lecture 5 – integration of network flow programming models topics min-cost flow problem (general...
TRANSCRIPT
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Lecture 5 – Integration of Network Flow Programming Models
Topics
• Min-cost flow problem (general model)
• Mathematical formulation and problem characteristics
• Pure vs. generalized networks
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GAINS
8
ATL
5
NY
6
DAL
4
CHIC
2
AUS
7
LA
3
PHOE
1
(6)
(3)
(5)
(7)
(4)
(2)
(4)
(5)
(5)(6)
(4)
(7)
(6)
(3)
[–150]
[200]
[–300]
[200]
[–200]
[–200]
(2)
(2)
(7)
[–250][700]
[supply / demand](shipping cost)
arc lower bounds = 0 arc upper bounds = 200
Distribution Problem
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•Warehouses store a particular commodity in Phoenix, Austin and Gainesville.
• Customers - Chicago, LA, Dallas, Atlanta, & New York
Supply [ si ] at each warehouse i
Demand [ dj ] of each customer j• Shipping links depicted by arcs, flow on each arc is
limited to 200 units.• Dallas and Atlanta - transshipment hubs• Per unit transportation cost (cij ) for each arc
Problem: Determine optimal shipping plan that minimizes transportation costs
Example: Distribution problem
Min-Cost Flow Problem
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In general:[supply/demand] on nodes(shipping cost per unit) on arcs
In example:all arcs have an upper bound of 200nodes labeled with a number 1,...,8
• Must indicate notation that is included in model:
(cij ) unit flow cost on arc (i, j )
(uij ) capacity (or simple upper bound) on arc (i,
j )
(gij ) gain or loss on arc (i, j )
• All 3 could be included: (cij , uij , gij )
Notation for Min-Cost Flow Problem
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arcname
terminationnode cost gain
originnode
lowerbound
upperbound
xij i j lij
The origin node is the arc’s tail
The termination node is called the head
Supplies are positive and demands are negative
External flow balance: total supply = total demand
i j
uij cij gij
externalflow
si or -di
Spreadsheet Input Data
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And here is the solution ...
Data Entry Using Jensen Network Solver
Network Model Name: Net1 Solver: Excel Solver Ph. 1 Iter. 13
5300 Type: Net Type: Linear Total Iter. 1517 Change Goal: Min Sens.: Yes Comp. Time 00:06
TRUE Cost: 5300 Side: No Status OptimalTRUE SolveTRUE100 Vary
Arc Data and Flows Node Data and Balance ConstraintsNum. Name Flow Origin Term. Upper Cost Red. Cost Num. Name Fixed Balance Dual Values Basis
1 Phoe-Chi 200 1 2 200 6 -3 1 Phoe 700 0 -11 -42 Phoe-LA 200 1 3 200 3 -7 2 Chi -200 0 -2 63 Phoe-Dal 200 1 4 200 3 -2 3 LA -200 0 -1 124 Phoe-Atl 100 1 5 200 7 0 4 Dal -300 0 -6 -75 Dal-LA 0 4 3 200 5 0 5 Atl -150 0 -4 86 Dal-Chi 0 4 2 200 4 0 6 NY -250 0 0 277 Dal-NY 50 4 6 200 6 0 7 Aus 200 0 -8 -138 Dal-Atl 50 4 5 200 2 0 8 Gain 200 0 -8 -169 Atl-NY 0 5 6 200 5 1
