slides by john loucks st. edward’s university
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1 Slide
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Slides by
JohnLoucks
St. Edward’sUniversity
2 Slide
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 6, Part ADistribution and Network Models
Transportation Problem• Network Representation• General LP Formulation
Transshipment Problem• Network Representation• General LP Formulation
3 Slide
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Transportation, Assignment, and Transshipment Problems
Network model • 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.
Types of network models• Transportation• assignment• Transshipment• shortest-route• maximal flow problems
4 Slide
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Transportation Problem
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.
Network representation• Graphical depiction of the problem• Consists of
• Nodes (dots)• Arcs (lines)
5 Slide
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Transportation Problem
Network Representation
2
c1
1 c12
c13c21
c22c23
d1
d2
d3
s1
s2
Sources Destinations
3
2
1
1
6 Slide
© 2011 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Transportation Problem Linear Programming Formulation Using the notation: xij = number of units shipped from origin i to destination j cij = cost per unit of shipping from origin i to destination j si = supply or capacity in units at origin i dj = demand in units at destination j
i = number assigned to each origin j = number assigned to each destination
7 Slide
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Transportation Problem Linear Programming Formulation (continued)
1 1Min
m n
ij iji j
c x
1 1,2, , Supply
n
ij ij
x s i m
1 1,2, , Demand
m
ij ji
x d j n
xij > 0 for all i and j
8 Slide
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LP Formulation Special Cases• Total supply exceeds total demand:
• Total demand exceeds total supply: Add a dummy origin with supply equal to the shortage amount. Assign a zero shipping cost per unit. The amount “shipped” from the dummy origin (in the solution) will not actually be shipped. *
* Since there is not an example of this in the book we will do one next class.
Assign a zero shipping cost per unit
• Maximum route capacity from i to j: xij < Li
Remove the corresponding decision variable.
Transportation Problem
No modification of LP formulation is necessary.
9 Slide
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LP Formulation Special Cases (continued)• The objective is maximizing profit or
revenue:
• 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.
Transportation Problem
Solve as a maximization problem.
10 Slide
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Transportation Problem: Example #1
Acme Block Company has orders for 80 tons ofconcrete blocks at three suburban locations as follows: Northwood -- 25 tons, Westwood -- 45 tons, andEastwood -- 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 fillthe above orders?
11 Slide
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Delivery Cost Per Ton
Northwood Westwood Eastwood Plant 1 24 30 40 Plant 2 30 40 42
Transportation Problem: Example #1
12 Slide
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NorthWood
WestWood
EastWood
Plant2
Plant1
25
45
10
Demand (t)Dest.
50
50
Supply (t) Orig.
24
3040
30
42
40
Costs
13 Slide
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Transportation Problem: Example #1
MIN• SHIPPING COSTS
ST• SUPPLY• DEMAND• NON-NEGATIVE
14 Slide
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Transportation Problem: Example #1 LET
xij = number of tons shipped from origin i to destination j
MIN• 24x11 + 30x12 + 40x13 + (Shipping cost from
P1)30x21 + 40x22 + 42x23 (Shipping cost from P2)
ST• x11 + x12 + x13 <= 50 (Supply of Plant 1)• x21 + x22 + x23 <= 50 (Supply of Plant 2)• x11 + x21 = 25 (Northwood Demand)• x12 + x22 = 45 (Westwood Demand)• x13 + x23 = 10 (Eastwood Demand)• Xij >= 0 (Nonnegative)
15 Slide
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Partial Spreadsheet Showing Optimal Solution
Transportation Problem: Example #1
A B C D E F G10 X11 X12 X13 X21 X22 X2311 Dec.Var.Values 5 45 0 20 0 1012 Minimized Total Shipping Cost 24901314 LHS RHS15 50 <= 5016 30 <= 5017 25 = 2518 45 = 4519 10 = 10Eastwood Demand
Westwood DemandNorthwood Demand
ConstraintsPlant 1 CapacityPlant 2 Capacity
16 Slide
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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 = $2,490
Transportation Problem: Example #1
17 Slide
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Partial Sensitivity Report (first half)
Transportation Problem: Example #1
Adjustable CellsFinal Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease$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$12 X23 10.000 0.000 42 4 1E+30
Adjustable CellsFinal Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease$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$12 X23 10.000 0.000 42 4 1E+30
- How much would the shipping cost from plant 1 to dest. 1 have to increase before a new optimal solution is met
- What about from plant 1 to dest 3?
