IEOR 4004 Final Review part I

IEOR 4004 Final Review part I.

April 28, 2014

Mathematical programming

• Help decision-maker(s) make decisions

(what and how much to produce, which path is fastest, how to allocate limited resources most efficiently)

• Problem description– decisions– goal (objective)– constraints

• Mathematical model– decision variables

(and variable domains)– objective function– constraint equations

Mathematical programming

Problem

Model

Solution

Problemsimplification

Modelformulation

Algorithmselection

Numericalcalculation

InterpretationSensitivity analysis

Model selection

• trade-off betweenand

Simple ComplexModel

Predictive power(quality of prediction) HighLow

Easyto computea solution(seconds)

Hard(if not impossible)

to compute a solution(lifetime of

the universe)

SolvingLinear programming

Network models Integer programmingDynamic programming

accuracy (predictive power)model simplicity (being able to solve it)

Today’s agenda

• Transportation Problem• Shortest path• Spanning tree• Maximum Flow (Ford Fulkerson)• Minimum-cost Flow (Network Simplex)

Transportation problem

Factories Customers

Requirementfor goods

Productioncapacity

... ...

Minimum cost of transportation satisfying the demand of customers.

ai bj

i-th factory

delivers to j-th customer at cost cij

necessary condition

a1

a2

an bm

b1

b2

Transportation tableau

7x11

3x12

4x13

u1

84

x21

2x22

2x23

u2

62

x31

1x32

5x33

u3

3v1

4v2

2v3

3

cost cij of delivering from ith factory to jth customer

supply ai of ith factory

demand bj of jth customershadow customer’s “price”

shadowfactory “price”

shadow prices are relative to some baseline

amounttransported

Transportation problem• Overproduction: If produced more then needed ()Dummy customer at zero delivery cost• Shortage: If required more than produced ()Dummy producer at market delivery cost(what it would cost to deliver it from a third party) a1 b1

a2 b2

b3a3

a4 b4

7x11

3x12

4x13

u1

84

x21

2x22

2x23

u2

62

x31

1x32

5x33

u3

3v1

4v2

2v3

3

7x11

3x12

4x13

0x14

u1

84

x21

2x22

2x23

0x24

u2

62

x31

1x32

5x33

0x34

u3

3v1

4v2

2v3

3v4

8

7x11

3x12

4x13

0x14

u1

84

x21

2x22

2x23

0x24

u2

62

x31

1x32

5x33

0x34

u3

39

x41

8x42

11x43

0x44

u4

0v1

4v2

2v3

3v4

8

Balanced transportation problem

7x11

3x12

4x13

0x14

u1

8

4x21

2x22

2x23

0x24

u2

6

2x31

1x32

5x33

0x34

u3

3

v1

4v2

2v3

3v4

8

Transportation Simplex

• Applying the Simplex method to the problem• Basic solution– min-cost method

• Pivoting– shadow prices

set u1 = 0, then ui + vj=cij

– reduced costpivot if ui + vj > cij

• Finding a loop

0

20

6

take the smaller of the two

2 6

0

10

3

1 2

0

3 3

0

0

must mark exactly m + n – 1 = 6 cells

cost = 3×7 + 3×4 + 2×0 + 1×2 + 2×1 + 6×0 = 37

z = 37

7x11

3x12

4x13

0x14

u1

8

4x21

2x22

2x23

0x24

u2

6

2x31

1x32

5x33

0x34

u3

3

v1

4v2

2v3

3v4

8






• Finding a loop

6

2

1 2

3 3

z = 37

0

4

ui =0 and vj must sum up to cij = 4 vj = 4

7 0

-5

6

0

7x11

3x12

4x13

0x14

u1

8

4x21

2x22

2x23

0x24

u2

6

2x31

1x32

5x33

0x34

u3

3

v1

4v2

2v3

3v4

8






• Finding a loop

6

2

1 2

3 3

z = 37

0

47 0

-5

6

0

-1

calculate ui + vj

-5

7 6 4

6

> >

>

>

≤ ≤

7x11

3x12

4x13

0x14

u1

8

4x21

2x22

2x23

0x24

u2

6

2x31

1x32

5x33

0x34

u3

3

v1

4v2

2v3

3v4

8






• Finding a loop• New basis

6

22+Δ

11+Δ 22-Δ

33-Δ 3

z = 37

0

47 0

-5

6

0

-1 -5

7 6 4

6

> >

>

>

≤ ≤

+Δ 6-Δ

Largest feasible Δ = 2

2 4

41

3

z = 29

Shortest path Problem

93

4

1086

6

9

7

59

10

85

6

8

9

8

s

b

g

c

h i

f

d

e

Network G = (V,E) where V is the set of nodes and E the set of edges• each edge (i,j) can have cost/weight/length cij

• path is a sequence of nodes connected by edgesv1,v2,...,vt where (vi,vi+1) in E for i=1,2,...,t-1

• length of a path is sum of lengths of its edgeslength of a shortest path from s to e?

