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Solving Fleet Assignment Problem with Multicommodity Network Flow ModelPowerPoint Presentation

Solving Fleet Assignment Problem with Multicommodity Network Flow Model

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Solving Fleet Assignment Problem with Multicommodity Network Flow Model

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Solving Fleet Assignment Problem with Multicommodity Network Flow Model

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Solving Fleet Assignment Problem with Multicommodity Network Flow Model

September 2, 2009

[s1]

[d1]

ci, cost per unit at arc

S1

D1

cSD1

uSD1

cST1

cTD1

Transshipment Node

si, supply at node Si

di, demand at node Di

uST1

uTD1

T1

Supply Nodes

Demand Nodes

Objective Function:

cST2

cTD2

[s2]

[d2]

uST2

uTD2

S2

D2

cSD2

uSD2

ui, capacity at arc

[s11, s12,… s1K]

cik, cost per unit at arc of commodity k

[d11, d12,… d1K]

S1

D1

cSD11, cSD12,… cSD1K

uSD11, uSD12,… uSD1K

cST11…

cTD11…

Transshipment Node

sik, supply at node Si of commodity k

dik, demand at node Di of commodity k

uST11…

uTD1…1

T1

Supply Nodes

Demand Nodes

Objective Function:

cST21…

cTD21…

[s21, s22,… s2K]

[d21, d22,… d2K]

uST21…

uTD21…

S2

D2

cSD21…

uSD21…

uik, capacity at arcarc of commodity k

- In multicommodity network flow problem, each arc has a finite capacity that applies to the sum of all commodities flowing over.
- Also, each node must receive a particular amount of each type of commodity.
- eg. A metal supplier ships aluminum bars, stainless steel rings, steel beams, etc. using a single limited capacity fleet of trucks.
- In general form, the multicommodity network problem is defined as
- Subject to:
- For each commodity k and node t:
- For each link i, j:

Dik = demand for commodity k at node i, with negative values denoting supply;

Cijk= cost per unit of shipping commodity k from node i to node j;

Uij= capacity of the link from node i to node j;

Xijk= amount of commodity k shipped from node i to node j.

- Indices:
- k, index for fleet types
- v, index for nodes
- i, index for arcs
- h, index for airports
- Input Parameters:
- K, number of fleet types
- V, set of all nodes
- I, set of all arcs
- IF, set of flight arcs
- nk, number of available aircraft for fleet k
- H, set of all airports
- sh, the node associated with airport h and the beginning of the day
- th, the node associated with airport h and the end of the day
- cki, cost of assigning fleet k to flight arc I
- Decision Variable:
- xki, indicating whether fleet type k is assigned to arc i

Decision Variable

For the objective function:

subject to

Cost

F221

F105

F223

F259

F228

F230

F412

F238

GC2400

CHI

8:00

9:00

11:00

14:00

14:23

15:21

19:59

21:21

GC1100

GC1423

GC1521

GC1959

GC0800

GC0900

GC1400

F293

F766

F221

F228

F238

F223

F274

F230

GD2400

DEN

9:34

10:39

11:00

11:26

12:00

14:00

18:00

19:12

GD1100

GD0934

GD1039

GD1126

GD1200

GD1420

GD1800

F274

F105

F412

F293

F766

F259

GL2400

LAX

8:00

13:14

14:00

15:10

16:00

16:09

GL1400

GL1314

GL0800

GL1510

GL1600

- Diagonal lines (with the connection to its partner) constitute flight variables, Fxxx, where xxx represents flight number.
- At each departure and arrival, including the end of the day (midnight), we suppose that there are decision variables of the numbers of aircraft on the ground, G(C,D, or L)hhmm,where hhmm represents time instance.

