Readings
Readings
Chapter 6
Distribution and Network Models
BA 452 Lesson B.2 Transshipment and Shortest Route
1
Overview
Overview
BA 452 Lesson B.2 Transshipment and Shortest Route
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Overview
Transshipment Problems are Transportation Problems extended so that a
shipment may move through intermediate nodes (transshipment nodes) before
reaching a particular destination node.
Transshipment Problems with Transshipment Origins are Transshipment
Problems where goods from one origin may move through other origins before
reaching a destination.
Shortest Route Problems are Transshipment Problems where there is one
origin, one destination, one unit supply, and one unit demand, and where that
unit is indivisible, as in driving through cities to work.
BA 452 Lesson B.2 Transshipment and Shortest Route
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Overview
Tool Summary
 Define decision variable xij = units moving from origin i to
destination j.
 Write origin constraints (with < or =):
n
x
ij
 si
i 1, 2,
,m
Supply
j 1

Write destination constraints (with < or =):
m
x
ij
 dj
j 1, 2,
,n
Demand
i 1

Write transshipment constraints (with < or =):

arcs out
xij 

xij  0
Transhipment nodes
arcs in
BA 452 Lesson B.2 Transshipment and Shortest Route
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Overview
Tool Summary
 Identify implicit assumptions needed to complete a
formulation, such as all agents having an equal value of
time.
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
Transshipment
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
Overview
Transshipment Problems are Transportation Problems
extended so that a shipment may move through
intermediate nodes (transshipment nodes) before reaching
a particular destination node.
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
This is the network representation for a transshipment
problem with two sources, three intermediate nodes, and
two destinations:
3
c13
s1
1
c37
c14
s2
d1
6
c46
c47
4
c15
Supply
c36
Demand
c23
2
Sources
c56
c24
c25
5
c57
7
d2
Destinations
Intermediate Nodes
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
Notation:
xij = number of units shipped from node i to node j
cij = cost per unit of shipping from node i to node j
si = supply at origin node i
dj = demand at destination node j
Min

c ij xij
all arcs
s.t.

xij 
arcs out


Origin nodes i
xij  0
Transhipment nodes
xij  d j
Destination nodes j
arcs in
xij 

arcs out
arcs in


arcs in
xij  si
xij 
arcs out
xij > 0 for all i and j
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
Problem Variations
• 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.
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
Question: 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 up to 75 units to its customers.
Because of long-standing contracts based on past orders, unit costs
from the manufacturers to the suppliers are:
Zeron N Zeron S
Arnold
5
8
Supershelf
7
4
The costs to install the shelving at the various locations are:
Zrox Hewes Rockrite
Zeron N
1
5
8
Zeron S
3
4
4
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
Formulate and solve a transshipment linear programming
problem for Zeron Industries.
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
Answer: Since demands by the three customer firms (Zrox,
Hewes, Rockrite) are fixed, revenue for Zeron Industries is
fixed, and so profit maximization is the same as cost
minimization. There is data on transportation costs, but
there is no data on the cost of ordering from suppliers.
Nevertheless, if the unit cost from each supplier is the
same, say P per unit, then the cost of ordering supply equal
to the fixed demand of 50, 60, and 40 is fixed at 150P.
Hence, to minimize cost, we just have to minimize
transportation cost.
To that end, it may help to draw a network model:
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
75
ARNOLD
Arnold
5
Zeron
N
8
75
4
50
Hewes
HEWES
60
RockRite
40
1
5
8
3
7
Super
Shelf
Zrox
Zeron
WASH
BURN
S
4
4
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment

Next, Define decision variables:
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)
Problem Features:
 There will be 1 variable for
each manufacturer-supplier
pair and each suppliercustomer pair, so
10 variables all together.
 There will be 1 constraint
for each manufacturer, 1 for
each supplier, and 1 for
each customer, so
7 constraints all together.
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment








