Transcript Transportation, Assignment, and Transshipment Problems

Slides Prepared by
JOHN S. LOUCKS
St. Edward’s University
© 2003 ThomsonTM/South-Western
Slide 1
Chapter 7
Transportation, Assignment, and
Transshipment Problems



Transportation Problem
•Network Representation and LP Formulation
•Transportation Simplex Method
Assignment Problem
•Network Representation and LP Formulation
•Hungarian Method
The Transshipment Problem
•Network Representation and LP Formulation
© 2003 ThomsonTM/South-Western
Slide 2
Transportation, Assignment, and
Transshipment Problems


A network model is one which can be represented by
a set of nodes, a set of arcs, and functions (e.g. costs,
supplies, demands, etc.) associated with the arcs
and/or nodes.
Transportation, assignment, and transshipment
problems of this chapter, as well as the shortest route,
minimal spanning tree, and maximal flow problems
(Chapter 9) and PERT/CPM problems (Chapter 10)
are all examples of network problems.
© 2003 ThomsonTM/South-Western
Slide 3
Transportation, Assignment, and
Transshipment Problems



Each of the three models of this chapter
(transportation, assignment, and transshipment
models) can be formulated as linear programs and
solved by general purpose linear programming
codes.
For each of the three models, if the right-hand side of
the linear programming formulations are all integers,
the optimal solution will be in terms of integer values
for the decision variables.
However, there are many computer packages
(including The Management Scientist) which contain
separate computer codes for these models which take
advantage of their network structure.
© 2003 ThomsonTM/South-Western
Slide 4
Transportation Problem


The transportation problem seeks to minimize the
total shipping costs of transporting goods from m
origins (each with a supply si) to n destinations (each
with a demand dj), when the unit shipping cost from
an origin, i, to a destination, j, is cij.
The network representation for a transportation
problem with two sources and three destinations is
given on the next slide.
© 2003 ThomsonTM/South-Western
Slide 5
Transportation Problem

Network Representation
s1
s2
1
c11
c13
c21
2
1
d1
2
d2
3
d3
c12
c22
c23
SOURCES
© 2003 ThomsonTM/South-Western
DESTINATIONS
Slide 6
Transportation Problem

LP Formulation
The LP formulation in terms of the amounts
shipped from the origins to the destinations, xij , can
be written as:
Min cijxij
ij
s.t.
xij < si for each origin i
j
xij = dj for each destination j
i
xij > 0 for all i and j
© 2003 ThomsonTM/South-Western
Slide 7
Transportation Problem

LP Formulation Special Cases
The following special-case modifications to the
linear programming formulation can be made:
•Minimum shipping guarantee from i to j:
xij > Lij
•Maximum route capacity from i to j:
xij < Lij
•Unacceptable route:
Remove the corresponding decision variable.
© 2003 ThomsonTM/South-Western
Slide 8
Example: BBC
Building Brick Company (BBC) has orders for 80
tons of bricks at three suburban locations as follows:
Northwood -- 25 tons, Westwood -- 45 tons, and
Eastwood -- 10 tons. BBC has two plants, each of which
can produce 50 tons per week. Delivery cost per ton
from each plant to each suburban location is shown on
the next slide.
How should end of week shipments be made to fill
the above orders?
© 2003 ThomsonTM/South-Western
Slide 9
Example: BBC

Delivery Cost Per Ton
Plant 1
Plant 2
Northwood
24
30
© 2003 ThomsonTM/South-Western
Westwood
30
40
Eastwood
40
42
Slide 10
Example: BBC

Partial Spreadsheet Showing Problem Data
A
B
1
2 Constraint
X11
3
#1
1
4
#2
5
#3
1
6
#4
7
#5
8 Obj.Coefficients 24
C
D
E
F
G
H
X22
X23
RHS
LHS Coefficients
X12
X13
1
1
X21
50
1
1
1
1
1
25
1
1
30
© 2003 ThomsonTM/South-Western
40
30
50
40
45
1
10
42
30
Slide 11
Example: BBC

Partial Spreadsheet Showing Optimal Solution
A
B
C
D
10
X11
X12
X13
11 Dec.Var.Values
5
45
0
12
Minimized Total Shipping Cost
13
14
LHS
Constraints
15
P1.Cap.
50
16
P2.Cap.
30
17
N.Dem.
25
18
W.Dem.
45
19
E.Dem.
10
© 2003 ThomsonTM/South-Western
E
X21
20
2490
<=
<=
=
=
=
F
X22
0
G
X23
10
RHS
50
50
25
45
10
Slide 12
Example: BBC

