D 100
$5
X
$4
X
$3
100 1
E
$8
200
$4 $3
300 1
F
$9
X
$7 $5
300 2
Table 10.13

TO
FROM
A B C TOTAL
AVAILABLE
D 100
$5
X
$4
X
$3
100
E X
$8
200
$4
100
$3
300
F
$9
X
$7 $5
300
Table 10.14
 Third VAM assignment with E’s requirements
satisfied

TO
FROM
A B C TOTAL
AVAILABLE
D 100
$5
X
$4
X
$3
100
E X
$8
200
$4
100
$3
300
F 200
$9
X
$7
100
$5
300
Table 10.15
 Final assignments to balance column and row
requirements

Unbalanced Transportation Problems
 In real-life problems, total demand is frequently
not equal to total supply
 These unbalanced problems
unbalanced problems can be handled
easily by introducing dummy sources
dummy sources or dummy
dummy
destinations
destinations
 If total supply is greater than total demand, a
dummy destination (warehouse), 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

Unbalanced Transportation Problems
 In either case, shipping cost coefficients of zero
are assigned to each dummy location or route as
no goods will actually be shipped
 Any units assigned to a dummy destination
represent excess capacity
 Any units assigned to a dummy source represent
unmet demand

Demand Less Than Supply
 Suppose that the Des Moines factory increases its
rate of production from 100 to 250 desks
 The firm is now able to supply a total of 850 desks
each period
 Warehouse requirements remain the same (700) so
the row and column totals do not balance
 We add a dummy column that will represent a fake
warehouse requiring 150 desks
 This is somewhat analogous to adding a slack
variable
 We use the northwest corner rule and either
stepping-stone or MODI to find the optimal
solution

Demand Less Than Supply
 Initial solution to an unbalanced problem where
demand is less than supply
TO
FROM
A B C DUMMY
WAREHOUSE
TOTAL
AVAILABLE
D 250
$5 $4 $3 0
250
E 50
$8
200
$4
50
$3 0
300
F
$9 $7
150
$5
150
0
300
WAREHOUSE
REQUIREMENTS
300 200 200 150 850
Table 10.16
New Des Moines
capacity
Total cost = 250($5) + 50($8) + 200($4) + 50($3) + 150($5) + 150(0) = $3,350

Demand Greater than Supply
 The second type of unbalanced condition occurs
when total demand is greater than total supply
 In this case we need to add a dummy row
representing a fake factory
 The new factory will have a supply exactly equal
to the difference between total demand and total
real supply
 The shipping costs from the dummy factory to
each destination will be zero

 Unbalanced transportation table for Happy
Sound Stereo Company
TO
FROM
WAREHOUSE
A
WAREHOUSE
B
WAREHOUSE
C
PLANT
SUPPLY
PLANT W
$6 $4 $9
200
PLANT X
$10 $5 $8
175
PLANT Y
$12 $7 $6
75
WAREHOUSE
DEMAND
250 100 150
450
500
Table 10.17
Totals
do not
balance

 Initial solution to an unbalanced problem in
which demand is greater than supply
TO
FROM
WAREHOUSE
A
WAREHOUSE
B
WAREHOUSE
C
PLANT
SUPPLY
PLANT W 200
$6 $4 $9
200
PLANT X 50
$10
100
$5
25
$8
175
PLANT Y
$12 $7
75
$6
75
PLANT Y
0 0
50
0
50
WAREHOUSE
DEMAND
250 100 150 500
Table 10.18
Total cost of initial solution = 200($6) + 50($10) + 100($5) + 25($8) + 75($6)
+ $50(0) = $2,850

Degeneracy in Transportation
Problems
 Degeneracy
Degeneracy occurs when the number of occupied
squares or routes in a transportation table
solution is less than the number of rows plus the
number of columns minus 1
 Such a situation may arise in the initial solution
or in any subsequent solution
 Degeneracy requires a special procedure to
correct the problem since there are not enough
occupied squares to trace a closed path for each
unused route and it would be impossible to apply
the stepping-stone method or to calculate the R
and K values needed for the MODI technique

Degeneracy in Transportation
Problems
 To handle degenerate problems, create an
artificially occupied cell
 That is, place a zero (representing a fake
shipment) in one of the unused squares and then
treat that square as if it were occupied
 The square chosen must be in such a position as
to allow all stepping-stone paths to be closed
 There is usually a good deal of flexibility in
selecting the unused square that will receive the
zero