s, g, c, d, i, e is a path of length 6+8+5+3+9 = 31

Shortest path Problem (Dijkstra)

93

4

1086

6

9

7

59

10

85

6

8

9

8

s

b

g

c

h i

f

d

e

s b c d e f g h i0* ∞ ∞ ∞ ∞ ∞ ∞ ∞ ∞0 8 ∞ ∞ ∞ ∞ 6* ∞ ∞0 8* 14 ∞ ∞ ∞ 6 15 ∞0 8 14* ∞ ∞ ∞ 6 15 ∞0 8 14 19 ∞ ∞ 6 15* ∞0 8 14 19* ∞ ∞ 6 15 230 8 14 19 ∞ 27 6 15 22*0 8 14 19 31 27* 6 15 220 8 14 19 31* 27 6 15 22

We calculate distance from s to all vertices (not just e)• keep distance estimate d(x) for each node x• d(x) is an upper bound on the length of a shortest path from s to xstart by setting d(s)=0 and d(x)=∞ for each x≠s; then iterativelypick smallest d(x) and for each (x,y) in E, update d(y)=min(d(y),d(x)+cxy)

Minimum spanning tree (Prim)

93

4

1086

6

9

7

59

10

85

6

8

9

8

Collection of edges touching all vertices but with no cycles

Tree growing1. start with s (or any other node)2. attach a cheapest edge sticking out of the tree already built3. repeat as long as possible

s

b

g

c

h i

f

d

e

Minimum spanning tree (Kruskal)

93

4

1086

6

9

7

59

10

85

6

8

9

8

Order edges by cost

1. start with empty forest (no nodes, no edges)2. add the cheapest edge that does not create a cycle3. repeat as long as possible

3 4 5 6 7 8 9 10

di ef bg, cd gs, cf, ch fh df, hi, cg, bs fi, ei, bc, gh de, cs

s

b

g

c

i

d

e

h

f

Maximum flow

4

1

3

3

34

2

4

2

3

2

s

b

g

c

h i

f

d

t1

42

1

2

3

2

capacitynot cost!

flow across an edge (i,j) = amount xij of goods transported from i to j• non-negative and at most the capacity uij

• what flows into a node flows out (except for source s and sink t)(conservation of flow)

How to find a maximum possible flow? Ford-Fulkerson

0/44

Maximum flow (Ford-Fulkerson)

0/4

0/1

0/3

0/3

0/30/4

0/2

0/2

0/3

0/2

s

b

g

c

h i

f

d

t0/1

0/40/2

0/1

0/2

0/3

0/2

start with azero flow

4

1

3

3

34

2

2

3

2

1

42

1

2

3

2

s

b

g

c

h i

f

d

t

residualnetwork

edges:• if not saturated• if positive flow

(put reverse edge)

representsremaining

capacity

can improve iffwe find a path

from s to t

can increaseup to Δ=2 units

2/4 2/3

2/3

2/3 2/4

2/2

0/41/4


2/4

0/1

2/3

0/3

2/32/4

0/2

2/2

2/3

0/2

s

b

g

c

h i

f

d

t0/1

0/40/2

0/1

0/2

0/3

0/2


s

b

g

c

h i

f

d

t

residualnetwork


(put reverse edge)

representsremaining

capacity


from s to t


3/41/1

3/3

1/2

1/42/4


3/4

0/1

2/3

0/3

3/32/4

1/2

2/2

2/3

0/2

s

b

g

c

h i

f

d

t1/1

0/40/2

0/1

0/2

0/3

0/2


s

b

g

c

h i

f

d

t

residualnetwork


(put reverse edge)

representsremaining

capacity


from s to t


4/4

1/3 1/2


3/4

0/1

2/3

0/3

3/32/4

1/2

2/2

2/3

0/2

s

b

g

c

h i

f

d

t1/1

0/40/2

0/1

0/2

0/3

0/2


s

b

g

c

h i

f

d

t

residualnetwork


(put reverse edge)

representsremaining

capacity


from s to t

4/4

1/3

2/4

1/2

-Δ+Δ

+Δ

+Δ+Δ

+Δ


1/1

3/3

0/1

2/3 2/2

3/4


4/4

1/1

3/3

0/3

3/32/4

1/2

1/3

2/2

2/3

0/2

s

b

g

c

h i

f

d

t0/1

0/40/2

0/1

0/2

0/3

0/2


s

b

g

c

h i

f

d

t

residualnetwork


(put reverse edge)

representsremaining

capacity


from s to t

2/3

3/4

2/2

nodes reachable from s

t cannot be reached from s minimum cut (of capacity 2+3=5)

Minimum-cost flow (Network Simplex)

3/4

1/1

3/3

1/3

3/32/4

1/2

2/3

2/2

2/3

1/2

s

b

g

c

h i

f

d

t

total cost: 2×$6+2×$8+2×$9+1×$8+1×$9+3×$5+3×$8+1×$8+2×$9+1×$4+3×$9 = $159

$9

$3 $4

$10$8$6$6

$9

$7

$5$9

$10

$8$5

$6

$8

$9

$8

Start by constructing a basis:1. Include all edges with nonzero flow but not saturated2. add any additional edges until a spanning tree obtained(all nodes touching at least one edge, no cycles)

Is this minimum possible cost(among all flows meeting the supplies/demands bis)?

netsupplybs = +4

netsupplybt = –4

Network Simplex algorithmwhat if you get a cycle in step 1.?

