02 - network flows
TRANSCRIPT
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IE252Networkflows 1
NETWORK FLOWS
TRANSPORTATION PROBLEM
Supply Demand
xcz ij
N
j ij
M
i
Min
11
S.t.
a) Supply:
Sx iijN
j
1
i=1,..,M
b)Demand
Dx jijN
j
1
j=1,..,N
c) Nonnegativity: 0xij i=1,..,M; j=1,..,N
1
i
N
1
M
cijxij
Si Dj
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IE252Networkflows 2
UNIMODULARITY PROPERTY: For the Transportation
problem if all RHS values (i.e. Siand Dj) are INTEGERS for
any OPTIMAL solution all VARIABLES (xij) are INTEGERS
ASSIGNMENT PROBLEM
Supply Demand
xcz ijN
jij
M
i
Min
11
S.t.
a) Supply:
1
1
N
ijj
x
i=1,..,M
b)Demand
1
1
N
ijj
x
j=1,..,M
c) 0 /1ijx i=1,..,M; j=1,..,M
1
i
M
1
M
cijxij
Si=1 Dj=1
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Number of Supply points (M) = Number of Demand Points (N)
Si =1 and Dj=1 i(j)=1,,M
xij=0/1 i(j)=1,,M
Special type of Transportation Problem
Solve by TP algorithm
Find the initial solution by Min. Cost method:
JOBS
1 2 3 4 S
14 5 8 7
1
1 1
2 12 6 5
2
1 1
MC 7 8 3 93
1 1
2 4 6 10
4
1 1
D 1 1 1 1
How many basic variables?
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IE252Networkflows 4
2M-1 7
JOBS
1 2 3 4 S u
14 5 8 7
1
1 0 0 1 0
2 12 6 5
2
1 1 -2
MC 7 8 3 9
3
1 1 -5
2 4 6 10
4
1 0 1 -1
D 1 1 1 1
v 3 4 8 7
Z=15
Find uiand vj
compute reduced cost coefficients for nonbasic cells
If may improve the current solution.
x43enters the basis
which of the current basic variable will leave the basis?
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Step1-1 : (Row Reduction) Find the minimum element of each
row i (ui) of the cost matrix and subtract it from every element
of that row.
{} for all i=1,..,M
for all i/j=1,..,M
Step1-2 : (Column Reduction) Find the minimum element of
each column of the reduced cost matrix (vj) and subtract it
from every element of that row.
{} for all j=1,..,M
for all i/j=1,..,M
Step 2: Draw minimum number of horizontal or vertical lines
to cover zero cost elements of the reduced matrix. If the
number of such lines equal to M stop, optimal assignments are
found.
Otherwise go to step 3.
Step 3: Find the smallest uncoverd nonzero element (call its
value k).
Subract k from each uncovered element
Add k to each twice covered element.
Go to Step 2.
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IE252Networkflows 7
U
14 5 8 7 5
2 12 6 5 2
7 8 3 9 3
2 4 6 10 2
9 0 3 2
0 10 4 3
4 5 0 6
0 2 4 8
v 0 0 0 2
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IE252Networkflows 8
9 0 3 0
0 10 4 1
4 5 0 4
0 2 4 6
u+v=14 =W
Minimization:
Obj. Funct. value of :
dual feasibleoptimal primal feasible
W=14 Z* Z=15
Draw minimum number of lines to cover zero cells:
9 0 3 0
0 10 4 1
4 5 0 4
0 2 4 6
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IE252Networkflows 9
k=1
10 0 3 0
0 9 3 0
5 5 0 4
0 1 3 5
10 0 3 0
0 9 3 0
5 5 0 4
0 1 3 5
4 lines =M
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10 0 3 0
0 9 3 0
5 5 0 4
0 1 3 5
x41= x12 + x24+ x33=1
Z= 15
W= u+v+k =15 =Z
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IE252Networkflows 11
TRANSSHIPMENT PROBLEM
Supply Transshipment Demand
1
i
N
1
M
ctjxtj
