1 1 modeling flows john h. vande vate spring, 2006

11

Modeling Flows

John H. Vande Vate

Spring, 2006

22

New Project Opportunity

• Sponsor: CARE and Shelter First

• Design temporary shelters (tents)

• Design delivery mechanism

• Design logistics network

• Evaluate inventory levels, locations and cost

33

Agenda

• Network Flows– Review definition

– Simple Transportation Model

– Minimum Cost Flow Model

• Adding Realism– Multiple commodities

– Weight, Cube, Linear Cube

– Concave Costs

– Modeling Time

44

Reference

Network Flows: Theory, Algorithms, and Applications (Hardcover)by Ravindra K. Ahuja, Thomas L. Magnanti, James B.

Orlin

55

What Are They?

• Specially Structured Linear Programs

• Each variable appears in at most two constraints– At most one constraint with a +1 coefficient– At most one constraint with a -1 coefficient

• Each variable may be constrained by bounds

• Integer Data => Integer Solutions

66

Transportation Model

• Move Goods from Plants to Warehouses

• Single Commodity!

• Plants have supplies

• Warehouses have demands

• Costs are proportional to volume shipped (this usually isn’t very realistic)

77

Example 2-3 (p 36)

• Single Product

• Two plants with identical production costs

• Two warehouses

• Markets allocated to warehouses for now

• Proportional transportation costs

88

Example 2-3 cont’dSimplified Version of Example 2-3 in Simchi-Levi

Plant1 Plant2 DemandWarehouse1 -$ 4$ 100Warehouse2 5$ 2$ 100Capacity 200 60

Plant1 Plant2 Total To DemandWarehouse1 100 0 100 >= 100Warehouse2 40 60 100 >= 100Total From 140 60

<= <=Capacity 200 60

Plant1 Plant2 TotalWarehouse1 -$ -$ -$ Warehouse2 200$ 120$ 320$ Total 200$ 120$ 320$

Unit Transport Costs

Flows

Costs

99

Transportation Model (AMPL)

• Set Plants;

• Set Warehouses;

• Param Supply {Plants};

• Param Demand {Warehouses};

• Param Cost{Plants, Warehouses};

• Var Flow{Plants, Warehouses}>= 0;

• minimize TotalCost: sum{p in Plants, w in Warehouses} Cost[p,w]*Flow[p,w];

• s.t. WithinSupply{p in Plants}:sum{w in Warehouses} Flow[p, w] <= Supply[p];

• s.t. MeetDemand{w in Warehouses}:sum{p in Plants} Flow[p, w] >= Demand[w];

1010

Minimum Cost Flow• Move Goods from Plants to Markets via

Warehouses

• Single Commodity!

• Plants have supplies

• Markets have demands

• Costs are proportional to volume shipped

1111

Example 2-3 Cont’d

Simplified Version of Example 2-3 in Simchi-Levi

Plant1 Plant2 Market1 Market2 Market3Warehouse1 -$ 4$ Warehouse1 3$ 4$ 5$ Warehouse2 5$ 2$ Warehouse2 2$ 1$ 2$ Capacity (000) 200 60 Demand (000) 50 100 50

Plant1 Plant2 Total To Total From Market1 Market2 Market3Warehouse1 - = - Warehouse1Warehouse2 - = - Warehouse2Total From - - - - - Total to

<= <= >= >= >=Capacity 200 60 50 100 50 Demand

Plant1 Plant2 Total Market1 Market2 Market3Warehouse1 -$ -$ -$ Warehouse1 -$ -$ -$ -$ Warehouse2 -$ -$ -$ Warehouse2 -$ -$ -$ -$ Total -$ -$ -$ -$ -$ -$ -$

Total Transportation Cost ($000)-$

Costs

Flows

Unit Transport Costs

1212

Minimum Cost Flow• Set Locs;

• Set Edges in Locs cross Locs;

• param Cost{Edges};

• param NetSupply{Locs};

• var FlowVol{Edges} >= 0;

• minimize TotalCost:sum{(f,t) in Edges} Cost[f,t]*FlowVol[f,t];

• s.t. PathDefinition{loc in Locs}: sum{(loc, t) in Edges} FlowVol[loc,t]

- sum{(f, loc) in Edges} FlowVol[f, loc] = NetSupply[loc];

1313

Homogenous Product

Must be able to interchange

positions of product anywhere

1414

Another New Project

• Milliken & Co. growing business in Asia looking to explore distribution strategy: Where should it be positioning inventories to serve that region?

