Integer Programming. 2 Ardavan Asef-Vaziri June-2013Integer Programming Computationally speaking, we can partition problems into two categories. Easy

Integer Programming

2Ardavan Asef-Vaziri June-2013Integer Programming

Computationally speaking, we can partition problems into two categories. Easy Problems and Hard Problems

We can say that easy problem ( or in some languages polynomial problems) are those problems with their solution time proportional to nk. Where n is the number of variables in the problem, and k is a constant, say 2, 3, 4 … Let’s assume that k = 2.

Therefore, easy problems are those problems with their solution time proportional to n2. Where n is the number of variables in the problem.

Difficult problems are those problems with their solution time proportional to kn, or in our case 2n.

Computational Complexity


Now suppose we have an easy problem with n variables. Therefore, the solution time is proportional to n2 . We have solved this problem in 1 hour using a computer with 5 GHz CPU.

Suppose we have a new computer with its processing capabilities 10000 times of a 5 GHz computer, and we have one century time. What is the size of the largest problem that we can solve with this revolutionary computer in one century (100 years and about 10000 hours per year).

1 n2

(10000) (100)(10000) (n+x)2



1 n2

(10)10 (n+x)2

1 (n+x)2 = (10)10 n2

(n+x)2 / n2 = (10)10

(n+x / n ) 2 = (10)10

n+x / n = (10)5 = 100000

n+x = 100,000nThe number of variables in the new problem is 100,000 times greater that the number of variables in the old problem.



Now supposed we have a hard problem with n variables. Therefore, the solution time is proportional to 2n . We have solved this problem in one hour using a computer with 900 MHz CPU.

Suppose we have a new computer with 900,000 MHz CPU, and we have one century time. What is the size of the largest problem that we can solve in one century using a 900, 000 MHz CPU.

1 2n

(10)10 2(n+x)



1 2n

(10)10 2(n+x)

(1) 2(n+x) = (10)102n

2(n+x) / 2n = (10)10

2x = (10)10

log 2x = log (10)10

x log 2 = 10 log (10)x ( .301) = 10(1)x = 33The number of variables in the new problem is 33 variables greater that the number of variables in the old problem.



1,02,1

021

2

011

1

XX

Otherwiseselectedisprojectif

X


X

Working with Binary Variables


1- Exactly one of the two projects is selectedX1+X2 =1 2- At least one of the two projects is selectedX1+X2 ≥1 3-At most one of the two projects is selectedX1+X2 ≤1 4- None of the projects should be selectedX1+X2 =05- Both projects must be selectedX1+X2 =26- none, or one or both of projects are selected7- If project 1 is selected then project 2 must be selectedX2 ≥ X18- If project 1 is selected then project 2 could not be selectedX1+X2 ≤1 !!!

Binary Variables


1,05,4,3,2,10

515

041

4

031

3

021

2

011

1

XXXXXOtherwise

selectedisprojectifX


X


X


X


X

More Practice


Either project 1&2 or projects 3&4&5 are selectedX1=X2X3=X4=X5X1+X3=1Suppose Y1 is our production of product 1 .

Due to technical considerations, we want one of the following two constraints to be satisfied.

If the other one is also satisfied it does not create any problem.

If the other one is violated it does not create any problem.

More Practice


We can produce 3 products but due to managerial difficulties, we want to produce only two of them.We can produce these products either in plant 1 or plant 2 but not in both. In other words, we should also decide whether we produce them in plant one or plant two.Other information are given below

Required hrs Available hrsProduct 1 Product 2 Product 3

Plant 1 3 4 2 30Plant 2 4 6 2 40

Unit profit 5 7 3Sales potential 7 5 9

One constraint out of two

Xij volume of production of product i in plant j, i=1,2,3, j=1,2..Yi is 1 if plant I produces and 0 otherwise.Tij w volume of production of product i, i=1,2,3.


Required hrs Available hrsProduct 1 Product 2 Product 3

Plant 1 3 4 2 30Plant 2 4 6 2 40

Unit profit 5 7 3Sales potential 7 5 9

One constraint out of two

X1+X2+X3 ≤ 2Y1+Y2 ≤ 1

Yi Plant i produces, i=1,2


Consider a linear program with the following set of constraints:

12x1 + 24x2 + 18x3 ≤ 2400

15x1 + 32x2 + 12x3 ≤ 1800

20x1 + 15x2 + 20x3 ≤ 2000

18x1 + 21x2 + 15x3 ≥ 1600

Suppose that meeting 3 out of 4 of these constraints is “good enough”.