Optimal Solution
From
To
Amount Cost
Plant 1 Northwood
5
120
Plant 1 Westwood
45
1,350
Plant 2 Northwood
20
600
Plant 2 Eastwood
10
420
Total Cost = $2,490
© 2003 ThomsonTM/South-Western
Slide 13
Example: BBC

Partial Sensitivity Report (first half)
Adjustable Cells
Final Reduced Objective Allowable Allowable
Cell Name Value
Cost
Coefficient Increase Decrease
$C$12 X11
5
0
24
4
4
$D$12 X12
45
0
30
4
1E+30
$E$12 X13
0
4
40
1E+30
4
$F$12 X21
20
0
30
4
4
$G$12 X22
0
4
40
1E+30
4
$H$12 X23 10.000
0.000
42
4
1E+30
© 2003 ThomsonTM/South-Western
Slide 14
Example: BBC

Partial Sensitivity Report (second half)
Constraints
Cell
$E$17
$E$18
$E$19
$E$20
$E$16
Name
P2.Cap
N.Dem
W.Dem
E.Dem
P1.Cap
Final Shadow Constraint Allowable Allowable
Value
Price
R.H. Side
Increase Decrease
30.0
0.0
50
1E+30
20
25.0
30.0
25
20
20
45.0
36.0
45
5
20
10.0
42.0
10
20
10
50.0
-6.0
50
20
5
© 2003 ThomsonTM/South-Western
Slide 15
Transportation Simplex Method



The transportation simplex method requires that the
sum of the supplies at the origins equal the sum of
the demands at the destinations.
If the total supply is greater than the total demand, a
dummy destination is added with demand equal to
the excess supply, and shipping costs from all origins
are zero. (If total supply is less than total demand, a
dummy origin is added.)
When solving a transportation problem by its special
purpose algorithm, unacceptable shipping routes are
given a cost of +M (a large number).
© 2003 ThomsonTM/South-Western
Slide 16
Transportation Simplex Method

A transportation tableau is given below. Each cell
represents a shipping route (which is an arc on the
network and a decision variable in the LP
formulation), and the unit shipping costs are given in
an upper right hand box in the cell.
D1
S1
S2
Demand
D2
D3
Supply
15
30
20
30
40
35
25
© 2003 ThomsonTM/South-Western
45
50
30
10
Slide 17
Transportation Simplex Method



The transportation problem is solved in two phases:
•Phase I -- Obtaining an initial feasible solution
•Phase II -- Moving toward optimality
Phase I: The Minimum-Cost Procedure can be used
to establish an initial basic feasible solution without
doing numerous iterations of the simplex method.
Phase II: The Stepping Stone Method - using the
MODI Method for evaluating the reduced costs - may
be used to move from the initial feasible solution to
the optimal one.
© 2003 ThomsonTM/South-Western
Slide 18
Transportation Simplex Method

Phase I - Minimum-Cost Method
•Step 1: Select the cell with the least cost. Assign to
this cell the minimum of its remaining row supply or
remaining column demand.
•Step 2: Decrease the row and column availabilities
by this amount and remove from consideration all
other cells in the row or column with zero
availability/demand. (If both are simultaneously
reduced to 0, assign an allocation of 0 to any other
unoccupied cell in the row or column before deleting
both.) GO TO STEP 1.
© 2003 ThomsonTM/South-Western
Slide 19
Transportation Simplex Method

Phase II - Stepping Stone Method
•Step 1: For each unoccupied cell, calculate the
reduced cost by the MODI method described on an
upcoming slide.
Select the unoccupied cell with the most
negative reduced cost. (For maximization problems
select the unoccupied cell with the largest reduced
cost.) If none, STOP.
© 2003 ThomsonTM/South-Western
Slide 20
Transportation Simplex Method

Phase II - Stepping Stone Method (continued)
•Step 2: For this unoccupied cell generate a stepping
stone path by forming a closed loop with this cell
and occupied cells by drawing connecting
alternating horizontal and vertical lines between
them.
Determine the minimum allocation where a
subtraction is to be made along this path.
© 2003 ThomsonTM/South-Western
Slide 21
Transportation Simplex Method