10 Atl-Dal 0 5 4 200 2 411 Atl-Chi 0 5 2 200 4 212 Aus-LA 0 7 3 200 7 013 Aus-Dal 200 7 4 200 2 014 Aus-Atl 0 7 5 200 5 115 Gain-Dal 0 8 4 200 6 416 Gain-Atl 0 8 5 200 4 017 Gain-NY 200 8 6 200 7 -1
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GAINS
ATL
NY
DAL
CHIC
AUS
LA
PHOE
(200)
(200)
(200)
(200)
(50)
(200)
[-150]
[200]
[-300]
[200]
[-200]
[-200]
(50)
(100)
[-250]
[700]
[supply / demand] (flow)
Solution to Distribution Problem
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Sensitivity Report for Distribution Problem
123456789
10111213141516171819202122232425262728293031323334
A B C D E F G HMicrosoft Excel 11.4 Sensitivity Report
Adjustable CellsFinal Reduced Objective Allowable Allowable
Cell Name Value Cost Coefficient Increase Decrease$E$11 Phoe-Chi Flow 200 0 6 2.999999974 1E+30$E$12 Phoe-LA Flow 200 -7.000000003 3 7.000000003 1E+30$E$13 Phoe-Dal Flow 200 -2 3 2 1E+30$E$14 Phoe-Atl Flow 100 0 7 13.00800361 2$E$15 Dal-LA Flow 0 0 5.000000056 1E+30 0$E$16 Dal-Chi Flow 0 2.999999975 3.99999999 1E+30 2.999999975$E$17 Dal-NY Flow 50 0 6 1.000000054 0.999999989$E$18 Dal-Atl Flow 50 0 2 0.999999989 1.000000054$E$19 Atl-NY Flow 0 1.000000054 5.000000056 1E+30 1.000000054$E$20 Atl-Dal Flow 0 3.999999964 1.999999949 1E+30 3.999999964$E$21 Atl-Chi Flow 0 4.99999999 3.99999999 1E+30 4.99999999$E$22 Aus-LA Flow 0 0 7.000000005 0 7.000000005$E$23 Aus-Dal Flow 200 0 2 1.000000054 0$E$24 Aus-Atl Flow 0 1.000000054 5.000000056 1E+30 1.000000054$E$25 Gain-Dal Flow 0 4.000000055 6.00000003 1E+30 4.000000055$E$26 Gain-Atl Flow 0 0 3.99999999 4.000000055 0.999999989$E$27 Gain-NY Flow 200 -0.999999989 7 0.999999989 1E+30
ConstraintsFinal Shadow Constraint Allowable Allowable
Cell Name Value Price R.H. Side Increase Decrease$N$11 Phoe Balance 0 0 0 0 1E+30$N$12 Chi Balance 0 -6 0 200 0$N$13 LA Balance 0 -10 0 0 0$N$14 Dal Balance 0 -5 0 100 0$N$15 Atl Balance 0 -7 0 100 0$N$16 NY Balance 0 -11 0 50 0$N$17 Aus Balance 0 -3 0 0 0$N$18 Gain Balance 0 -3.000000011 0 100 0
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Conservation of flow at nodes. At each node flow in = flow out. At supply nodes there is an external inflow (positive)At demand nodes there is an external outflow (negative).
Flows on arcs must obey the arc bounds; i.e., lower bound & upper bound (capacity)
Each arc has a per unit cost & the goal is to minimize total cost.
Characteristics of Network Flow Problems
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Distribution Network Used in Formulation
8
5
6
4
2
7
3
1
(6)
(3)
(5)
(7)
(4)
(2)
(4)(5)
(5) (6)(4)
(7)
(6)
(3)
[-150]
[200]
[-300]
[200]
[-200]
[-200]
(2)
(2)
(7)
[-250]
[700]
[external flow] (cost)
lower = 0, upper = 200
Notation
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G = (N, A) network with node set N and arc set A
Indices i, j N denote nodes and (i, j ) A denote arcs
Originating set of arcs for node i (tails are i ) is the forward star of i
FS(i ) = { (i, j ) : (i, j ) A }
Terminating set of arcs for node i is the reverse star of i
RS(i ) = { (j,i ) : (j,i ) A }.
