Matakuliah
Tahun
: K0442-Metode Kuantitatif
: 2009
Transportation, Assignment,
and Transshipment Problems
Pertemuan 7
Material Outline
• Transportation Problem
Network Representation
General LP Formulation
• Assignment Problem
Network Representation
General LP Formulation
• Transshipment Problem
Network Representation
General LP Formulation
Bina Nusantara University
3
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 PERT/CPM
problems (in another chapter) are all examples of
network problems.
Transportation, Assignment, and
Transshipment Problems
• Each of the three models of this chapter 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) that contain
separate computer codes for these models which
take advantage of their network structure.
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.
Transportation Problem
• Network Representation
1
d1
2
d2
3
d3
c11
s1
c12
1
c13
c21
s2
2
Sources
c23
c22
Destinations
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
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.
Example: Acme Block Co.
Acme Block Company has orders for 80 tons of
concrete blocks at three suburban locations
as follows: Northwood -- 25 tons,
Westwood -- 45 tons, and
Eastwood -- 10 tons. Acme
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?
Example: Acme Block Co.

Delivery Cost Per Ton
Northwood
Plant 1
24
Plant 2
30
Westwood
30
40
Eastwood
40
42
Example: Acme Block Co.

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
Example: Acme Block Co.

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
Example: Acme Block Co.

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
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.
Assignment Problem
1
Agents
c11
1
c12
c13
c21
2
Tasks
c22
2
c23
c31
3
c32
c33
• Network Representation
3
Assignment Problem
• LP Formulation
Min cijxij
ij
s.t.
xij = 1
j
for each agent i
xij = 1
i
xij = 0 or 1
for each task j
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.
Assignment Problem

LP Formulation Special Cases
• Number of agents exceeds the number of tasks:
xij < 1
for each agent i
j
• 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.
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
xij < a
j
tasks:
for each agent i
Example: Who Does What?
An electrical 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
mileage costs?
to minimize total
Example: Who Does What?
50
West.
Subcontractors
A
36
16
Projects
28
30
B
Fed.
18
35
Gol.
C
20
25
Univ.
32
25
14

Network Representation
Example: Who Does What?

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
Example: Who Does What?
• The optimal assignment is:
Subcontractor Project Distance
Westside
C
16
Federated
A
28
Goliath
(unassigned)
Universal
B
25
Total Distance = 69 miles
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.
Transshipment Problem
3
c13
s1
1
c37
c14
c15
4
Supply
s2
c36
6
c46
c47
d1
Demand
c23
2
Sources
c56
c24
c25
5
c57
Intermediate Nodes
• Network Representation
7
d2
Destinations
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
i
j
for each intermediate
node k
xij = dj
i
xij > 0
for each destination j
for all i and j
Example: Zeron Shelving
The Northside and Southside facilities of
Zeron Industries supply three firms (Zrox, Hewes,
Rockrite) 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 Rockrite.
Both Arnold and Supershelf can supply at most 75
units to its customers.
Additional data is shown on the next slide.
Example: Zeron Shelving
Because of long standing contracts based on past
orders, unit costs from the manufacturers to the
suppliers are:
Arnold
Supershelf
Zeron N
5
7
Zeron S
8
4
The costs to install the shelving at the various
locations are:
Thomas
Washburn
Zrox
1
3
Hewes
5
4
Rockrite
8
4
Example: Zeron Shelving
ZROX
Zrox
Arnold
75 ARNOLD
5
Zeron
N
8
50
1
5
8
Hewes 60
HEWES
7
75
Super
Shelf 4
• Network Representation
3
Zeron
S
WASH
BURN
4
4
RockRite 40
Example: Zeron Shelving
• 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 (Zeron N), 4 (Zeron S)
k = 5 (Zrox), 6 (Hewes), 7 (Rockrite)
– Objective Function Defined
Minimize Overall Shipping Costs:
Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37
+ 3x45 + 4x46 + 4x47
Example: Zeron Shelving
• Constraints Defined
Amount Out of Arnold:
Amount Out of Supershelf:
Amount Through Zeron N:
0
Amount Through Zeron S:
0
Amount Into Zrox:
Amount Into Hewes:
Amount Into Rockrite:
x13 + x14 < 75
x23 + x24 < 75
x13 + x23 - x35 - x36 - x37 =
x14 + x24 - x45 - x46 - x47 =
x35 + x45 = 50
x36 + x46 = 60
x37 + x47 = 40
Non-negativity of Variables: xij > 0, for all i and j.
Example: Zeron Shelving

Optimal Solution (from The Management Scientist )
Objective Function Value =
Variable
X13
X14
X23
X24
X35
X36
X37
X45
X46
X47
Value
75.000
0.000
0.000
75.000
50.000
25.000
0.000
0.000
35.000
40.000
1150.000
Reduced Costs
0.000
2.000
4.000
0.000
0.000
0.000
3.000
3.000
0.000
0.000
Example: Zeron Shelving

Optimal Solution
ZROX
Zrox
Arnold
75 ARNOLD
5
75
Zeron
N
8
50
1
5
8
Hewes 60
HEWES
3 4
7
75
Super
Shelf 4
Zeron
S
WASH
BURN
4
RockRite 40
Example: Zeron Shelving

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
Dual Prices
0.000
2.000
-5.000
-6.000
-6.000
-10.000
-10.000
Example: Zeron Shelving

Optimal Solution (continued)
OBJECTIVE COEFFICIENT RANGES
Variable
X13
X14
X23
X24
X35
X36
X37
X45
X46
X47
Lower Limit Current Value Upper Limit
3.000
5.000
7.000
6.000
8.000
No Limit
3.000
7.000
No Limit
No Limit
4.000
6.000
No Limit
1.000
4.000
3.000
5.000
7.000
5.000
8.000
No Limit
0.000
3.000
No Limit
2.000
4.000
6.000
No Limit
4.000
7.000
Example: Zeron Shelving

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
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