transport student

7/29/2019 Transport Student

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Slide 2008 Thomson South-Western. All Rights Reserved

Distribution and Network Models

Transportation ProblemNetwork RepresentationGeneral LP Formulation

Transshipment Problem

Network Representation

General LP Formulation

Shortest route method

Network RepresentationGeneral LP Formulation


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Transportation, Assignment, andTransshipment 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 thearcs and/or nodes.

Transportation, assignment, transshipment,

shortest-route, and maximal flow problems of thischapter as well as the minimal spanning tree andPERT/CPM problems (in others chapter) are allexamples of network problems.


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Transportation, Assignment, andTransshipment Problems

Each of the five models of this chapter can be

formulated as linear programs and solved bygeneral purpose linear programming codes.

For each of the five models, if the right-hand sideof the linear programming formulations are all

integers, the optimal solution will be in terms ofinteger values for the decision variables.

However, there are many computer packages(including The Management Scientist) that contain

separate computer codes for these models whichtake advantage of their network structure.


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4


Transportation Problem

The transportation problem seeks to minimize the

total shipping costs of transporting goods from morigins (each with a supply si) to n destinations(each with a demand dj), when the unit shippingcost from an origin, i, to a destination,j, is cij.

The network representation for a transportationproblem with two sources and three destinations isgiven on the next slide.


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5




2

c11

c12

c13

c21

c22

c23

d1

d2

d3

s1

s2

Sources Destinations

3

2

1

1


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Linear Programming Formulation

Using the notation:

xij = number of units shipped from

origin i to destinationj

cij= cost per unit of shipping fromorigin i to destinationj

si = supply or capacity in units at origin i

dj = demand in units at destinationj

continued


7/337Slide 2008 Thomson South-Western. All Rights Reserved


Linear Programming Formulation (continued)

1 1

Min

m n

ij iji j

c x

1

1,2, , Supplyn

ij ij

x s i m

1

1,2, , Demandm

ij ji

x d j n

xij > 0 for all i andj



LP Formulation Special Cases

The objective is maximizing profit or revenue:

Minimum shipping guarantee from i toj:

xij > Lij

Maximum route capacity from i toj:

xij < LijUnacceptable route:

Remove the corresponding decision variable.


Solve as a maximization problem.




Transshipment problems are transportation problems

in which a shipment may move through intermediatenodes (transshipment nodes)before reaching aparticular destination node.

Transshipment problems can be converted to largertransportation problems and solved by a specialtransportation program.

Transshipment problems can also be solved bygeneral 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.





2

3

4

5

6

7

1

c13

c14

c23

c24c25

c15

s1

c36

c37

c46c47

c56

c57

d1

d2

Intermediate Nodes

Sources Destinations

s2

DemandSupply





Using the notation:

xij = number of units shipped from node i to nodej

cij = cost per unit of shipping from node i to nodej

si= supply at origin node idj= demand at destination node j

continued




all arcs

Min ij ijc x

arcs out arcs in

s.t. Origin nodesij ij ix x s i

xij > 0 for all i andj

arcs out arcs in

0 Transhipment nodesij ijx x

arcs in arcs out

Destination nodesij ij jx x d j


continued




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 are

identical to those required for the special cases in

the transportation problem.


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14Slide 2008 Thomson South-Western. All Rights Reserved

The Northside and Southside facilities

of Zeron Industries supply three firms(Zrox, Hewes, Rockrite) with customizedshelving for its offices. They both ordershelving 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 forRockrite. Both Arnold and Supershelf can

supply at most 75 units to its customers.Additional data is shown on the next

slide.

Transshipment Problem: Example


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Because of long standing contracts based on

past orders, unit costs from the manufacturers to thesuppliers are:

Zeron N Zeron SArnold 5 8

Supershelf 7 4

The costs to install the shelving at the variouslocations are:

Zrox Hewes RockriteThomas 1 5 8

Washburn 3 4 4



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ARNOLD

WASH

BURN

ZROX

HEWES

75

75

50

60

40

5

8

7

4

15

8

3

44

Arnold

SuperShelf

Hewes

Zrox

Zeron

N

ZeronS

Rock-Rite



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Decision Variables Defined

xij = amount shipped from manufacturer i to supplierj

xjk = amount shipped from supplierj 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



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Constraints Defined

Amount Out of Arnold: x13 + x14 < 75

Amount Out of Supershelf: x23 + x24 < 75

Amount Through Zeron N: x13 + x23 - x35 - x36 - x37 = 0

Amount Through Zeron S: x14

+ x24

- x45

- x46

- x47

= 0

Amount Into Zrox: x35 + x45 = 50

Amount Into Hewes: x36 + x46 = 60

Amount Into Rockrite: x37 + x47 = 40

Non-negativity of Variables: xij > 0, for all i andj.