Phase II - Stepping Stone Method (continued)
•Step 3: Add this allocation to all cells where
additions are to be made, and subtract this allocation
to all cells where subtractions are to be made along
the stepping stone path.
(Note: An occupied cell on the stepping
stone path now becomes 0 (unoccupied). If more
than one cell becomes 0, make only one unoccupied;
make the others occupied with 0's.)
GO TO STEP 1.
© 2003 ThomsonTM/South-Western
Slide 22
Transportation Simplex Method

MODI Method (for obtaining reduced costs)
Associate a number, ui, with each row and vj with
each column.
•Step 1: Set u1 = 0.
•Step 2: Calculate the remaining ui's and vj's by
solving the relationship cij = ui + vj for occupied cells.
•Step 3: For unoccupied cells (i,j), the reduced cost =
cij - ui - vj.
© 2003 ThomsonTM/South-Western
Slide 23
Example: BBC

Initial Transportation Tableau
Since total supply = 100 and total demand = 80, a
dummy destination is created with demand of 20 and 0
unit costs.
Northwood Westwood
Eastwood
Dummy
Supply
Plant 1
24
30
40
0
50
Plant 2
30
40
42
0
50
Demand
25
45
© 2003 ThomsonTM/South-Western
10
20
Slide 24
Example: BBC

Phase I: Minimum-Cost Procedure
•Iteration 1: Tie for least cost (0), arbitrarily select x14.
Allocate 20. Reduce s1 by 20 to 30 and delete the
Dummy column.
•Iteration 2: Of the remaining cells the least cost is 24
for x11. Allocate 25. Reduce s1 by 25 to 5 and
eliminate the Northwood column.
© 2003 ThomsonTM/South-Western
Slide 25
Example: BBC

Phase I: Minimum-Cost Procedure (continued)
•Iteration 3: Of the remaining cells the least cost is 30
for x12. Allocate 5. Reduce the Westwood column to
40 and eliminate the Plant 1 row.
•Iteration 4: Since there is only one row with two
cells left, make the final allocations of 40 and 10 to x22
and x23, respectively.
© 2003 ThomsonTM/South-Western
Slide 26
Example: BBC

Phase II – Iteration 1
•MODI Method
1. Set u1 = 0
2. Since u1 + vj = c1j for occupied cells in row 1, then
v1 = 24, v2 = 30, v4 = 0.
3. Since ui + v2 = ci2 for occupied cells in column 2,
then u2 + 30 = 40, hence u2 = 10.
4. Since u2 + vj = c2j for occupied cells in row 2, then
10 + v3 = 42, hence v3 = 32.
© 2003 ThomsonTM/South-Western
Slide 27
Example: BBC

Phase II – Iteration 1
•MODI Method (continued)
Calculate the reduced costs (circled numbers on the
next slide) by cij - ui + vj.
Unoccupied Cell
(1,3)
(2,1)
(2,4)
© 2003 ThomsonTM/South-Western
Reduced Cost
40 - 0 - 32 = 8
30 - 24 -10 = -4
0 - 10 - 0 = -10
Slide 28
Example: BBC

Phase II – Iteration 1
•MODI Method (continued)
Northwood Westwood
Eastwood
Dummy
30
40
0
Plant 1
25
Plant 2
-4
vj
24
24
30
5
40
30
© 2003 ThomsonTM/South-Western
40
+8
10
32
42
20
-10
0
ui
0
10
0
Slide 29
Example: BBC

Phase II – Iteration 1
•Stepping Stone Method
The stepping stone path for cell (2,4) is (2,4), (1,4),
(1,2), (2,2). The allocations in the subtraction cells are 20
and 40, respectively. The minimum is 20, and hence
reallocate 20 along this path. Thus for the next tableau:
x24 = 0 + 20 = 20 (0 is its current allocation)
x14 = 20 - 20 = 0 (blank for the next tableau)
x12 = 5 + 20 = 25
x22 = 40 - 20 = 20
The other occupied cells remain the same.
© 2003 ThomsonTM/South-Western
Slide 30
Example: BBC

