The Minimum Cost Network Flow (MCNF) Problem

• Extremely useful model in IEOR

• Important Special Cases of the MCNF Problem– Transportation and Assignment Problems

– Maximum Flow Problem

– Minimum Cut Problem

– Shortest Path Problem

• Network Structure– BFS for MCNF LPs always have integer values !!!

– Problems can be formulated graphically

General Form of the MCNF Problem

• Defined on a network N = (V,A)

• V is a set of vertices (nodes)– Each node i has an assoicated value bi

• bi < 0 => node i is a demand node with a demand of -bi

• bi = 0 => node i is a transshipment node

• bi > 0 => node i is a supply node with a supply of bi

• A is a set of arcs– arc (i,j) from node i to node j has

• cost cij per unit of flow on arc (i,j)

• upper bound on flow of uij (capacity)

• lower bound on flow of lij (usually 0)

General Form of the MCNF Problem Continued

• A flow is feasible if– Flow on all arcs is within the allowable bounds

– Flow is balanced (conserved) • total flow going out of node i - total flow coming into node i = b i

• We want to find a minimum cost feasible flow

• LP Formulation– Let xij be the units of flow on arc (i,j)

Aji

Vi

uxlbxx

xc

ijijij

iAijjji

Ajijij

Ajiijij

),(

st

min

),|(),|(

),(

Example: Medley Relay TeamFreestyle Butterfly Backstroke Breaststroke

John 5.88 6.91 9.10 7.36Paul 9.30 6.53 7.62 3.28George 8.46 4.16 2.62 6.23Ringo 5.27 7.01 2.47 7.56

•Supply Nodes 1, 2, 3 and 4 for John, Paul, George and Ringo•Each has a supply of one swimmer, so bi = 1

•Demand Nodes 5, 6, 7 and 8 for Freestyle, Butterfly, Backstroke and Breaststroke

• Each even needs a swimmer, so bi = -1 for i=6,7,8,9•Arcs from each supply node to each demand node

•cij = swimmer i’s time in event j•uij = 1, only one swimmer can be in any event•lij = 0

John

Paul

George

Ringo

Freestyle

Butterfly

Backstroke

Breaststorke

1

1

1

1

-1

-1

-1

-1

5.88

9.30

8.46

5.27

6.91

9,10

7.36

LP Formulation

• Let xij = 1 if swimmer i swims event j and 0 otherwise

MIN 5.88 X15 + 6.91 X16 + 9.1 X17 + 7.36 X18 + 9.3 X25 + 6.53 X26 + 7.62 X27 + 3.28 X28 + 8.46 X35 + 4.16 X36 + 2.62 X37 + 6.23 X38 + 5.27 X45 + 7.01 X46 + 2.47 X47 + 7.56 X48 SUBJECT TO 2) X15 + X16 + X17 + X18 = 1 ! John = swimmer 1 3) X25 + X26 + X27 + X28 = 1 ! Paul = swimmer 2 4) X35 + X36 + X37 + X38 = 1 ! George = swimmer 3 5) X45 + X46 + X47 + X48 = 1 ! Ringo = swimmer 3 6) - X15 - X25 - X35 - X45 = - 1 ! Freestyle = event 1 7) - X16 - X26 - X36 - X46 = - 1 ! Butterfly = event 2 8) - X17 - X27 - X37 - X47 = - 1 ! Backstroke = event 3 9) - X18 - X28 - X38 - X48 = - 1 ! Breaststroke = event 4 0 <= Xij <= 1 for all (i,j)

LP Solution

• X15 = X28 = X36 = X47 = 1 all other variables = 0

• John swims Freestyle, Paul swims Breaststroke, George swims Butterfly and Ring swims Backstroke.

• The total time is 15.79 minutes.

• Observe that the LP solution is integer valued.

Example 2 (From Bazarra and Jarvis)

• Transport 20,000,000 barrels of oil from Dhahran, Saudi Arabia to Rotterdam (4 Mil.), Marseilles (12 Mil) and Naples (4 Mil.) in Europe

• Routes– Ship oil around Africa to

• Rotterdam $1.20/barrel

• Marseilles $1.40/barrel

• Naples $1.40/barrel

– Dhahran -> Suez -> Suez Canal -> Port Said• $0.30/barrel from Dhahran to Suez

• $0.20/barrel through Suez Canal

• Port Said to

– Rotterdam $0.25

– Marseilles $0.20

– Naples $0.15

Example 2 Continued

– Dhahran to Suez then pipeline to Alexandria• $0.15/barrel through pipeline

• Alexandria to

– Rotterdam $0.22

– Marseilles $0.20

– Naples $0.15

• 30% of oil in Dhahran shipped in large tankers that can’t go through the Suez Canal

• Pipeline from Suez to Alexandria has a capacity of 10 million barrels of oil

• Formulate as a MCNF Problem

Network Formulation

S

A

M

N

D

PS

R

Network Formulation Continued

• Supply Node– Dharhan has a supply of 20 M

• Demand Nodes– Rotterdam, Marseilles and Naples have demands of 4 M, 12 M

and 4M, respectively

• Transshipment Nodes– Suez, Alexandria and Port Said

Arcs

i j cij i j cij i j cij

D R 1.2 S PS 0.2 PS N 0.15D M 1.4 S A 0.15 A R 0.22D N 1.4 PS R 0.25 A M 0.20D S 0.3 PS M 0.20 A N 0.15

Lower bound = 0 for all arcs

Upper bound = infinity for all arcs except(S,A) = 10 million(S,PS) = 14 million

LP Formulation

minimize 1.2 XDR + 1.4 XDM + 1.4 XDN + 0.3 XDS +0.2 XSPS + 0.15 XSA +0.25 XPSR + 0.2 XPSM + 0.15 XPSN +0.25 XAR + 0.2 XAM + 0.15 XANsubject toXDR + XDM + XDN + XDS = 20000000 ! DXSPS + XSA - XDS = 0 ! SXPSR + XPSM + XPSN - XSPS = 0 ! PSXAR + XAM + XAN - XSA = 0 ! PA-XPSR - XAR - XDR = - 4000000 ! R-XPSM - XAM - XDM = - 12000000 ! M-XPSN - XAN - XDN = - 4000000 ! NXSPS <= 14000000 ! At most 70% through canalXSA <= 10000000 ! Pipeline capacityend

Solution

OBJECTIVE FUNCTION VALUE

Cost $13,500,000

VARIABLE VALUE XDS 20,000,000 XSPS 10,000,000 XSA 10,000,000 XPSR 4,000,000 XPSM 6,000,000 XAM 6,000,000 XAN 4,000,000