18 Slide
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Partial Sensitivity Report (second half)
Transportation Problem: Example #1
ConstraintsFinal Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease$E$17 P2.Cap 30.0 0.0 50 1E+30 20$E$18 N.Dem 25.0 30.0 25 20 20$E$19 W.Dem 45.0 36.0 45 5 20$E$20 E.Dem 10.0 42.0 10 20 10$E$16 P1.Cap 50.0 -6.0 50 20 5
ConstraintsFinal Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease$E$17 P2.Cap 30.0 0.0 50 1E+30 20$E$18 N.Dem 25.0 30.0 25 20 20$E$19 W.Dem 45.0 36.0 45 5 20$E$20 E.Dem 10.0 42.0 10 20 10$E$16 P1.Cap 50.0 -6.0 50 20 5
- How much unused capacity is there at plant 1? plant 2?
19 Slide
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Transshipment Problem
Transshipment problems • transportation problems in which a shipment
may move through intermediate 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.
20 Slide
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Transshipment Problem
Network Representation
2
3
4
5
6
7
1c13
c14
c23
c24c25
c15
s1
c36
c37
c46c47
c56
c57
d1
d2
Intermediate NodesSources Destinationss2
DemandSupply
21 Slide
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Transshipment Problem
Linear Programming Formulation
Using the 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 i = number assigned to each origin
j = number assigned to each destination
* In this case, each transshipment node is both an origin and a destination!
22 Slide
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Transshipment Problem
all arcsMin ij ijc x
arcs out arcs ins.t. Origin nodes ij ij ix x s i
xij > 0 for all i and j
arcs out arcs in0 Transhipment nodesij ijx x
arcs in arcs out Destination nodes ij ij jx x d j
Linear Programming Formulation (continued)
continued
23 Slide
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Transshipment Problem
LP Formulation Special Cases• Total supply not equal to total demand• Maximization objective function• Route capacities or route minimums• Unacceptable routesThe LP model modifications required here areidentical to those required for the special
cases inthe transportation problem.
24 Slide
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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.
Transshipment Problem: Example
25 Slide
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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 Thomas 1 5 8
Washburn 3 4 4
Transshipment Problem: Example
26 Slide
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Network Representation
ARNOLD
WASHBURN
ZROX
HEWES
75
75
50
60
40
5
8
7
4
15
8
34
4
Arnold
SuperShelf
Hewes
Zrox
ZeronN
ZeronS
Rock-Rite
Transshipment Problem: Example
27 Slide
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Transshipment Problem: Example
How to model transhipment nodes• Flow out equal to flow in• (Unless a transshipment node has some sort
of demand)
Flow out = Flow in Flow out – flow in = 0 OR Flow in – flow out = 0
28 Slide
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LETxij = 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)
Transshipment Problem: Example
Transshipment Problem: Example Cont’d MIN
5x13 + 8x14 + 7x23 + 4x24 + (From supply to xship)1x35 + 5x36 + 8x37 + 3x45 + 4x46 + 4x47 (From xship to demand)
SUBJECT TOAmount Out of Arnold: x13 + x14 < 75Amount Out of Supershelf: x23 + x24 < 75Amount Through Zeron N: x13 + x23 - x35 - x36 - x37 = 0Amount Through Zeron S: x14 + x24 - x45 - x46 - x47 = 0Amount Into Zrox: x35 + x45 = 50Amount Into Hewes: x36 + x46 = 60Amount Into Rockrite: x37 + x47 = 40Non-negativity of Variables: xij > 0, for all i and j.