Phase II - Iteration 2
•MODI Method
The reduced costs are found by calculating
the ui's and vj's for this tableau.
1. Set u1 = 0.
2. Since u1 + vj = cij for occupied cells in row 1, then
v1 = 24, v2 = 30.
3. Since ui + v2 = ci2 for occupied cells in column 2,
then u2 + 30 = 40, or u2 = 10.
4. Since u2 + vj = c2j for occupied cells in row 2, then
10 + v3 = 42 or v3 = 32; and, 10 + v4 = 0 or v4 = -10.
© 2003 ThomsonTM/South-Western
Slide 31
Example: BBC

Phase II - Iteration 2
•MODI Method (continued)
Calculate the reduced costs (circled numbers on the
next slide) by cij - ui + vj.
Unoccupied Cell
(1,3)
(1,4)
(2,1)
© 2003 ThomsonTM/South-Western
Reduced Cost
40 - 0 - 32 = 8
0 - 0 - (-10) = 10
30 - 10 - 24 = -4
Slide 32
Example: BBC

Phase II - Iteration 2
•MODI Method (continued)
Plant 1
Plant 2
vj
Northwood Westwood
Eastwood
Dummy
30
40
0
25
-4
24
24
30
25
20
30
© 2003 ThomsonTM/South-Western
40
+8
10
36
42
+10
20
0
ui
0
10
-6
Slide 33
Example: BBC

Phase II - Iteration 2
•Stepping Stone Method
The most negative reduced cost is = -4 determined
by x21. The stepping stone path for this cell is
(2,1),(1,1),(1,2),(2,2). The allocations in the subtraction
cells are 25 and 20 respectively. Thus, the new solution
is obtained by reallocating 20 on the stepping stone
path.
© 2003 ThomsonTM/South-Western
Slide 34
Example: BBC

Phase II - Iteration 2
•Stepping Stone Method (continued)
Thus, for the next tableau:
x21 = 0 + 20 = 20 (0 is its current allocation)
x11 = 25 - 20 = 5
x12 = 25 + 20 = 45
x22 = 20 - 20 = 0 (blank for the next tableau)
The other occupied cells remain the same.
© 2003 ThomsonTM/South-Western
Slide 35
Example: BBC

Phase II - Iteration 3
•MODI Method
The reduced costs are found by calculating
the ui's and vj's for this tableau.
1. Set u1 = 0
2. Since u1 + vj = c1j for occupied cells in row 1, then
v1 = 24 and v2 = 30.
3. Since ui + v1 = ci1 for occupied cells in column 2,
then u2 + 24 = 30 or u2 = 6.
4. Since u2 + vj = c2j for occupied cells in row 2, then
6 + v3 = 42 or v3 = 36, and 6 + v4 = 0 or v4 = -6.
© 2003 ThomsonTM/South-Western
Slide 36
Example: BBC

Phase II - Iteration 3
•MODI Method (continued)
Calculate the reduced costs (circled numbers on the
next slide) by cij - ui + vj.
Unoccupied Cell
(1,3)
(1,4)
(2,2)
© 2003 ThomsonTM/South-Western
Reduced Cost
40 - 0 - 36 = 4
0 - 0 - (-6) = 6
40 - 6 - 30 = 4
Slide 37
Example: BBC

Phase II - Iteration 3
•MODI Method (continued)
Since all the reduced costs are non-negative, this is
the optimal tableau.
Plant 1
Plant 2
vj
Northwood Westwood
Eastwood
Dummy
30
40
0
5
20
24
24
30
45
+4
30
© 2003 ThomsonTM/South-Western
40
+4
10
36
42
+6
20
0
ui
0
6
-6
Slide 38
Example: BBC

Optimal Solution
From
Plant 1
Plant 1
Plant 2
Plant 2
To
Amount Cost
Northwood
5
120
Westwood
45
1,350
Northwood
20
600
Eastwood
10
420
Total Cost = $2,490
© 2003 ThomsonTM/South-Western
Slide 39
Assignment Problem




An assignment problem seeks to minimize the total cost
assignment of m workers to m jobs, given that the cost
of worker i performing job j is cij.
It assumes all workers are assigned and each job is
performed.
An assignment problem is a special case of a
transportation problem in which all supplies and all
demands are equal to 1; hence assignment problems
may be solved as linear programs.
The network representation of an assignment problem
with three workers and three jobs is shown on the next
slide.
© 2003 ThomsonTM/South-Western
Slide 40
Assignment Problem