Degeneracy in an Initial Solution
 The Martin Shipping Company example illustrates
degeneracy in an initial solution
 They have three warehouses which supply three
major retail customers
 Applying the northwest corner rule the initial
solution has only four occupied squares
 This is less than the amount required to use
either the stepping-stone or MODI method to
improve the solution (3 rows + 3 columns – 1 = 5)
 To correct this problem, place a zero in an
unused square, typically one adjacent to the last
filled cell

Degeneracy in an Initial Solution
 Initial solution of a degenerate problem
TO
FROM
CUSTOMER 1 CUSTOMER 2 CUSTOMER 3 WAREHOUSE
SUPPLY
WAREHOUSE 1 100
$8 $2 $6
100
WAREHOUSE 2
$10
100
$9
20
$9
120
WAREHOUSE 3
$7 $10
80
$7
80
CUSTOMER
DEMAND
100 100 100 300
Table 10.19
0
0
0
0
Possible choices of
cells to address the
degenerate solution

Degeneracy During
Later Solution Stages
 A transportation problem can become degenerate
after the initial solution stage if the filling of an
empty square results in two or more cells
becoming empty simultaneously
 This problem can occur when two or more cells
with minus signs tie for the lowest quantity
 To correct this problem, place a zero in one of the
previously filled cells so that only one cell
becomes empty

Degeneracy During
 Bagwell Paint Example
 After one iteration, the cost analysis at Bagwell
Paint produced a transportation table that was
not degenerate but was not optimal
 The improvement indices are
factory A – warehouse 2 index = +2
factory A – warehouse 3 index = +1
factory B – warehouse 3 index = –15
factory C – warehouse 2 index = +11
Only route with
a negative index

Degeneracy During
 Bagwell Paint transportation table
TO
FROM
WAREHOUSE
1
WAREHOUSE
2
WAREHOUSE
3
FACTORY
CAPACITY
FACTORY A 70
$8 $5 $16
70
FACTORY B 50
$15
80
$10 $7
130
FACTORY C 30
$3 $9
50
$10
80
WAREHOUSE
REQUIREMENT
150 80 50 280
Table 10.20

Degeneracy During
 Tracing a closed path for the factory B –
warehouse 3 route
TO
FROM
WAREHOUSE 1 WAREHOUSE 3
FACTORY B 50
$15 $7
FACTORY C 30
$3
50
$10
Table 10.21
+
+ –
–
 This would cause two cells to drop to zero
 We need to place an artificial zero in one of these
cells to avoid degeneracy

More Than One Optimal Solution
 It is possible for a transportation problem to have
multiple optimal solutions
 This happens when one or more of the
improvement indices zero in the optimal solution
 This means that it is possible to design
alternative shipping routes with the same total
shipping cost
 The alternate optimal solution can be found by
shipping the most to this unused square using a
stepping-stone path
 In the real world, alternate optimal solutions
provide management with greater flexibility in
selecting and using resources

Maximization Transportation Problems
 If the objective in a transportation problem is to
maximize profit, a minor change is required in the
transportation algorithm
 Now the optimal solution is reached when all the
improvement indices are negative or zero
 The cell with the largest positive improvement
index is selected to be filled using a stepping-
stone path
 This new solution is evaluated and the process
continues until there are no positive improvement
indices

Unacceptable Or Prohibited Routes
 At times there are transportation problems in
which one of the sources is unable to ship to one
or more of the destinations
 When this occurs, the problem is said to have an
unacceptable
unacceptable or prohibited route
prohibited route
 In a minimization problem, such a prohibited
route is assigned a very high cost to prevent this
route from ever being used in the optimal
solution
 In a maximization problem, the very high cost
used in minimization problems is given a
negative sign, turning it into a very bad profit

Facility Location Analysis
 The transportation method is especially useful in
helping a firm to decide where to locate a new
factory or warehouse
 Each alternative location should be analyzed
within the framework of one overall
overall distribution
system
 The new location that yields the minimum cost
for the entire system
entire system is the one that should be
chosen

