Management Science –
MNG221
Lecture 1: Transportation,
Transshipment, and Assignment
Transportation, Transshipment,
and Assignment
• Transportation,
transshipment,
and
assignment problems are special types of
linear programming model formulations.
• They are part of a larger class of linear
programming problems known as network
flow problems.
• There solution approaches are variations of
the traditional simplex solution procedure.
Transportation, Transshipment, and Assignment
Transportation Model
Transportation Model
• Transportation models are formulated to
solve problems where:
A product is transported from a number of
sources to a number of destinations at the
minimum possible cost.
Each source is able to supply a fixed
number of units of the product, and each
destination has a fixed demand for the
product.
Transportation Model
• In a transportation problem, items are
allocated from sources to destinations
at a minimum cost.
• The linear programming model for a
transportation problem has constraints
for supply at each source and demand at
each destination.
Transportation Model
• A Transportation can be a:
1. Balanced transportation model in which supply
equals demand, all constraints are equalities.
Supply = Demand
2. Unbalanced transportation model in which supply
exceeds demand or demand exceeds supply:
Supply < Demand
Supply > Demand
Transportation problems are usually solved manually
within a tableau format.
Transportation, Transshipment, and Assignment
The Transshipment Model
Transshipment Model
• The transshipment model is an extension of
the transportation model in which
intermediate transshipment points are
added
between
the
sources
and
destinations.
• An example of a transshipment point is a
distribution center or warehouse located
between plants and stores.
Transshipment Model
• In a transshipment problem, items may
be transported:
1. From sources through transshipment
points on to destinations,
2. From one source to another,
3. From one transshipment point to
another,
Transshipment Model
4. From one destination to another,
5. Or directly from sources to
destinations,
6. Or some combination of these
alternatives.
• The transshipment model includes
intermediate points between sources
and destinations.
Transportation, Transshipment, and Assignment
The Assignment Model
Assignment Model
• The Assignment Model is a special form
of a linear programming model that is
similar to the transportation model.
• The differences, however, is that the
supply at each source and the demand
at each destination are each limited to
one unit.
Assignment Model
• The linear programming formulation of
the assignment model is similar to the
formulation of the transportation
model, except all the supply values for
each source equal one, and all the
demand values at each destination
equal one.
Transportation, Transshipment, and Assignment
Transportation: Worked Example
Transportation: Worked Example
• Wheat is harvested in the Midwest and stored in grain
elevators in three different cities - Kansas City, Omaha,
and Des Moines.
• These grain elevators supply three flour mills, located in
Chicago, St. Louis, and Cincinnati.
• The cost of transporting one ton of wheat from each grain
elevator (source) to each mill (destination) differs
according to the distance and rail system.
• The problem is to determine how many tons of wheat to
transport from each grain elevator to each mill on a
monthly basis in order to minimize the total cost of
transportation.
Transportation: Worked Example
Grain Elevator
Supply
Mill
Demand
1. Kansas City
150
A. Chicago
200
2. Omaha
175
B. St. Louis
100
3. Des Moines 275
C. Cincinnati
300
Total
Total
600
600
Mills
Grain Elevator
A. Chicago
B. St. Louis
C. Cincinnati
1. Kansas City
$6
$8
$10
2. Omaha
$7
$11
$11
3. Des Moines
$4
$5
$12
Transportation, Transshipment, and Assignment
Transportation: Worked Example
Transportation
Transportation: Worked Example
Grain Elevator
Supply
Mill
Demand
1. Kansas City
150
A. Chicago
200
2. Omaha
175
B. St. Louis
100
3. Des Moines 275
C. Cincinnati
300
Total
Total
600
600
Mills
Grain Elevator
A. Chicago
B. St. Louis
C. Cincinnati
1. Kansas City
$6
$8
$10
2. Omaha
$7
$11
$11
3. Des Moines
$4
$5
$12
Transportation: Worked Example
Minimize Z = $61A + $81B + $101C + $72A + $112B + $112C + $43A + $53B + $123C
Subject to:
Supply Constraints
• x1A + x1B + x1C = 150
• x2A + x2B + x2C = 175
• x3A + x3B + x3C = 275
Xij - the number of tons of wheat
transported from each grain elevator,
i - (where i = 1, 2, 3), to each mill,
j -(where j = A, B, C).
Demand Constraints
• x1A + x2A + x3A = 200
• x1B + x2B + x3B = 100
• x1C + x2C + x3C = 300
Example
Supply Constraint, x1A + x1B + x1C = 150
Kansas City to all three mills: Chicago (x1A),
St. Louis (x1B), and Cincinnati (x1C).