1515

Multi-Commodity Flows

• Several single commodity models • You can’t to turn lead into gold

– Conserve flow of each product through each facility

• Joined by common constraints– Often capacity on lanes– Or modeling cost (through capacity)– …

1616

Multiple ProductsShipping Costs

Heavy DC Cust 1 Cust 2Plant H 300$ 500$ 600$ Plant L 100$ 300$ 400$

DC 300$ 400$

Light DC Cust 1 Cust 2Plant H 200$ 600$ 400$ Plant L 50$ 400$ 400$

DC 300$ 400$

FlowsHeavy DC Cust 1 Cust 2 Total Supply

Plant H - 600 DC - -

Total - 0 0Demand - 400 180

Light DC Cust 1 Cust 2 Total Supply Plant L - 600

DC - - Total - 0 0

Demand - 200 300

Transportation Costs ($/Unit)

1717

Conveyance Capacity

• Weight Limits, e.g., 40,000 lbs• Cubic Capacity (53’ trailer)

– 3,970 cubic ft.– Length 52' 6" – Width 99" – Height 110"

• Linear Cube (53’ trailer)– If load is not stackable, just floor space

1818

Conveyance Capacity

Product H Product L Total- - - Limit/Truck

Unit Weight 500 100 0 40,000 Unit Cube 10 20 0 3,900

1919

Multiple Products Weight & Cube

Shipping Costs Conveyances Shipping Costs Flows Conveyances Total CostWeight Cube

From\To DC Cust 1 Cust 2 Weight (Lbs) Cube (cu. ft) From\To DC Cust 1 Cust 2 From\To DC Cust 1 Cust 2Plant H 300$ 500$ 600$ 500 10 Plant H - - - Plant H - - - Plant L 100$ 300$ 400$ 100 20 Plant L - - - Plant L - - -

DC 300$ 400$ DC - - DC - -

Conveyance Capacity 40,000 3,900 Shipping Costs Conveyance to Carry the Weight Conveyance to Carry the CubeFlows From\To DC Cust 1 Cust 2 From\To DC Cust 1 Cust 2Conveyances Plant H - - - Plant H - - - Total Cost Plant L - - - Plant L - - -

DC - - DC - -

Flows Total Cost Shipping Costs Flows Conveyances Total CostHeavy DC Cust 1 Cust 2 Total Supply Conveyances Costs

Plant H - 600 From\To DC Cust 1 Cust 2 From\To DC Cust 1 Cust 2DC - - Plant H - - - Plant H -$ -$ -$

Total - 0 0 Shipping Costs Plant L - - - Plant L -$ -$ -$ Demand - 400 180 Flows DC - - DC -$ -$

Conveyances Total -$ -$ -$ Light DC Cust 1 Cust 2 Total Supply Total Cost Total

Plant L - 600 -$ DC - -

Total - 0 0Demand - 200 300

Cross Dock with Weight & Cube

Transportation Costs ($/Conveyance)

2020

Note

• No Longer a Network Model

• Solutions not integral

• Adding integrality constraints

2121

Multiple Products Weight & Cube

Shipping Costs Conveyances Shipping Costs Flows Conveyances Total CostWeight Cube

From\To DC Cust 1 Cust 2 Weight (Lbs) Cube (cu. ft) From\To DC Cust 1 Cust 2 From\To DC Cust 1 Cust 2Plant H 300$ 500$ 600$ 500 10 Plant H - - - Plant H - - - Plant L 100$ 300$ 400$ 100 20 Plant L - - - Plant L - - -

DC 300$ 400$ DC - - DC - -

Conveyance Capacity 40,000 3,900 Shipping Costs Conveyance to Carry the Weight Conveyance to Carry the CubeFlows From\To DC Cust 1 Cust 2 From\To DC Cust 1 Cust 2Conveyances Plant H - - - Plant H - - - Total Cost Plant L - - - Plant L - - -

DC - - DC - -

Flows Total Cost Shipping Costs Flows Conveyances Total CostHeavy DC Cust 1 Cust 2 Total Supply Conveyances Costs

Plant H - 600 From\To DC Cust 1 Cust 2 From\To DC Cust 1 Cust 2DC - - Plant H - - - Plant H -$ -$ -$

Total - 0 0 Shipping Costs Plant L - - - Plant L -$ -$ -$ Demand - 400 180 Flows DC - - DC -$ -$

Conveyances Total -$ -$ -$ Light DC Cust 1 Cust 2 Total Supply Total Cost Total

Plant L - 600 -$ DC - -

Total - 0 0Demand - 200 300

Cross Dock with Weight & Cube

Transportation Costs ($/Conveyance)

2222

Concave Cost

Shipment Size

Cos

tCost per unit decreasing

Without special constraints, what will solver do?