Locating a New Factory for
Hardgrave Machine Company
 Hardgrave Machine produces computer
components at three plants and they ship to four
warehouses
 The plants have not been able to keep up with
demand so the firm wants to build a new plant
 Two sites are being considered, Seattle and
Birmingham
 Data has been collected for each possible
location
 Which new location will yield the lowest cost for
the firm in combination with the existing plants
and warehouses

 Hardgrave’s demand and supply data
WAREHOUSE
MONTHLY
DEMAND
(UNITS)
PRODUCTION
PLANT
MONTHLY
SUPPLY
COST TO PRODUCE
ONE UNIT ($)
Detroit 10,000 Cincinnati 15,000 48
Dallas 12,000 Salt Lake 6,000 50
New York 15,000 Pittsburgh 14,000 52
Los Angeles 9,000 35,000
46,000
Supply needed from new plant = 46,000 – 35,000 = 11,000 units per month
ESTIMATED PRODUCTION COST
PER UNIT AT PROPOSED PLANTS
Seattle $53
Birmingham $49
Table 10.22

 Hardgrave’s shipping costs
TO
FROM DETROIT DALLAS NEW YORK
LOS
ANGELES
CINCINNATI $25 $55 $40 $60
SALT LAKE 35 30 50 40
PITTSBURGH 36 45 26 66
SEATTLE 60 38 65 27
BIRMINGHAM 35 30 41 50
Table 10.23

 Optimal solution for the Birmingham location
TO
LOS
ANGELES
FACTORY
CAPACITY
CINCINNATI 10,000
73 103
1,000
88
4,000
108
15,000
SALT LAKE
85
1,000
80 100
5,000
90
6,000
PITTSBURGH
88 97
14,000
78 118
14,000
BIRMINGHAM
84
11,000
79 90 99
11,000
WAREHOUSE
REQUIREMENT
10,000 12,000 15,000 9,000 46,000
Table 10.24

 Optimal solution for the Seattle location
TO
LOS
ANGELES
FACTORY
CAPACITY
CINCINNATI 10,000
73
4,000
103
1,000
88 108
15,000
SALT LAKE
85
6,000
80 100 90
6,000
PITTSBURGH
88 97
14,000
78 118
14,000
SEATTLE
113
2,000
91 118
9,000
80
11,000
WAREHOUSE
REQUIREMENT
10,000 12,000 15,000 9,000 46,000
Table 10.25

 By comparing the total system costs of the two
alternatives, Hardgrave can select the lowest cost
option
 The Birmingham location yields a total system
cost of $3,741,000
 The Seattle location yields a total system cost of
$3,704,000
 With the lower total system cost, the Seattle
location is favored
 Excel QM can also be used as a solution tool

 Excel input screen
Program 10.2A

 Output from Excel QM analysis
Program 10.2A

Assignment Model Approach
 The second special-purpose LP algorithm is the
assignment method
 Each assignment problem has associated with it
a table, or matrix
 Generally, the rows contain the objects or people
we wish to assign, and the columns comprise the
tasks or things we want them assigned to
 The numbers in the table are the costs associated
with each particular assignment
 An assignment problem can be viewed as a
transportation problem in which the capacity
from each source is 1 and the demand at each
destination is 1

 The Fix-It Shop has three rush projects to repair
 They have three repair persons with different
talents and abilities
 The owner has estimates of wage costs for each
worker for each project
 The owner’s objective is to assign the three
project to the workers in a way that will result in
the lowest cost to the shop
 Each project will be assigned exclusively to one
worker

 Estimated project repair costs for the Fix-It shop
assignment problem
PROJECT
PERSON 1 2 3
Adams $11 $14 $6
Brown 8 10 11
Cooper 9 12 7
Table 10.26

 Summary of Fix-It Shop assignment alternatives
and costs
PRODUCT ASSIGNMENT
1 2 3
LABOR
COSTS ($)
TOTAL
COSTS ($)
Adams Brown Cooper 11 + 10 + 7 28
Adams Cooper Brown 11 + 12 + 11 34
Brown Adams Cooper 8 + 14 + 7 29
Brown Cooper Adams 8 + 12 + 6 26
Cooper Adams Brown 9 + 14 + 11 34
Cooper Brown Adams 9 + 10 + 6 25
Table 10.27