- Assumes that there are two types of aircraft , aircraft A and B, and that all profit contributions for each combination of aircraft and flight are known, we get
- Thus, the C-plex code for the objective function of this example is:
- For conservation of flow constraints, each instant of either an arrival or a departure constitutes a node, so:
- (no. of aircraft on ground at this city at this instant) + (arrivals at this instant)
- = (no. of departures from this city at this instant) + (no. of aircraft on the ground after this instant).

maximize 105*F221A + 121*F221B + 109*F223A + 108*F223B + 110*F274A + 115*F274B + 130*F105A + 140*F105B + 106*F228A + 122*F228B + 112*F230A +

115*F230B + 132*F259A + 129*F259B + 115*F293A + 123*F293B + 133*F412A +

135*F412B + 108*F766A + 117*F766B + 116*F238A + 124*F238B;

- The previous statement gives 22 constraints and 33 variables for one type of aircraft. However, the number of constraints and variables can be reduced if
- a) Each arrival is delayed until the first departure after that arrival
- b) Each departure is advanced (made earlier) to the most recent departure just after an arrival.

GC2400

GC1400

GC2400

2

1

F221

F105

F223

F259

F228

F230

F412

F238

CHI

4

GD2400

8:00

9:00

11:00

14:00

14:23

15:21

19:59

21:21

GD1100

GD1800

3

F293

F766

F221

F228

F238

F223

F274

F230

5

GD2400

DEN

9:34

10:39

11:00

11:26

12:00

14:00

18:00

19:12

7

GL1600

GL2400

GL1400

GL2400

GL0800

F274

F105

F412

F293

F766

F259

6

9

8

LAX

8:00

13:14

14:00

15:10

16:00

16:09

- The reduced flow conservation constraints can be described in C-plex as
- T

/*Conservation of flow constraints for type A aircraft*/

/*Chicago at 8am, sources - uses = 0*/

-F221A - F223A - F105A - F259A - GC1400A + GC2400A == 0;

/*Chicago at midnight*/

F228A + F230A + F412A + F238A + GC1400A - GC2400A == 0;

/*Denver at 11 AM*/

F221A + F223A - F228A - GD1100A + GD2400A == 0;

/*Denver at high noon*/

F274A - F230A - F293A - F238A + GD1100A - GD1800A == 0;

/*Denver at midnight*/

F766A - GD2400A + GD1800A == 0;

/*LA at 8 AM*/

- F274A - GL0800A + GL2400A == 0;

/*LA at 1400*/

F105A - F412A + GL0800A - GL1400A == 0;

/*LA at 1600*/

F293A - F766A + GL1400A - GL1600A == 0;

/*LA at midnight*/

F259A - GL2400A + GL1600A == 0;

Must be repeated for type B aircraft

- Then, in order to guarantee that each flight will be covered by at most one plane, we impose the following constraints:
- which is converted to C-plex code as

F221A + F221B <= 1;

F223A + F223B <= 1;

F274A + F274B <= 1;

F105A + F105B <= 1;

F228A + F228B <= 1;

F230A + F230B <= 1;

F259A + F259B <= 1;

F293A + F293B <= 1;

F412A + F412B <= 1;

F766A + F766B <= 1;

F238A + F238B <= 1;

- If we set a limit on number of aircraft type B as 2 aircrafts, then the constraint
- can be converted to C-plex code as
- Finally, the 0-1 constraint on each flight arc can be carried out by declaring the flight decision variables as Boolean type:

GC2400B + GD2400B + GL2400B <= 2;

dvar boolean F221A;

dvar boolean F221B;

dvar boolean F223A;

dvar boolean F223B;

.

.

.

- Running the C-plex code with ILOG, the solution is obtained as

// solution (optimal) with objective 1325

F223A = 1; F274A = 1; F105A = 1; F230A = 1; F259A = 1; F412A = 1; F238A = 1;

F221B = 1; F228B = 1; F293B = 1; F766B = 1;

GC2400A = 3; GD1100A = 1; GL2400A = 1; GC2400B = 1; GD1100B = 1; GD2400B = 1;