Define objective function: Minimize total shipping costs.
Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37 + 3x45 + 4x46 + 4x47
Constrain amount out of Arnold:
x13 + x14 < 75
Constrain amount out of Supershelf: x23 + x24 < 75
Constrain amount through Zeron N: x13 + x23 - x35 - x36 - x37 = 0
Constrain amount through Zeron S: x14 + x24 - x45 - x46 - x47 = 0
Constrain amount into Zrox:
x35 + x45 = 50
Constrain amount into Hewes:
x36 + x46 = 60
Constrain amount into Rockrite:
x37 + x47 = 40
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
Indicies: i = 1 (Arnold), 2 (Supershelf)
j = 3 (Zeron N), 4 (Zeron S)
k = 5 (Zrox), 6 (Hewes), 7
(Rockrite)
Minimized shipping costs
Out of Arnold through Zeron N
Out of Supershelf through Zeron S
Through Zeron N into Zrox
Through Zeron S into Hewes
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment
Indicies: i = 1 (Arnold), 2 (Supershelf)
j = 3 (Zeron N), 4 (Zeron S)
k = 5 (Zrox), 6 (Hewes), 7
(Rockrite)
ZROX
75
ARNOLD
Arnold
5
75
Zeron
N
8
75
4
50
Hewes
HEWES
60
RockRite
40
1
5
8
3 4
7
Super
Shelf
Zrox
Zeron
WASH
BURN
S
4
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment with Transshipment Origins
Transshipment with Transshipment
Origins
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment with Transshipment Origins
Overview
Transshipment Problems with Transshipment Origins
seek 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), where goods from
one origin may move through other origins (transshipment
nodes) before reaching a particular destination node.
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment with Transshipment Origins
Question: Index cities
i = 1 (Newbury Park),
i = 2 (Thousand Oaks),
i = 3 (Westlake Hills),
i = 4 (Agoura Hills),
i = 5 (Calabasas).
Suppose you run rental car lots in each city. Newbury Park
has a surplus of 3 cars (it has 3 more cars than it needs),
Westlake Hills has a surplus of 2 cars (it has 2 more cars
than it needs), and Calabasas has a deficit of 5 cars (it
needs 5 more cars than it has).
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment with Transshipment Origins
Suppose you calculate the following costs per car of transporting cars
between the cities:
• transporting between 1 and 2 (that is, either 1 to 2, or 2 to 1) costs $2
• transporting between 1 and 3 costs $3
• transporting between 1 and 4 costs $4
• transporting between 1 and 5 costs $5
• transporting between 2 and 3 costs $2
• transporting between 2 and 4 costs $3
• transporting between 2 and 5 costs $4
• transporting between 3 and 4 costs $2
• transporting between 3 and 5 costs $3
• transporting between 4 and 5 costs $2
How should you move cars between cities? Formulate your rental-car
problem as a linear program, but you need not solve for the optimum.
Tip: Your written answer should define the decision variables, and
formulate the objective and constraints.
BA 452 Lesson B.2 Transshipment and Shortest Route
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Transshipment with Transshipment Origins
Answer:
Define decision variables:
xij = amount of cars moved from City i to City j
Define objective function: Minimize total costs.
Min 2(x12+x21) + 3(x13+x31) + 4(x14+x41) + 5(x15+x51) + 2(x23+x32)
+ 3(x24+x42) + 4(x25+x52) + 2(x34+x43) + 3(x35+x53) + 2(x45+x54)
Constrain cars from City 1:
x12 + x13 + x14 + x15 = 3 + x21 + x31 + x41 + x51
Constrain cars from City 2: x21 + x23 + x24 + x25 = x12 + x32 + x42 + x52
Constrain cars from City 3:
x31 + x32 + x34 + x35 = 2 + x13 + x23 + x43 + x53
Constrain cars from City 4: x41 + x42 + x43 + x45 = x14 + x24 + x34 + x54
Constrain cars from City 5:
x51 + x52 + x53 + x54 = -5 + x15 + x25 + x35 + x45
(Or constraints can be written with < rather than =. It does not matter
since excess supply exactly cancels excess demand.)
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route
Shortest Route
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route
Overview
Shortest Route Problems are Transshipment Problems
where there is one origin, one destination, one unit
supplied, and one unit demanded, and where that unit is
indivisible. Shortest Route Problems find the shortest path
in a network from one node (or set of nodes) to another
node (or set of nodes). The criterion to be minimized in the
shortest-route problem is not limited to distance, but can be
minimum time or cost.
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route
Notation:
xij =
1 if the arc from node i to node j
is on the shortest route
0 otherwise
cij = distance, time, or cost associated
with the arc from node i to node j
Min

cij xij
all arcs
s.t.