The Hungarian Method
(Flood’s Technique)
 The Hungarian method
Hungarian method is an efficient method of
finding the optimal solution to an assignment
problem without having to make direct
comparisons of every option
 It operates on the principle of matrix reduction
matrix reduction
 By subtracting and adding appropriate numbers
in the cost table or matrix, we can reduce the
problem to a matrix of opportunity costs
opportunity costs
 Opportunity costs show the relative penalty
associated with assigning any person to a project
as opposed to making the best
best assignment
 We want to make assignment so that the
opportunity cost for each assignment is zero

Three Steps of the Assignment Method
1.
1. Find the opportunity cost table by
Find the opportunity cost table by:
(a) Subtracting the smallest number in each row
of the original cost table or matrix from every
number in that row
(b) Then subtracting the smallest number in
each column of the table obtained in part (a)
from every number in that column
2.
2. Test the table resulting from step 1 to see
Test the table resulting from step 1 to see
whether an optimal assignment can be made
whether an optimal assignment can be made by
drawing the minimum number of vertical and
horizontal straight lines necessary to cover all
the zeros in the table. If the number of lines is
less than the number of rows or columns,
proceed to step 3.

Three Steps of the Assignment Method
3.
3. Revise the present opportunity cost table
Revise the present opportunity cost table by
subtracting the smallest number not covered by
a line from every other uncovered number. This
same number is also added to any number(s)
lying at the intersection of horizontal and
vertical lines. Return to step 2 and continue the
cycle until an optimal assignment is possible.

Test opportunity cost table to
see if optimal assignments are
possible by drawing the
minimum possible lines on
columns and/or rows such that
all zeros are covered
Find opportunity cost
(a) Subtract smallest number in
each row from every number
in that row, then
(b) subtract smallest number in
each column from every
number in that column
Steps in the Assignment Method
Set up cost table for problem Revise opportunity cost table
in two steps:
(a) Subtract the smallest
number not covered by a line
from itself and every other
uncovered number
(b) add this number at every
intersection of any two lines
Optimal solution at zero
locations. Systematically make
final assignments.
(a) Check each row and
column for a unique zero and
make the first assignment in
that row or column
(b) Eliminate that row and
column and search for another
unique zero. Make that
assignment and proceed in a
like manner.
Step 1
Not
optimal
Step 2
Optimal
Figure 10.3

 Step 1: Find the opportunity cost table
 We can compute row
row opportunity costs and
column
column opportunity costs
 What we need is the total
total opportunity cost
 We derive this by taking the row opportunity
costs and subtract the smallest number in that
column from each number in that column

 Cost of each person-
project assignment
 Row opportunity
cost table
PROJECT
PERSON 1 2 3
Adams $11 $14 $6
Brown 8 10 11
Cooper 9 12 7
Table 10.28
PROJECT
PERSON 1 2 3
Adams $5 $8 $0
Brown 0 2 3
Cooper 2 5 0
Table 10.29
 The opportunity cost of assigning Cooper to
project 2 is $12 – $7 = $5

 We derive the total opportunity costs by taking
the costs in Table 29 and subtract the smallest
number in each column from each number in that
column
 Row opportunity
cost table
PROJECT
PERSON 1 2 3
Adams $5 $8 $0
Brown 0 2 3
Cooper 2 5 0
Table 10.29
 Total opportunity
cost table
PROJECT
PERSON 1 2 3
Adams $5 $6 $0
Brown 0 0 3
Cooper 2 3 0
Table 10.30

 Step 2: Test for the optimal assignment
 We want to assign workers to projects in such
a way that the total labor costs are at a
minimum
 We would like to have a total assigned
opportunity cost of zero
 The test to determine if we have reached an
optimal solution is simple
 We find the minimum
minimum number of straight lines
necessary to cover all the zeros in the table
 If the number of lines equals the number of
rows or columns, an optimal solution has been
reached

 Test for optimal solution
PROJECT
PERSON 1 2 3
Adams $5 $6 $0
Brown 0 0 3
Cooper 2 3 0
Table 10.31
Covering line 1
Covering line 1
Covering line 2
Covering line 2
 This requires only two lines to cover the zeros so
the solution is not optimal

 Step 3: Revise the opportunity-cost table
 We subtract
subtract the smallest number not covered
by a line from all numbers not covered by a
straight line
 The same number is added to every number
lying at the intersection of any two lines
 We then return to step 2 to test this new table