Network Representation
1
AGENTS
c11
c13
c21
2
1
c12
TASKS
c22
2
c23
c31
3
c32
c33
© 2003 ThomsonTM/South-Western
3
Slide 41
Assignment Problem

LP Formulation
Min cijxij
ij
s.t. xij = 1
j
xij = 1
i
for each agent i
for each task j
xij = 0 or 1 for all i and j
•Note:
A modification to the right-hand side of the
first constraint set can be made if a worker is
permitted to work more than 1 job.
© 2003 ThomsonTM/South-Western
Slide 42
Assignment Problem

LP Formulation Special Cases
•Number of agents exceeds the number of tasks:
xij < 1
j
for each agent i
•Number of tasks exceeds the number of agents:
Add enough dummy agents to equalize the
number of agents and the number of tasks.
The objective function coefficients for these
new variable would be zero.
© 2003 ThomsonTM/South-Western
Slide 43
Assignment Problem

LP Formulation Special Cases (continued)
•The assignment alternatives are evaluated in terms
of revenue or profit:
Solve as a maximization problem.
•An assignment is unacceptable:
Remove the corresponding decision variable.
•An agent is permitted to work a
tasks:
xij < a for each agent i
j
© 2003 ThomsonTM/South-Western
Slide 44
Example: Hungry Owner
A contractor pays his subcontractors a fixed fee
plus mileage for work performed. On a given day the
contractor is faced with three electrical jobs associated
with various projects. Given below are the distances
between the subcontractors and the projects.
Subcontractor
Westside
Federated
Goliath
Universal
Projects
A B C
50 36 16
28 30 18
35 32 20
25 25 14
How should the contractors be assigned to minimize
total costs?
© 2003 ThomsonTM/South-Western
Slide 45
Example: Hungry Owner

Network Representation
West.
Subcontractors
50
36
16
28
Fed.
18
35
Gol.
Univ.
20
25
A
Projects
30
B
32
C
25
14
© 2003 ThomsonTM/South-Western
Slide 46
Example: Hungry Owner

Linear Programming Formulation
Min
s.t.
50x11+36x12+16x13+28x21+30x22+18x23
+35x31+32x32+20x33+25x41+25x42+14x43
x11+x12+x13 < 1
x21+x22+x23 < 1
Agents
x31+x32+x33 < 1
x41+x42+x43 < 1
x11+x21+x31+x41 = 1
x12+x22+x32+x42 = 1
Tasks
x13+x23+x33+x43 = 1
xij = 0 or 1 for all i and j
© 2003 ThomsonTM/South-Western
Slide 47
Hungarian Method


The Hungarian method solves minimization
assignment problems with m workers and m jobs.
Special considerations can include:
•number of workers does not equal the number of
jobs -- add dummy workers or jobs with 0
assignment costs as needed
•worker i cannot do job j -- assign cij = +M
•maximization objective -- create an opportunity loss
matrix subtracting all profits for each job from the
maximum profit for that job before beginning the
Hungarian method
© 2003 ThomsonTM/South-Western
Slide 48
Hungarian Method




Step 1: For each row, subtract the minimum number in
that row from all numbers in that row.
Step 2: For each column, subtract the minimum number
in that column from all numbers in that column.
Step 3: Draw the minimum number of lines to cover all
zeroes. If this number = m, STOP -- an assignment can be
made.
Step 4: Subtract d (the minimum uncovered number)
from uncovered numbers. Add d to numbers covered by
two lines. Numbers covered by one line remain the
same. Then, GO TO STEP 3.
© 2003 ThomsonTM/South-Western
Slide 49
Hungarian Method

Finding the Minimum Number of Lines and
Determining the Optimal Solution
•Step 1: Find a row or column with only one unlined
zero and circle it. (If all rows/columns have two or
more unlined zeroes choose an arbitrary zero.)
•Step 2: If the circle is in a row with one zero, draw a
line through its column. If the circle is in a column
with one zero, draw a line through its row. One
approach, when all rows and columns have two or
more zeroes, is to draw a line through one with the
most zeroes, breaking ties arbitrarily.
•Step 3: Repeat step 2 until all circles are lined. If this
minimum number of lines equals m, the circles
provide the optimal assignment.
© 2003 ThomsonTM/South-Western
Slide 50
Example: Hungry Owner