Non-negativity
• xij ≥ 0
Constraints are an equation ( = ) and not a
inequality (≤), because all of wheat available
is needed to meet demand of 600 tons.
Transportation
Minimize Z = $61A + $81B + $101C + $72A + $112B + $112B + $43A + $53B + $123C
Subject to:
x1A + x1B + x1C = 150
x1A + x2A + x3A = 200
xij ≥ 0
x2A + x2B + x2C = 175
x1B + x2B + x3B = 100
x3A + x3B + x3C = 275
x1C + x2C + x3C = 300
To
From
•Each cell in a transportation
tableau is analogous to a
decision variable that indicates
the amount allocated from a
source to a destination.
•The supply and demand
values along the outside of the
rim of the tableau are called
rim requirements.
Transportation: Worked Example
• 2 -Methods for solving a Transportation Model –
1. The Stepping-Stone Method
2. The Modified Distribution Method (also known
as MODI)
BUT
• Must be given an Initial Solution
1. The Northwest Corner Method
2. The Minimum Cell Cost Method
3. Vogel’s Approximation Method
Transportation
To
From
150
50
100
25
Northwest Corner Method
the
largest
possible
allocation is made in the
cell in the upper left hand
corner of the tableau,
followed by allocations to
adjacent feasible cells.
275
The initial solution is
complete when all rim
requirements are satisfied.
The cost of the initial cost is obtained by substituting the allocations in the objective
function
Z = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C
= 6(150) + 8(0) + 10(0) + 7(50) + 11(100) + 11(25) + 4(0) + 5(0) + 12(275) = $5,925
Transportation
To
From
25
125
175
200
75
The Minimum Cost
Method as much as
possible is allocated to
the cell with the
minimum cost.
The initial solution is
complete when all rim
requirements
are
satisfied.
The cost of the initial cost is obtained by substituting the allocations in the objective
function
Z = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C
= 6(0) + 8(25) + 10(125) + 7(0) + 11(0) + 11(175) + 4(200) + 5(75) + 12(0) = $4,550
Transportation
• Vogel's Approximation Model is based on the
concept of penalty cost or regret.
• A Penalty Cost is the difference between the
largest and next largest cell cost in a row (or
column).
• VAM allocates as much as possible to the
minimum cost cell in the row or column with the
largest penalty cost.
Transportation
• Vogel's Approximation Model Steps
1. Determine the penalty cost for each row and
column by subtracting the lowest cell cost in
the row or column from the next lowest cell
cost in the same row or column.
•
Column A: L C -4, Next LC -6. As such 6-4 = 2
2. Select the row or column with the highest
penalty cost (breaking ties arbitrarily or
choosing the lowest-cost cell).
Transportation
• Vogel's Approximation Model Steps
3. Allocate as much as possible to the
feasible
cell
with
the
lowest
transportation cost in the row or
column with the highest penalty cost.
4. Repeat steps 1, 2, and 3 until all rim
requirements have been met.
Transportation
To
From
2
4
175
1
2
3
1
Transportation
To
From
2
175
81
100
2
3
1
After each VAM cell allocation, all row and column penalty costs are
recomputed
Transportation
To
From
150
2
175
25
2
100
150
8
1
After each VAM cell allocation, all row and column penalty costs are
recomputed
Transportation
To
From
150
175
25
100
150
The cost of the initial cost is obtained by substituting the allocations in the objective
function
Z = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C
= 6(0) + 8(0) + 10(150) + 7(175) + 11(0) + 11(0) + 4(25) + 5(100) + 12(150) = $5,125
Transportation
• After obtaining the initial solution, the problem may
be solved using:
1. Stepping-Stone Method
2. Modified Distribution Method (MODI)
Stepping Stone Method – determines if there is a cell with
no allocation that would reduce cost if used.
Modified Distribution Method (MODI)– a modified version
of the stepping-stone method. However, the individual cell
cost changes are determined mathematically.
Transportation
• Stepping Stone Method
• Step 1: Determine the stepping-stone path and cost
changes for each empty cell in the tableau.
• Step 2: Allocate as much as possible to the empty cell
with the greatest net decrease in cost.
• Step 3: Repeat steps 1 and 2 until an empty cell have
positive cost changes that indicates an optimal
solution.