2323

Modeling Economies of Scale

• Linear Programming– Greedy– Takes the High-Range Unit Cost first!

• Integer Programming

– Add constraints to ensure first things first

– Several Strategies

2424

Convex Combination• Weighted Average

Tot

al C

ost

0

First Break Point

Second Break Point

10 20

$22

$27

Mid Point

What will the costbe?

1/5th of the way

2525

Conclusion

• If the Volume of Activity is a fraction of the way from one breakpoint to the next, the cost will be that same fraction of the way from the cost at the first breakpoint to the cost at the next

• If Volume = 10 + 20(1-)

• Then Cost = 22 + 27(1-)

2626

Idea• Express Volume of Activity as a Weighted

Average of Breakpoints• Express Cost as the same Weighted Average of

Costs at the Breaks• Activity = Min Level 0 + Break 1 1 +

Break 2 2 + Max Level 3

• Cost = Cost at Min Level 0 + Cost at Break 1 1 +

Cost at Break 2 2 + Cost at Max Level 3

• 1 = 0 + 1 + 2 + 3

2727

Does that Do It?• What can go wrong?

Volume of Activity

Tot

al C

ost

0

Minimum Sustainable

Level

Low R

ange

Cos

t/Unit

Mid-Range Cost/Unit High-Range Cost/Unit

First Break Point

Second Break Point

Maximum Operating

Level

X

2828

Role of Integer Variables

• Ensure we express Activity as a combination of two consecutive breakpoints

• var InRegion{1..NBreaks} binary;

Tot

al C

ost

0

Minimum Sustainable

Level

First Break Point

Second Break Point

Maximum Operating

Level

InRegion[1] InRegion[2] InRegion[3]

2929

Constraints• Lambda[2] = 0 unless activity is between

– BreakPoint[1] and BreakPoint[2] (Region[2]) or – BreakPoint[2] and BreakPoint[3] (Region[3])

• Lambda[2] InRegion[2] + InRegion[3];

Tot

al C

ost

Minimum Sustainable

Level

First Break Point

Second Break Point

Maximum Operating

Level

InRegion[1] InRegion[2] InRegion[3]

BreakPoint[0] BreakPoint[1] BreakPoint[2] BreakPoint[3]

3030

We can’t go wrong

Volume of Activity0

Minimum Sustainable

Level

Low R

ange

Cos

t/Unit

Mid-Range Cost/Unit High-Range Cost/Unit

First Break Point

Second Break Point

Maximum Operating

Level

X

3131

Concave CostsFixed Cost Unit Cost From To

1 5 0 1

3 3 1 2

5 2 2 3

8 1 3 4Range Select One Limits Weighting Break Cost at Break Activity Cost0 to 1 (0.00) (0.00) (0.00) 0 1.00$ 2.25 9.50$ 1 to 2 - (0.00) (0.00) 1 6.00$ 2 to 3 1.00 1.00 0.75 2 9.00$ 3 to 4 - 1.00 0.25 3 11.00$

0 - 4 12.00$ Sum 1 1 Minimum 2.25

3232

In AMPL Speak• param NBreaks;• param BreakPoint{0..NBreaks};• param CostAtBreak{0..NBreaks};• var Lambda{0..NBreaks} >= 0;• var Activity;• var Cost;• s.t. DefineCost:• Cost = sum{b in 0..NBreaks} CostAtBreak[b]*Lambda[b];• s.t. DefineActivity:• Activity = sum{b in 0..NBreaks} BreakPoint[b]*Lambda[b];• s.t. ConvexCombination:• 1 = sum{b in 0..NBreaks}Lambda[b];

3333

And Activity in One Region

• InRegion[1] + InRegion[2] + InRegion[3] 1• Why 1?• If it is in Region[2]:

– Lambda[1] InRegion[1] + InRegion[2] = 1– Lambda[2] InRegion[2] + InRegion[3] = 1– Other Lambda’s are 0

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AMPL Speakparam NBreaks;param BreakPoint{0..NBreaks};param CostAtBreak{0..NBreaks};var Lambda{0..NBreaks} >= 0;var Activity;var Cost;s.t. DefineCost:Cost = sum{b in 0..NBreaks} CostAtBreak[b]*Lambda[b];s.t. DefineActivity:Activity = sum{b in 0..NBreaks} BreakPoint[b]*Lambda[b];s.t. ConvexCombination:1 = sum{b in 0..NBreaks}Lambda[b];