Initial Tableau Setup
Since the Hungarian algorithm requires that there
be the same number of rows as columns, add a Dummy
column so that the first tableau is:
Westside
Federated
Goliath
Universal
© 2003 ThomsonTM/South-Western
A
50
28
35
25
B
36
30
32
25
C Dummy
16
0
18
0
20
0
14
0
Slide 51
Example: Hungry Owner


Step 1: Subtract minimum number in each row from all
numbers in that row. Since each row has a zero, we
would simply generate the same matrix above.
Step 2: Subtract the minimum number in each column
from all numbers in the column. For A it is 25, for B it
is 25, for C it is 14, for Dummy it is 0. This yields:
Westside
Federated
Goliath
Universal
© 2003 ThomsonTM/South-Western
A
25
3
10
0
B
11
5
7
0
C Dummy
2
0
4
0
6
0
0
0
Slide 52
Example: Hungry Owner

Step 3: Draw the minimum number of lines to cover all
zeroes. Although one can "eyeball" this minimum, use
the following algorithm. If a "remaining" row has only
one zero, draw a line through the column. If a
remaining column has only one zero in it, draw a line
through the row.
Westside
Federated
Goliath
Universal
© 2003 ThomsonTM/South-Western
A
25
3
10
0
B
11
5
7
0
C Dummy
2
0
4
0
6
0
0
0
Slide 53
Example: Hungry Owner

Step 4: The minimum uncovered number is 2. Subtract
2 from uncovered numbers; add 2 to all numbers
covered by two lines. This gives:
Westside
Federated
Goliath
Universal
© 2003 ThomsonTM/South-Western
A
23
1
8
0
B
9
3
5
0
C Dummy
0
0
2
0
4
0
0
2
Slide 54
Example: Hungry Owner

Step 3: Draw the minimum number of lines to cover all
zeroes.
Westside
Federated
Goliath
Universal
© 2003 ThomsonTM/South-Western
A
23
1
8
0
B
9
3
5
0
C Dummy
0
0
2
0
4
0
0
2
Slide 55
Example: Hungry Owner

Step 4: The minimum uncovered number is 1. Subtract
1 from uncovered numbers. Add 1 to numbers covered
by two lines. This gives:
Westside
Federated
Goliath
Universal
A
23
0
7
0
© 2003 ThomsonTM/South-Western
B
9
2
4
0
C Dummy
0
1
1
0
3
0
0
3
Slide 56
Example: Hungry Owner

Step 3: The minimum number of lines to cover all 0's is
four. Thus, there is a minimum-cost assignment of 0's
with this tableau. The optimal assignment is:
Subcontractor Project Distance
Westside
C
16
Federated
A
28
Goliath
(unassigned)
Universal
B
25
Total Distance = 69 miles
© 2003 ThomsonTM/South-Western
Slide 57
Transshipment Problem




Transshipment problems are transportation problems
in which a shipment may move through intermediate
nodes (transshipment nodes)before reaching a
particular destination node.
Transshipment problems can be converted to larger
transportation problems and solved by a special
transportation program.
Transshipment problems can also be solved by general
purpose linear programming codes.
The network representation for a transshipment
problem with two sources, three intermediate nodes,
and two destinations is shown on the next slide.
© 2003 ThomsonTM/South-Western
Slide 58
Transshipment Problem

Network Representation
3
c13
s1
1
c36
c37
c14
6
d1
c46
c15
Supply
Demand
c47
4
c23
s2
2
c56
c24
c25
SOURCES
5
d2
c57
INTERMEDIATE
NODES
© 2003 ThomsonTM/South-Western
7
DESTINATIONS
Slide 59
Transshipment Problem

Linear Programming Formulation
xij represents the shipment from node i to node j
Min cijxij
ij
s.t.
xij < si
j
for each origin i
xik - xkj = 0 for each intermediate
i
j
node k
xij = dj
i
xij > 0
© 2003 ThomsonTM/South-Western
for each destination j
for all i and j
Slide 60
Example: Transshipping
Thomas Industries and Washburn Corporation
supply three firms (Zrox, Hewes, Rockwright) with
customized shelving for its offices. They both order
shelving from the same two manufacturers, Arnold
Manufacturers and Supershelf, Inc.
Currently weekly demands by the users are 50 for
Zrox, 60 for Hewes, and 40 for Rockwright. Both
Arnold and Supershelf can supply at most 75 units to
its customers.
Additional data is shown on the next slide.
© 2003 ThomsonTM/South-Western
Slide 61
Example: Transshipping
Because of long standing contracts based on past
orders, unit costs from the manufacturers to the
suppliers are:
Thomas Washburn
Arnold
5
8
Supershelf
7
4
The costs to install the shelving at the various
locations are:
Zrox
Thomas
1
Washburn
3
© 2003 ThomsonTM/South-Western
Hewes Rockwright
5
8
4
4
Slide 62
Example: Transshipping