Objective Function Value = 1150.000
Variable Value Reduced Costs
X13 75.000 0.000
X14 0.000 2.000
X23 0.000 4.000
X24 75.000 0.000
X35 50.000 0.000
X36 25.000 0.000
X37 0.000 3.000
X45 0.000 3.000
X46 35.000 0.000
X47 40.000 0.000
Transshipment Problem: Example
How much would the cost to ship from node 1 to 4 have to be reducedIn order for this route to be a factor in the solution?
31 Slide
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Solution
ARNOLD
WASHBURN
ZROX
HEWES
75
75
50
60
40
5
8
7
4
15
8
3 4
4
Arnold
SuperShelf
Hewes
Zrox
ZeronN
ZeronS
Rock-Rite
75
75
50
25
35
40
Transshipment Problem: Example
32 Slide
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Shortest-Route Problem Shortest-route problem
• finding the shortest path in a network from one node (or set of nodes) to another node (or set of nodes).
Can be used to find minimized cost or time as well as distance
Flow through a node is represented by “1”• that mea
33 Slide
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Linear Programming Formulation Using the 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 continued
Shortest-Route Problem
34 Slide
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all arcsMin ij ijc x
arcs outs.t. 1 Origin node ijx i
arcs out arcs in0 Transhipment nodesij ijx x
arcs in1 Destination node ijx j
Linear Programming Formulation (continued)
Shortest-Route Problem
35 Slide
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Susan Winslow has an important business meetingin Paducah this evening. She has a number of alternateroutes 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 routeshould she take to minimize the total travel cost?
Example: Shortest Route
36 Slide
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6
AB
C
DE
F
G
H I
J
K L
M
Example: Shortest Route
PaducahLewisburg1
2 5
3
4
Network Representation
37 Slide
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Example: Shortest Route
Transport Time TicketRoute Mode (hours) Cost A Train 4
$ 20 B Plane 1
$115 C Bus 2 $ 10 D Taxi 6
$ 90 E Train 3 1/3
$ 30 F Bus 3
$ 15 G Bus 4 2/3
$ 20 H Taxi 1
$ 15 I Train 2 1/3 $ 15 J Bus 6 1/3
$ 25 K Taxi 3 1/3 $ 50 L Train 1 1/3 $ 10 M Bus 4 2/3
$ 20
38 Slide
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Example: Shortest Route
Transport Time Time Ticket TotalRoute Mode (hours) Cost Cost Cost A Train 4 $60 $ 20 $ 80 B Plane 1 $15 $115 $130 C Bus 2 $30 $ 10 $ 40 D Taxi 6 $90 $ 90 $180 E Train 3 1/3 $50 $ 30 $ 80 F Bus 3 $45 $ 15 $ 60 G Bus 4 2/3 $70 $ 20 $ 90 H Taxi 1 $15 $ 15 $ 30 I Train 2 1/3 $35 $ 15 $ 50 J Bus 6 1/3 $95 $ 25 $120 K Taxi 3 1/3 $50 $ 50 $100 L Train 1 1/3 $20 $ 10 $ 30 M Bus 4 2/3 $70 $ 20 $ 90
39 Slide
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Example: Shortest Route
LP Formulation• Objective Function
Min 80x12 + 40x13 + 80x14 + 130x15 + 180x16 + 60x25
+ 100x26 + 30x34 + 90x35 + 120x36 + 30x43 + 50x45
+ 90x46 + 60x52 + 90x53 + 50x54 + 30x56 • 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)
40 Slide
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Example: Shortest Route
Solution Summary
Minimum total cost = $150x12 = 0 x25 = 0 x34 = 1 x43 =
0 x52 = 0 x13 = 1 x26 = 0 x35 = 0 x45 =
1 x53 = 0 x14 = 0 x36 = 0 x46 = 0
x54 = 0 x15 = 0 x56 = 1 x16 = 0
41 Slide
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End of Chapter 6, Part A
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