 Revised opportunity cost table (derived by
subtracting 2 from each cell not covered by a line
and adding 2 to the cell at the intersection of the
lines)
PROJECT
PERSON 1 2 3
Adams $3 $4 $0
Brown 0 0 5
Cooper 0 1 0
Table 10.32

 Optimality test on the revised opportunity cost
table
PROJECT
PERSON 1 2 3
Adams $3 $4 $0
Brown 0 0 5
Cooper 0 1 0
Table 10.33
Covering line 2
Covering line 2
Covering line 3
Covering line 3
 This requires three lines to cover the zeros so the
solution is optimal
Covering line 1
Covering line 1

Making the Final Assignment
 The optimal assignment is Adams to project 3,
Brown to project 2, and Cooper to project 1
 But this is a simple problem
 For larger problems one approach to making the
final assignment is to select a row or column that
contains only one zero
 Make the assignment to that cell and rule out its
row and column
 Follow this same approach for all the remaining
cells

 Total labor costs of this assignment are
ASSIGNMENT COST ($)
Adams to project 3 6
Brown to project 2 10
Cooper to project 1 9
Total cost 25

 Making the final assignments
(A) FIRST
ASSIGNMENT
(B) SECOND
ASSIGNMENT
(C) THIRD
ASSIGNMENT
1 2 3 1 2 3 1 2 3
Adams 3 4 0 Adams 3 4 0 Adams 3 4 0
Brown 0 0 5 Brown 0 0 5 Brown 0 0 5
Cooper 0 1 0 Cooper 0 1 0 Cooper 0 1 0
Table 10.34

Using Excel QM for the Fix-It Shop
Assignment Problem
 Excel QM assignment module
Program 10.3A

Using Excel QM for the Fix-It Shop
Assignment Problem
 Excel QM output screen
Program 10.3A

Unbalanced Assignment Problems
 Often the number of people or objects to be
assigned does not equal the number of tasks or
clients or machines listed in the columns, and the
problem is unbalanced
unbalanced
 When this occurs, and there are more rows than
columns, simply add a dummy column
dummy column or task
 If the number of tasks exceeds the number of
people available, we add a dummy row
dummy row
 Since the dummy task or person is nonexistent,
we enter zeros in its row or column as the cost or
time estimate

Unbalanced Assignment Problems
 The Fix-It Shop has another worker available
 The shop owner still has the same basic problem
of assigning workers to projects
 But the problem now needs a dummy column to
balance the four workers and three projects
PROJECT
PERSON 1 2 3 DUMMY
Adams $11 $14 $6 $0
Brown 8 10 11 0
Cooper 9 12 7 0
Davis 10 13 8 0
Table 10.35

Maximization Assignment Problems
 Some assignment problems are phrased in terms
of maximizing the payoff, profit, or effectiveness
 It is easy to obtain an equivalent minimization
problem by converting all numbers in the table to
opportunity costs
 This is brought about by subtracting every
number in the original payoff table from the
largest single number in that table
 Transformed entries represent opportunity costs
 Once the optimal assignment has been found, the
total payoff is found by adding the original
payoffs of those cells that are in the optimal
assignment

 The British navy wishes to assign four ships to
patrol four sectors of the North Sea
 Ships are rated for their probable efficiency in
each sector
 The commander wants to determine patrol
assignments producing the greatest overall
efficiencies

 Efficiencies of British ships in patrol sectors
SECTOR
SHIP A B C D
1 20 60 50 55
2 60 30 80 75
3 80 100 90 80
4 65 80 75 70
Table 10.36

 Opportunity cost of British ships
SECTOR
SHIP A B C D
1 80 40 50 45
2 40 70 20 25
3 20 0 10 20
4 35 20 25 30
Table 10.37

 First convert the maximization efficiency table
into a minimizing opportunity cost table by
subtracting each rating from 100, the largest
rating in the whole table
 The smallest number in each row is subtracted
from every number in that row and the smallest
number in each column is subtracted from every
number in that column
 The minimum number of lines needed to cover
the zeros in the table is four, so this represents
an optimal solution

 The overall efficiency
ASSIGNMENT EFFICIENCY
Ship 1 to sector D 55
Ship 2 to sector C 80
Ship 3 to sector B 100
Ship 4 to sector A 65
Total efficiency 300

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