Transportation
To
From
X
25
125
X
X
175
200
75
X
Step 1: Determine the
stepping-stone path
and cost changes for
each empty cell in the
tableau.
But First Identify
Empty Cells :
1A
2A
2B
3C
Transportation
Evaluation of Cell
1A
To
From
+1
-1
25
125
175
-1
+1
200
75
In evaluating the empty
cells the constraint of the
problems
cannot
be
violated, and feasibility
must be maintained.
Review
of
the
cost
increase/reduction of the
process.
1A
1B
3B
3A
$6 - $8 + $5 - $4 = -$1
Transportation
Evaluation of Cell
2A
To
From
-1
+1
25
125
Review of the cost
increase/reduction of
the process.
-1
+1
175
-1
200
+1
75
2A
2C
1C
1B
3B
+$7 - $11 + $10 - $8 +$5 - $4 = -$1
3A
Transportation
To
From
-1
+1
25
125
+1
-1
175
200
75
Evaluation of Cell
1A
2A
2B
3C
Review
of
the
cost
increase/reduction of the
process.
2B
2C
1C
1B
+$11 - $11 + $10 - $8 = +$2
Transportation
To
From
+1
-1
25
125
175
-1
200
Evaluation of Cell
1A
2A
2B
3C
+1
75
Review
of
the
cost
increase/reduction of the
process.
2B
2C
1C
1B
+$12 - $10 + $8 - $5 = +$5
Transportation
Summary
To
From
25
125
175
200
75
1A – reduction $1
2A – reduction $1
2B – increase $2
3C – increase $5
Transportation
To
From
+1
25
-1
25 -25= 0 125
175
-1
Evaluation of Cell
1A
2A
2B
3C
+1
200 -25 = 175 75 + 25 = 100
Identify the minimum in
the stepping stone path.
The cost of the initial cost is obtained by substituting the allocations in the objective function
Z = $61A + $81A + $101C + $72A + $112B + $112B + $43A + $53B + $123C
= 6(25) + 8(0) + 10(125) + 7(0) + 11(0) + 11(175) + 4(175) + 5(100) + 12(0) = $4,525
Transportation, Transshipment, and Assignment
The Assignment Model
Assignment Model
An assignment problem is a special form of
transportation problem where all supply and
demand values equal one.
Steps of the assignment solution method are:
1. Perform row reductions by subtracting the
minimum value in each row from all row
values.
Assignment Model
Steps of the assignment solution method (Continued)
2. Perform column reductions by subtracting
the minimum value in each column from all
column value.
3. In the completed Opportunity Cost Table,
cross out all zeros, using the minimum
number of horizontal or vertical lines.
Assignment Model
Steps of the assignment solution method (Continued)
4. If fewer than m lines are required (where m
= the number of rows or column), subtract
the minimum uncrossed value from all the
other uncrossed values, and add this same
minimum uncrossed value where two lines
intersect. Leave all other values unchanged.
Assignment Model
Steps of the assignment solution method
(Continued)
5. If m lines are required, the tableau contains
the optimal solution and m unique
assignments are made. If fewer than m lines
are required, repeat step 4.
Transportation, Transshipment, and Assignment
The Assignment Model
- A Worked Example
Assignment Model - Example
• The Atlantic Coast Conference has four
basketball games on a particular night.
• The conference office wants to assign four
teams of officials to the four games in a way
that will minimize the total distance traveled
by the officials.
• The distances in miles for each team of
officials to each game location are shown
below.
Assignment Model
Step 1: Perform row reductions by subtracting the minimum value in each
row from all row values.
Assignment Model
Step 1: Perform column reductions by subtracting the minimum value in
each column from all column value.
Assignment Model
Step 3: In the completed opportunity cost
table, cross out all zeros, using the minimum
number of horizontal or vertical lines.
Assignment Model
Step 4: If fewer than m lines are required (where m = the number of
rows or column), subtract the minimum uncrossed value from all the
other uncrossed values, and add this same minimum uncrossed value
where two lines intersect. Leave all other values unchanged.
Assignment Model
Step 4: If m lines are required, the tableau contains the
optimal solution and m unique assignments are made.
If fewer than m lines are required, repeat step 4.
Assignment
Team A-Atlanta
Team B-Raleigh
Team C-Durham
Team D-Clemson
Total
Distance
90
100
140
120
420
Assignment Model
Unbalanced Assignment Tableau with a dummy column
•When demand exceeds supply, a dummy row is added to the assignment
tableau.
•When supply exceeds demands, a dummy column is added to the
assignment tableau.