Pure Minimum Cost Flow Problem
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FS(1) = { (1,2), (1,3), (1,4), (1, 5) }
RS(1) = Ø
FS(4) = { (4,2), (4,3), (4,5), (4,6) }
RS(4) = { (1,4), (5, 4), (7,4), (8,4) }
xij – xji = bi
(i, j )FS(i )
where bi = positive for supply node i= negative for demand node i= 0 otherwise
(j,i )RS(i )
Flow balance equation for node i :
In our example:
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Indices/setsi, j N nodes
arcsforward star of ireverse star of i
(i, j ) A
FS(i )RS(i )
Data cij unit cost of flow on (i, j )
lower bound on flow (i, j )
upper bound on flow (i, j )
external flow at node i
lijuij
bi
Pure Min-Cost Flow Model
Total supply = total demand: i bi = 0
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Decision variables
xij = flow on arc (i, j )
Formulation for pure min-cost flow model
Min cijxij(i, j )A
s.t. xij xji = bi, i N (i, j )FS(i ) (j, i )RS(i )
lij xij uij, (i, j ) A
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Decision variables are the flow variables xij
By examining the flow balance constraints we see that xij appears in exactly two of them:
xij 0. . . 0
+1 node i
( or in the other order if i
> j )
0 . . . 0 1 node j0 . . . 0
i j
This structure is called total unimodularity and guarantees integer solutions
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• If we add the constraints we obtain zero on the left-hand side so the right-hand side must also be zero for feasibility.
• In particular, this meanssum of supplies = sum of demands.
• Mathematically, we have one redundant constraint.
• Must be careful in interpreting shadow prices on the flow balance constraints.
• Cannot change only a supply or demand and have model make sense.
Observations from LP Model
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• Only oneone modification to “pure” formulation
a possible gain (or loss) on each arc, denoted by gij
• If gij = 0.95 then 100 units of flow leaves node i and
95 units arrive at node j
Generalized Minimum Cost Network Flow Model
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Min cijxij(i, j )A
s.t.
lij xij uij, (i, j )A
Note that if gij =1 (i, j ) A, then we obtain the “pure” model
xij gjixji = bi, i N(i, j )FS(i ) (j, i )RS(i )
Generalized Formulation
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• Might experience 5% spoilage of a perishable good during
transportation on a particular arc.
gij = 0.95 for the associated arc (i,j).
• In production of manufacturing formulations we might incur losses due to production defects.
• In financial examples we can have gains due to currency exchange or gains due to returns on investments.
US $
Swissfrancs
Year 1 Year 2
Currencyexchange
15% returnon investment
Gain = 1.78
Gain = 1.15
Gains and Losses
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Fact: If bi, lij and uij are integer-valued then all
extreme points of the feasible region for a
pure network flow problem give integer
values for xij. (Same cannot be said for generalized network models.)
This integer property means that if we use the
simplex method to solve a pure network flow
problem then we are guaranteed that xij will be
integer at optimality.
Pure Network Problems vs. General Network Problems
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This is critical when we formulate the assignment, shortest path problems, and other network problems.
Special cases of the pure min-cost flow model:
• Transportation problem
• Assignment problem
• Shortest path problem
• Maximum flow problem
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US $ Yen(100) CHF D-Mark Brit £
1 US $ 1 1.05 1.45 1.72 .682 Yen(100) .95 1 1.41 1.64 .643 CHF .69 .71 1 1.14 .484 D-Mark .58 .61 0.88 1 .395 Brit £ 1.50 1.56 2.08 2.08 1
• The table is to be read as follows:
The 1.45 in row 1 column 3 means that $1 US will purchase 1.45 Swiss Francs (CHF).
• In addition, there is a 1% fee that is charged on each exchange.
Checking for Arbitrage Opportunities
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Each arc has a gain of gij. For example,g12 = (1.05)(0.99)g35 = (0.48)(0.99)
5
43
2
1
[-1]
Brit £
US $
D-MarkCHF
Yen
Arc costs:cij = $ equivalent
(first column of table)
For example:c12 = 1.05, c35 = 0.48
Arbitrage Network: Generalized Min-Cost Flow Problem
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1
4
5
3
30.473
0.674
34.986
Start with 13.801 £ 34.986 D-Mark 30.473 CHF 14.475 £
Remove 0.674 £ $1 leaving 13.801 £
Brit £
US $
CHF
D-Mark13
.801
g54 = 2.535
g43 = 0.871
g35 = 0.475
Note (£ $):g51 = 1.485
Arc gains in optimal cycle:
Total cycle gain: = 1.0488= 4.88%
Solution to Arbitrage Network
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What You Should Known About General Network Flow Problems
• How to view flow on an arch with a gain or loss.
• How to formulate a general network flow problem as a linear program.
• What the implications are for a network flow problem with gains.
• How to solve general network flow problems using the Excel add-ins.