1
xij
Origin node i
arcs out

arcs out

xij 

xij  0
Transhipment nodes
arcs in
xij  1
Destination node j
arcs in
xij > 1
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route
Question: Susan Winslow has an important business
meeting in Paducah this evening. She has a
number of alternate routes by which she can travel
from the company headquarters in Lewisburg to
Paducah. The network of alternate routes and their
respective travel time, ticket cost, and transport
mode appear on the next two slides.
If Susan earns a wage of $15 per hour, what route
should she take to minimize the total travel cost?
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route
F
2
5
K
A
L
B
C
1
Lewisburg
D
E
G
3
J
6
I
H
4
Paducah
M
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route
Route
(Arc)
A
B
C
D
E
F
G
H
I
J
K
L
M
Transport
Mode
Train
Plane
Bus
Taxi
Train
Bus
Bus
Taxi
Train
Bus
Taxi
Train
Bus
Time
(hours)
4
1
2
6
3 1/3
3
4 2/3
1
2 1/3
6 1/3
3 1/3
1 1/3
4 2/3
Ticket
Cost
$ 20
$115
$ 10
$ 90
$ 30
$ 15
$ 20
$ 15
$ 15
$ 25
$ 50
$ 10
$ 20
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route
Using the wage of $15 per hour, compute total cost.
Route
A
B
C
D
E
F
G
H
I
J
K
L
M
Transport
Mode
Train
Plane
Bus
Taxi
Train
Bus
Bus
Taxi
Train
Bus
Taxi
Train
Bus
Time
(hours)
4
1
2
6
3 1/3
3
4 2/3
1
2 1/3
6 1/3
3 1/3
1 1/3
4 2/3
Time
Cost
$60
$15
$30
$90
$50
$45
$70
$15
$35
$95
$50
$20
$70
Ticket
Cost
$ 20
$115
$ 10
$ 90
$ 30
$ 15
$ 20
$ 15
$ 15
$ 25
$ 50
$ 10
$ 20
BA 452 Lesson B.2 Transshipment and Shortest Route
Total
Cost
$ 80
$130
$ 40
$180
$ 80
$ 60
$ 90
$ 30
$ 50
$120
$100
$ 30
$ 90
30
Shortest Route


Define indices: Nodes 1 (origin), 2, …, 6 (destination)
Define decision variables:
xij = 1 if the route from node i to node j is on the shortest route
Problem Features:
 There is 1 decision variable
for each possible route.
 There is 1 constraint for
each node.
Cost of route from 1 to 4

Define objective function: Minimize total transportation costs.
Min 80x12 + 40x13 + 80x14 + 130x15 + 180x16 + 60x25 + 100x26 + 30x34 +
90x35 + 120x36 + 30x43 + 50x45 + 90x46 + 60x52 + 90x53 + 50x54 + 30x56
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route

Node flow-conservation constraints:
x12 + x13 + x14 + x15 + x16 = 1 (origin)
– x12 + x25 + x26 – x52 = 0 (node 2)
– x13 + x34 + x35 + x36 – x43 – x53 = 0 (node 3)
– x14 – x34 + x43 + x45 + x46 – x54 = 0 (node 4)
– x15 – x25 – x35 – x45 + x52 + x53 + x54 + x56 = 0 (node 5)
x16 + x26 + x36 + x46 + x56 = 1 (destination)
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route
Minimized
transportation
cost
Variables:
xij =
1 if the arc from node i to node j
is on the shortest route
0 otherwise
Possible arcs: x12, x13, x14, x15, x16, x25, x26, x34,
x35, x36, x43, x45, x46, x52, x53, x54, x56
From Origin to Node 3
From Node 3 to Node 4
From Node 4 to Node 5
From Node 5 to Destination
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route
BA 452 Lesson B.2 Transshipment and Shortest Route
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Shortest Route
Input window
Cost of Arc from Node 1 to Node 3
Output window
Minimized
transportation
costs
BA 452 Lesson B.2 Transshipment and Shortest Route
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BA 452
Quantitative Analysis
End of Lesson B.2
BA 452 Lesson B.2 Transshipment and Shortest Route
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