Si Dj
t
t
1
cityit
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IE252Networkflows 12
xcycz tjN
jtj
t
tit
T
tij
M
i
Min
1111
'
S.t.a) Supply:
Sy iitN
j
1
i=1,..,M
b) Transshipment:
1 1
M N
tjiti j
y x
for all t=,1..,T
c)Demand:
Dx jtjT
t
1
j=1,..,N
yit 0 and xtj0 i=1,..,M; t=1,..,T; j=1,..,N
S
D
ST
T
DD
S
T
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IE252Networkflows 13
A GRAPH or NETWORK: is defined by two symbols:
NODES (VERTICES) and ARCS
An ARC consists of an ordered pair of NODES (vertices) andrepresents a possible direction of FLOW (motion)
i j
Arc (i,j) - directed
i j
i j
Arc (i,j) - undirected
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CHAIN: A sequence of arcs such that every arc has exactly one
vertex (node) in common with the previous
PATH: is a chain in which the terminal node of each arc is
identical to the initial node of the next arc (directed chain)
CHAIN:
(1,3)(4,3)(4,5)
13456PATH:
(1,2)(2,6)(6,5)(5,4)
12654Directed chain
1
4
5
3
6
2
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MINIMUM COST NETWORK FLOW PROBLEMS
(MCNFP)
N = number of nodes
D.V.s:
xij : amount of units sent from node i to node j, flow on arc (i,j)
Parameters:
Si : amount of supply at node i
Di : amount of demand at node i
Cij: unit cost of transportinf from i to j
Lij: lower bound on flow from i to j
Uij: upper bound on flow from i to j
BALANCE OF FLOW AT EACH NODE:
ik j
Si
Di
INFLOW OUTFLOW
+v+v
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INFLOW(available) = OUTFLOW (use)
xkiN
k
1
+Si= xijN
j
1
+Di for all i=,1..,N
xkiN
k
1
- xijN
j
1
= Di - Si
xijN
j
1
- xkiN
k
1
= Si - Di = bi
bxx ikiN
kij
N
j 11 for all i=,1..,N
bi= Si-Di= Net supply or demand at node i
Si>Di Net supplier node (Si-Di >0)
bi= {Si=Di Transhipment node (Si-Di =0)Si< Di Net supplier node (Si-Di
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IE252Networkflows 17
GrainCO supplies corn from three silos to three poultry farms.
Supply amounts 100, 200, and 50 x1000 tons and demand at
the three farms are 150, 80, and 120 x1000 tons.
OUTFLOWINFLOW = b
x12 + x13+ x14 =100
x23 + x25- x11 =200
x34 + x35x13x23 =50
x45x14x34 =-150
x56x25x35 =-80
x46x56 =-120
x14 80 x46 100
1
3
2 5
6
4
S1=100
D5=80
D4=150
D6=120
S2=200
$3
$4
$6
$4 U=80
$1
$2 U=100
$2
$4
S3=50
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IE252Networkflows 18
UNIMODULARITY PROPERTY: For the MCNFP if all
RHS values (i.e. bi) are INTEGERS for any OPTIMAL
solution all VARIABLES (xij) are INTEGERS
So as:
- TRANSPORTATION- ASSIGNMENT- TRANSSHIPMENT- SHORTEST PATH- MAXIMUM FLOW- Which are all special type of MCNFP
SHORTEST PATH PROBLEM
1
42
53
6
c12=4
2
3
2
2
3
3
City 1
City 6
b1= 1 b2= b3= b4= b5= 0 b6= -1
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IE252Networkflows 19
1 if node i is ORIGIN
bi= {0 if node i is Transhipment-1 if node i is DESTINATION
xcz ijN
jij
N
i
Min
11
s.t.:
bxx ikiN
k
ij
N
j
11
for all i=,1..,N
xij= 0/1 i=1,..,N; j=1,..,N
Purchase car at time 0 for $12.000
Age Annual
maintanence cost
($)
Trade-in price-
salvage value($)
0 2000 7000
1 4000 6000
2 5000 2000
3 9000 1000
4 12000 0
What to do next 5 years?