3535

What We Added

• var InRegion{1..NBreaks} binary;

• s.t. InOneRegion:

• sum{b in 1..NBreaks} InRegion[b] <= 1;

• s.t. EnforceConsecutive{b in 0..NBreaks-1}:

• Lambda[b] <= InRegion[b] + InRegion[b+1];

• s.t. LastLambda:

– Lambda[NBreaks] <= InRegion[NBreaks];

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Good News!• AMPL offers syntax to “automate” this• Read Chapter 14 of Fourer for details• <<BreakPoint[1], BreakPoint[2]; Slope[1],

Slope[2], Slope[3]>> Variable;

– Slope[1] before BreakPoint[1]– Slope[2] from BreakPoint[1] to BreakPoint[2]– Slope[3] after BreakPoint[2]– Has 0 cost at activity 0

3737

Modeling Time

• Incorporating Schedules into Network Flow Models

• Example– Given several scheduled pick-ups and

deliveries (must pick-up and deliver on-time)

– Question: How many vehicles required?– Assumption: One load on a vehicle at a

time (No shared capacity)

3838

Example

• How many vehicles are required to meet a schedule of departures and returns…

• No shared capacity

Leg Departure Return1 8:00 10:252 9:30 11:453 14:00 16:534 11:31 15:215 8:12 9:526 15:03 17:137 12:24 14:228 13:33 16:439 8:00 10:34

10 10:56 14:25

3939

Network ModelFrom\To LegLeg 1 2 3 4 5 6 7 8 9 10

1 0 0 1 1 0 1 1 1 0 12 0 0 1 0 0 1 1 1 0 03 0 0 0 0 0 0 0 0 0 04 0 0 0 0 0 0 0 0 0 05 0 0 1 1 0 1 1 1 0 16 0 0 0 0 0 0 0 0 0 07 0 0 0 0 0 1 0 0 0 08 0 0 0 0 0 0 0 0 0 09 0 0 1 1 0 1 1 1 0 1

10 0 0 0 0 0 1 0 0 0 0

From\To Leg

Leg 1 2 3 4 5 6 7 8 9 10

Total Immediate Successors

1 0 0 0 0 0 0 0 0 0 1 12 0 0 0 0 0 0 1 0 0 0 13 0 0 0 0 0 0 0 0 0 0 04 0 0 0 0 0 0 0 0 0 0 05 0 0 1 0 0 0 0 0 0 0 16 0 0 0 0 0 0 0 0 0 0 07 0 0 0 0 0 1 0 0 0 0 18 0 0 0 0 0 0 0 0 0 0 09 0 0 0 1 0 0 0 0 0 0 1

10 0 0 0 0 0 0 0 0 0 0 0

Total Immediate

Predecessors 0 0 1 1 0 1 1 0 0 1

Total Routes

Required 5

Assignments of successor legs

Which Legs can be in the same route

4040

How to Construct the Routes?From\To Leg

Leg 1 2 3 4 5 6 7 8 9 10

Total Immediate Successors

1 0 0 0 0 0 0 0 0 0 1 12 0 0 0 0 0 0 1 0 0 0 13 0 0 0 0 0 0 0 0 0 0 04 0 0 0 0 0 0 0 0 0 0 05 0 0 1 0 0 0 0 0 0 0 16 0 0 0 0 0 0 0 0 0 0 07 0 0 0 0 0 1 0 0 0 0 18 0 0 0 0 0 0 0 0 0 0 09 0 0 0 1 0 0 0 0 0 0 1

10 0 0 0 0 0 0 0 0 0 0 0

Total Immediate

Predecessors 0 0 1 1 0 1 1 0 0 1

Total Routes

Required

Assignments of successor legs

Route ending with

3Who precedes

3?

5 does

Route: 5 => 3

Who precedes 5?