Network Representation
75 ARNOLD
Arnold
5
4
© 2003 ThomsonTM/South-Western
50
Hewes
HEWES
60
RockWright
40
5
Thomas
8
3
7
Super
Shelf
Zrox
1
8
75
ZROX
WashWASH
BURN
Burn
4
4
Slide 63
Example: Transshipping

Linear Programming Formulation
•Decision Variables Defined
xij = amount shipped from manufacturer i to supplier j
xjk = amount shipped from supplier j to customer k
where i = 1 (Arnold), 2 (Supershelf)
j = 3 (Thomas), 4 (Washburn)
k = 5 (Zrox), 6 (Hewes), 7 (Rockwright)
•Objective Function Defined
Minimize Overall Shipping Costs:
Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37
+ 3x45 + 4x46 + 4x47
© 2003 ThomsonTM/South-Western
Slide 64
Example: Transshipping

Constraints Defined
Amount Out of Arnold:
Amount Out of Supershelf:
Amount Through Thomas:
Amount Through Washburn:
Amount Into Zrox:
Amount Into Hewes:
Amount Into Rockwright:
x13 + x14 < 75
x23 + x24 < 75
x13 + x23 - x35 - x36 - x37 = 0
x14 + x24 - x45 - x46 - x47 = 0
x35 + x45 = 50
x36 + x46 = 60
x37 + x47 = 40
Non-negativity of Variables: xij > 0, for all i and j.
© 2003 ThomsonTM/South-Western
Slide 65
Example: Transshipping

Optimal Solution (from The Management Scientist )
Objective Function Value =
1150.000
Variable
Value
Reduced Costs
X13
X14
X23
X24
X35
X36
X37
X45
X46
X47
75.000
0.000
0.000
75.000
50.000
25.000
0.000
0.000
35.000
40.000
0.000
2.000
4.000
0.000
0.000
0.000
3.000
3.000
0.000
0.000
© 2003 ThomsonTM/South-Western
Slide 66
Example: Transshipping

Optimal Solution
75 ARNOLD
Arnold
5
75
4
© 2003 ThomsonTM/South-Western
50
Hewes
HEWES
60
RockWright
40
5
Thomas
8
3 4
7
Super
Shelf
Zrox
1
8
75
ZROX
WashWASH
BURN
Burn
4
Slide 67
Example: Transshipping

Optimal Solution (continued)
Constraint
1
2
3
4
5
6
7
Slack/Surplus
0.000
0.000
0.000
0.000
0.000
0.000
0.000
© 2003 ThomsonTM/South-Western
Dual Prices
0.000
2.000
-5.000
-6.000
-6.000
-10.000
-10.000
Slide 68
Example: Transshipping

Optimal Solution (continued)
OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value Upper Limit
X13
X14
X23
X24
X35
X36
X37
X45
X46
X47
3.000
6.000
3.000
No Limit
No Limit
3.000
5.000
0.000
2.000
No Limit
© 2003 ThomsonTM/South-Western
5.000
8.000
7.000
4.000
1.000
5.000
8.000
3.000
4.000
4.000
7.000
No Limit
No Limit
6.000
4.000
7.000
No Limit
No Limit
6.000
7.000
Slide 69
Example: Transshipping

Optimal Solution (continued)
RIGHT HAND SIDE RANGES
Constraint
1
2
3
4
5
6
7
Lower Limit
75.000
75.000
-75.000
-25.000
0.000
35.000
15.000
© 2003 ThomsonTM/South-Western
Current Value Upper Limit
75.000
No Limit
75.000
100.000
0.000
0.000
0.000
0.000
50.000
50.000
60.000
60.000
40.000
40.000
Slide 70
The End of Chapter 7
© 2003 ThomsonTM/South-Western
Slide 71