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The Management Scientist Solution

Objective Function Value = 1150.000

Variable Value Reduced CostsX13 75.000 0.000

X14 0.000 2.000X23 0.000 4.000X24 75.000 0.000X35 50.000 0.000X36 25.000 0.000

X37 0.000 3.000X45 0.000 3.000X46 35.000 0.000X47 40.000 0.000



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Solution

ARNOLD

WASH

BURN

ZROX

HEWES

75

75

50

60

40

5

8

7

4

1

58

3 4

4

Arnold

SuperShelf

Hewes

Zrox

Zeron

N

ZeronS

Rock-Rite

75



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The Management Scientist Solution (continued)

Constraint Slack/Surplus Dual Prices

1 0.000 0.000

2 0.000 2.000

3 0.000 -5.0004 0.000 -6.000

5 0.000 -6.000

6 0.000 -10.000

7 0.000 -10.000



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OBJECTIVE COEFFICIENT RANGES

Variable Lower Limit Current Value Upper LimitX13 3.000 5.000 7.000

X14 6.000 8.000 No LimitX23 3.000 7.000 No LimitX24 No Limit 4.000 6.000X35 No Limit 1.000 4.000X36 3.000 5.000 7.000

X37 5.000 8.000 No LimitX45 0.000 3.000 No LimitX46 2.000 4.000 6.000X47 No Limit 4.000 7.000



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RIGHT HAND SIDE RANGES

Constraint Lower Limit Current Value Upper Limit

1 75.000 75.000 No Limit

2 75.000 75.000 100.0003 -75.000 0.000 0.000

4 -25.000 0.000 0.000

5 0.000 50.000 50.000

6 35.000 60.000 60.0007 15.000 40.000 40.000



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Shortest-Route Problem

The shortest-route problem is concerned with

finding the shortest path in a network from onenode (or set of nodes) to another node (or set ofnodes).

If all arcs in the network have nonnegative valuesthen a labeling algorithm can be used to find theshortest paths from a particular node to all othernodes in the network.

The criterion to be minimized in the shortest-routeproblem is not limited to distance even though the

term "shortest" is used in describing the procedure.Other criteria include time and cost. (Neither timenor cost are necessarily linearly related to distance.)


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Using the notation:

xij = 1 if the arc from node i to nodej

is on the shortest route

0 otherwise

cij= distance, time, or cost associated

with the arc from node i to nodej

continued



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all arcs

Min ij ijc x

arcs out

s.t. 1 Origin nodeijx i

arcs out arcs in

0 Transhipment nodesij ijx x

arcs in

1 Destination nodeijx j




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Susan Winslow has an important business meeting

in 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 andtheir 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


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6

A

B C

D

E

F

G

H I

J

K L

M


PaducahLewisburg

1

2 5

3

4

Network Model


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Transport Time Ticket

Route Mode (hours) CostA Train 4 $ 20B Plane 1 $115C Bus 2 $ 10D Taxi 6 $ 90

E Train 3 1/3 $ 30F Bus 3 $ 15G Bus 4 2/3 $ 20H Taxi 1 $ 15I Train 2 1/3 $ 15

J Bus 6 1/3 $ 25K Taxi 3 1/3 $ 50L Train 1 1/3 $ 10M Bus 4 2/3 $ 20


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Transport Time Time Ticket Total

Route Mode (hours) Cost Cost CostA Train 4 $60 $ 20 $ 80B Plane 1 $15 $115 $130C Bus 2 $30 $ 10 $ 40D Taxi 6 $90 $ 90 $180

E Train 3 1/3 $50 $ 30 $ 80F Bus 3 $45 $ 15 $ 60G Bus 4 2/3 $70 $ 20 $ 90H Taxi 1 $15 $ 15 $ 30I Train 2 1/3 $35 $ 15 $ 50

J Bus 6 1/3 $95 $ 25 $120K Taxi 3 1/3 $50 $ 50 $100L Train 1 1/3 $20 $ 10 $ 30M Bus 4 2/3 $70 $ 20 $ 90


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LP Formulation

Objective FunctionMin 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 + x26x52 = 0 (node 2)

x13 + x34 + x35 + x36

x43

x53 = 0 (node 3) x14x34 + x43 + x45 + x46x54 = 0 (node 4)

x15 x25x35x45 + x52 + x53 + x54 + x56 = 0 (node 5)

x16 + x26 + x36 + x46 + x56 = 1 (destination)


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Solution Summary

Minimum total cost = $150

x12 = 0 x25 = 0 x34 = 1 x43 = 0 x52 = 0

x13 = 1 x26 = 0 x35 = 0 x45 = 1 x53 = 0x14 = 0 x36 = 0 x46 = 0 x54 = 0

x15 = 0 x56 = 1

x16 = 0


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33Slid 2008 Th S th W t All Ri ht R d

transport student

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