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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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IE252Networkflows 3

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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IE252Networkflows 6

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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IE252Networkflows 10

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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TRANSSHIPMENT PROBLEM

Supply Transshipment Demand

1

i

N

1

M

ctjxtj

Si Dj

t

t

1

cityit


12/32


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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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


14/32


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


16/32


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


17/32


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


18/32


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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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


23/32


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


24/32


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)


25/32


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


29/32


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.


31/32


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

02 - network flows

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