0/1

0/40/2

0/1

0/2

0/3

0/2


3/40/1

1/1

0/43/30/2

0/1

1/3

0/2

3/32/4

0/3

1/20/2

2/3

2/2

2/3

1/2

s

b

g

c

h i

f

d

t

$9

$3 $4

$10$8$6$6

$9

$7

$5$9

$10

$8$5

$6

$8

$9

$8

Steps:1. shadow prices: set ys = $0, then yu – yv = cuv for edge (u,v)2. reduced costs yu – yv – cuv

$0

-$6

-$14-$5

-$15 -$23

-$14

-$6

-$32

Economic interpretation: (-yu) is the price in node u the cost of transportation using green edges matches these prices;

perhaps using an edge outside the green network we can do better reduced costs (e.g. take nodes d and i with prices $6 and $23; transporting along (d,i) costs only $3...

it pays to buy at d (at $6), sell at i (at $23) and even pay the transportation ($3)

0 – $6 = -$6

$0 – (-$5) – $8= -$3

(-$5)(-$4)

(-$24)

(-$3)(-$13)

(-$6)

(-$6)

($14)

(-$36)

($14)

3. pivot on edge (u,v) if either• no flow on edge (u,v) and positive reduced cost• edge (u,v) saturated and negative reduced cost

(e.g. $32 at t and $0 at s)


3/40/1

1/1

0/43/30/2 (3-Δ)/3

0/1

1/3

0/2

3/32/4

0/3

1/20/2

2/3

2/2

2/3

1/2

s

b

g

c

h i

f

d

t

$9

$3 $4

$10$8$6$6

$9

$7

$5$9

$10

$8$5

$6

$8

$9

$8


$0

-$6

-$14-$5

-$15 -$23

-$14

-$6

-$32

(-$5)(-$4)

(-$24)

(-$3)(-$13)

(-$6)

(-$6)

($14)

(-$36)

($14)


choosethis edgeto enter(3-Δ)/3

(2-Δ)/3

(1+Δ)/2(1+Δ)/3

(1-Δ)/2

largest feasible Δ=1we choose thisedge to leave

total cost: 2×$6+2×$8+2×$9+0×$8+2×$9+2×$5+2×$8+2×$8+1×$9+1×$4+3×$9 = $146


3/40/1

1/1

0/42/30/2

0/1

2/3

0/2

2/32/4

0/3

0/20/2

2/3

2/2

1/3

2/2

s

b

g

c

h i

f

d

t

$9

$3 $4

$10$8$6$6

$9

$7

$5$9

$10

$8$5

$6

$8

$9

$8


$0

-$6

-$14-$5

-$15 -$23

-$14

-$6

-$32

(-$5)(-$4)

(-$24)

(-$3)(-$13)

(-$6)

($14)

(-$36)

($14)


choosethis edgeto enter

largest feasible Δ=1we choose thisedge to leave

(3-Δ)/3

(3-Δ)/3

(2-Δ)/3

(1+Δ)/3

(1-Δ)/2

(1+Δ)/2

(-$6)

New basis

What if Δ can be arbitrarily large?What does it mean for the network?

Network problemsas Minimum-cost flow

93

41086

6

97

59

1085

6

8

98

s

b

g

c

h i

f

d

e

Shortest path

$9$3

$4$10$8$6

$6

$9$7

$5$9

$10$8$5

$6

$8

$9$8

s

b

g

c

h i

f

d

e

Minimum-cost flow

netsupplybs = 1

netdemandbe = 1

all edges infinity

capacity


4

13

3

34

24

2

32

s

b

g

c

h i

f

d

t1

42

12

32

4

13

3

34

24

2

32

s

b

g

c

h i

f

d

t1

42

12

32

Maximum flow

Minimum-cost flow

∞

all edges cost $0except (t,s) has cost -$1

and infinity capacity

-$1


Transportation problem Minimum-cost flow

all edges infinity capacity(edge costs remain the same)

b1

b2

b3

b4

(a2)

net supply(a1)

(a3)

(a4)

(-b2)

net supply(-b1)

(-b3)

(-b4)

a1

a2

a3

a4


Minimum-cost flow Minimum-cost flowwith one source

and one sinknodes with

positivenet supply

nodes withnegative

net supply

net supply of s= the sum ofthe (positive)net supplies

net supply of t= the sum of

the (negative)net supplies

new edges have $0 costall nodes here now

have zero net supply

G

(200)

(500)

(300)

(100) (-400)

(-200)

(-200)

(-300)

s t

(1100) 100

200

500

300

400

200

200

300

(-1100)

Documents

IEOR 4004 Final Review part I