4141

How to Construct the Routes?From\To Leg

Leg 1 2 3 4 5 6 7 8 9 10

Total Immediate Successors

1 0 0 0 0 0 0 0 0 0 1 12 0 0 0 0 0 0 1 0 0 0 13 0 0 0 0 0 0 0 0 0 0 04 0 0 0 0 0 0 0 0 0 0 05 0 0 1 0 0 0 0 0 0 0 16 0 0 0 0 0 0 0 0 0 0 07 0 0 0 0 0 1 0 0 0 0 18 0 0 0 0 0 0 0 0 0 0 09 0 0 0 1 0 0 0 0 0 0 1

10 0 0 0 0 0 0 0 0 0 0 0

Total Immediate

Predecessors 0 0 1 1 0 1 1 0 0 1

Total Routes

Required

Assignments of successor legs

Route ending with

6

Who precedes 6?

7 does

Route: 2 => 7 => 6

Who precedes 7?

2 does

Who precedes 2?

4242

2nd Example of Time

• How many vehicles?

• Common: – Schedule provided– No shared capacity

• Different: Not just one terminal

Trip From Departure To Arrival From\To Port1 Port21 Port2 0 Refinery3 12 Refinery1 21 162 Port1 8 Refinery1 29 Refinery2 19 153 Port1 32 Refinery2 51 Refinery3 13 124 Port2 49 Refinery3 61

4343

As a Network Problem

Service RequirementsTrip From Departure To Arrival Deadhead Travel Times

1 Port2 0 Refinery3 12 From\To Port1 Port22 Port1 8 Refinery1 29 Refinery1 21 163 Port1 32 Refinery2 51 Refinery2 19 154 Port2 49 Refinery3 61 Refinery3 13 12

Which Trips can be in the same routeFrom\To TripTrip 1 2 3 4

1 0 0 1 1 These calculations determine whether the row and column are different and 2 0 0 0 1 if the arrival time of the row's trip plus the deadhead travel time from the row trip's destination 3 0 0 0 0 to the column trip's origin is less than or equal to the departure time of the column trip.4 0 0 0 0

Assignments of successor tripsFrom\To Trip

Trip 1 2 3 4

Total Immediate Successors

1 02 03 04 0

Total Immediate

Predecessors 0 0 0 0Total Routes

Required

4

4444

Singapore Electric Generator

Unit Costs Jan Feb Mar Apr. MayProduction 28.00$ 27.00$ 27.80$ 29.00$

Inventory 0.30$ 0.30$ 0.30$ 0.30$

Production Qty 0 0 0 0Production Limits 60 62 64 66

Beginning Inventory 15 -43 -79 -113Delivery Reqmts 58 36 34 59 MinimumEnding Inventory (43) (79) (113) (172) 7

Production Cost -$ -$ -$ -$ Inventory Cost (4.20)$ (18.30)$ (28.80)$ (42.75)$ Total

Total Cost (4.20)$ (18.30)$ (28.80)$ (42.75)$ (94.05)$

Singapore Electric Generator Production

4545

Average Balances

• Assuming Smooth Demands

Averages (Starting + Ending)/2

4646

Inventory

• Balancing Your Checkbook– Previous Balance + Income - Expenses = New Balance

• Modeling Dynamic Inventory– Starting Inv. + Production - Shipments = Ending Inv.

4747

Singapore Electric Generator

Unit Costs Jan Feb Mar Apr. MayProduction 28.00$ 27.00$ 27.80$ 29.00$

Inventory 0.30$ 0.30$ 0.30$ 0.30$

Production Qty 0 0 0 0Production Limits 60 62 64 66

Beginning Inventory 15 -43 -79 -113Delivery Reqmts 58 36 34 59 MinimumEnding Inventory (43) (79) (113) (172) 7

Production Cost -$ -$ -$ -$ Inventory Cost (4.20)$ (18.30)$ (28.80)$ (42.75)$ Total

Total Cost (4.20)$ (18.30)$ (28.80)$ (42.75)$ (94.05)$

Singapore Electric Generator Production

4848

Network Model

• Ending Inv = Calculated Ending Inv

• Prod. Qty <= Production Limits

• FinalInv >= MinimumEndingInv

Bounds

4949

Network Model

• Ending Inv = Calculated Ending Inv

• How Many constraints is this?

• Which constraints does Ending Inv for Jan appear in?

• With what coefficients?

• Which constraints does Production Qty in February appear in?

5050

Summary• Network Flows

– Transportation• Supply & Demand

– Minimum Cost Flows • Supply, Demand &Flow conservation

• Multicommodity Flows– Conserve flow of each commodity

• Weight, Cube & Conveyances– Not a network flow model– Non-integral solutions

• Non-Linear Costs– Requires integer variables

• Modeling Time

5151

Next

• Inventory– Deterministic (predominantly)

• Pipeline

• Cycle

– Stochastic (later)• Safety stock safety