slides by john loucks st. edward’s university

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


Chapter 6, Part ADistribution and Network Models

Transportation Problem• Network Representation• General LP Formulation

Transshipment Problem• Network Representation• General LP Formulation

3 Slide


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


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



Network Representation

2

c1

1 c12

c13c21

c22c23

d1

d2

d3

s1

s2

Sources Destinations

3

2

1

1

6 Slide


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


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


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.


No modification of LP formulation is necessary.

9 Slide


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.


Solve as a maximization problem.

10 Slide


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?

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Delivery Cost Per Ton

Northwood Westwood Eastwood Plant 1 24 30 40 Plant 2 30 40 42


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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



MIN• SHIPPING COSTS

ST• SUPPLY• DEMAND• NON-NEGATIVE

14 Slide


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


Partial Spreadsheet Showing Optimal Solution


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


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


17 Slide


Partial Sensitivity Report (first half)


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


Partial Sensitivity Report (second half)


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


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




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



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



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



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


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


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


26 Slide



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


27 Slide



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

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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 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


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


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


32 Slide


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


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

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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


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


6

AB

C

DE

F

G

H I

J

K L

M


PaducahLewisburg1

2 5

3

4


37 Slide



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

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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



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



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


End of Chapter 6, Part A

slides by john loucks st. edward’s university

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transportation problemsolve

total shipping costs

unit shipping cost

total demand

xij lij maximum route

xij lij unacceptable