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cij= total cost between years ij if car is bought at the beginingof year i and sold at the begining of year j
(at the end of year j-1)
c12= 12+2-7=7c13= 12+6-6=12
c14= 12+11-2=21
c15= 12+20-1=31
c16= 12+220=44
c23= 12+2-7=7
c24= 12+6-6=12
c25= 12+11-2=21
c26= 12+20-1=31
c34= 12+2-7=7
c35= 12+6-6=12
c36= 12+11-2=21
c45= 7 c56= 7
c46= 12
1 2 3 4 5 6
Year 1 Year 2 Year 3 Year 4 Year 5
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Shorest Path Network
1 32 4 5 7 7777
1212
12
21
21
12
21
31
31
44
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SHORTEST PATH PROBLEM
SOLVED AS A TRANSPORTATION PROBLEM
i/j 2 3 4 5 6 Supply
1 4 3 M M M 1
2 0 M 3 2 M 1
3 M 0 M 3 M 1
4 M M 0 M 2 1
5 M M M 0 2 1
Demand 1 1 1 1 1 1
1
42
53
6
c12=4
2
3
2
2
3
3
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i/j 2 3 4 5 6 Supply
1 4
1
3 M M M 1
2 0 M 3 2
1
M 1
3 M 0
1
M 3 M 1
4 M M 0
1
M 2 1
5 M M M 0 2
1
1
Demand 1 1 1 1 1 1
PROBLEM :?
DEGENERACY!!!
CYCLING
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DIJKSTRAs ALGORITHM
s -------------------------------> t
STEP 0. assign temporary label to all nodes, permanent
label to s.
Set p=s
STEP 1. Set
l(i) = Minimum { l(i) , l(p) +cpi}
for all i with temporary label.
Let
l(k) = Minimum { l(i) }iT
Make k permanent, let
p=k
STEP 2.
If node t is temporary go to STEP 1.
Otherwise STOP l(t) is the shortest path. (Backtract)
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p=1 l(1)=0
1 2 3 4 5 6
0*
0* 4 3*
0* 4* 3* 6
0* 4* 3* 7 6*
0* 4* 3* 7* 6* 8
1
42
53
6
4
2
3
2
2
3
3
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Alternative shortest paths
1
42
53
6
4
2
3
2
2
3
3
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MAXIMUM FLOW PROBLEM
Uij= capacity of arc (i,j)
1 ----------------------------------------NOrigin Destination
MAXIMUM AMOUNT OF FLOW =?
bi = F if node i is Origin
bi= -F if node i is Destination
bi= 0 other nodes
1
3
42 5
U12=5U24=1
U14=3
U35=4
U23=2
U45=3
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Max Z = F
s.t.:
Fxx kiN
kij
N
j
11
if i is Origin
0
11
xx kiN
kij
N
j if i is other
Fxx kiN
kij
N
j
11
if i is Destination
0 xijUij i=1,..,N; j=1,..,N
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MINIMUM SPANNING TREE
5 COMPUTERS
SPANNING TREE:
For a network of N nodes spanning tree is a group of N-1 arcs
- that connects all nodes to each other- contains no loops
A spanning tree of minimum length is a
MINIMUM SPANNING TREE
4
1
5
3
2
6
5
24
2
4
3
C12=1
2
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MINIMUM SPANNING TREE (MST)
ALGORITHM
(GREEDY)
Let:
C : set of connected nodes
C : set of unconnected nodes
C U C = N
STEP 1 :
Choose any j N
C = {j} C = N {j}
STEP 2:
Let dkr= Minimum { dij}iC ; jC
STEP 3:
C C + { r } C C - { r }
If C = STOP otherwise
GO TO STEP 2.
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C={1} C={2,3,4,5}
r=2 12C= {1,2} C={3,4,5}
r=5 15
C= {1,2,5} C={34}
r=3 53
C= {1,2,5} C={4}
r=4 45
C= {1,2,3,4,5} C=
4
1
5
3
2
6
5
24
2
4
3
C12=1
2
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4
1
5
3
2
6
5
24
2
4
3
1
2