Meeting a Subset of Constraints


We want two of the following 4 constraints to be satisfied.The other 2 are free, they may be automatically satisfied, but if they are not satisfied there is no problem

y1 + y2 100y1 +2 y3 160y2 + y3 50y1 + y2 + y3 170

In general we want to satisfy k constraints out of n constraints

Still Fun


A Constraint With k Possible Values

y1 + y2 = 10 or 20 or 100

Mastering Formulation of Binary Variables


We have 5 demand centers, referred to as 1, 2, 3, 4,5.We plan to open one or more Distribution Centers (DC) to serve these markets. There are 5 candidate locations for these DCs, referred to as A, B, C, D, and E. Annual cost of meeting all demand of each market from a DC located in each candidate location is given below

DC1 DC2 DC3 DC4 DC5Market 1 1 5 18 13 17Market 2 15 2 26 14 14Market 3 28 18 7 8 8Market 4 100 120 8 8 9Market 5 30 20 20 30 40

Each DC can satisfy the demand of one, two, three, four, or five market

A Location Allocation Problem


Depreciated initial investment and operating cost of a DC in location A, B, C, D, and E is 199, 177, 96, 148, 111.

The objective function is to minimize total cost (Investment &Operating and Distribution) of the system.

Suppose we want to open only one DC. Where is the optimal location.Suppose we don't impose any constraint on the number of DCs. What is the optimal number of DCs.

A Location Allocation Problem


Mercer Development is considering the potential of four different development projects. Each project would be completed in at most three years. The required cash outflow for each project is given in the table below, along with the net present value of each project to Mercer, and the cash that is available (from previous projects) each year.

Cash Outflo w Required ($mil lio n) CashAvailable

Project 1 Project 2 Project 3 Project 4 ($mi llio n)Year 1 10 8 6 12 30Year 2 8 5 4 0 15Year 3 8 0 6 0 12NPV 35 18 24 16

Example #1 (Capital Budgeting)


Maximum Flow Problem : D.V. and OF

5

122

24

1 43

2

5

3

jnodetoinodeFromFlowMaterial:t ij

origintheofoutgoingflowtotalimizemaxtoisobjectiveThe

1312 ttZMax


Maximum Flow Problem : Arc Capacity Constraints

5

122

24

1 43

2

5

3arcsallFor

ijij Tt

2t 23

2t5t3t1t2t4t

45

34

25

24

13

12


Maximum Flow Problem : Flow Balance Constraint

5

122

24

1 43

2

5

3nodesallFor

453424 ttt

flowoutgoingTotaltoequalisflowgminincoTotal

tij = tji i N \ O and D

4nodeFor

25242312 tttt

342313 ttt 3nodeFor

2nodeFor


Maximum Flow Problem : Flow Balance Constraint

5

122

24

1 43

2

5

3

453424 ttt

25242312 tttt

342313 ttt

2t5t3t1t2t2t4t

45

34

25

24

23

13

12

1312 ttZMax

2

122

23

1 43

2

5

3

1

121

24

1 43

2

5

3

2

121

14

1 43

2

5

3


Maximum Flow Problem with Restricted Number of Arcs

xij : The decision variable for the directed arc from node i to nod j.

xij = 1 if arc ij is on the flow path

xij = 0 if arc ij is not on the flow path

xij 4


Maximum Flow Problem with Restricted Number of Arcs

453424 ttt

25242312 tttt

342313 ttt

2t5t3t1t2t2t4t

45

34

25

24

23

13

12

1312 ttZMax

4xxxxxxx 45342524231312

5

122

24

1 43

2

5

3

1

121

24

1 43

2

5

3


Divisibility 1.5, 500.3, 111.11

Certainty cj, aij, bi

Linearity No x1 x2, x1

2, 1/x1, sqrt (x1)

aijxj

aijxj + aikxk

Nonnegativity

The relationship between flow and arc variables

2tx 2323

t23 could be greater than 0 while x23 is 0,


Relationship between Flow and Arcs

5

112

24

1 43

2

5

3

453424 ttt

25242312 tttt

342313 ttt

4545

3434

2525

2424

2323

1313

1212

x2tx5tx3tx1tx2tx2tx4t

1312 ttZMax

4xxxxxxx 45342524231312

5

122

24

1 43

2

5

3


On a Path

453424 xxx

25242312 xxxx

342313 xxx

5

122

24

1 43

2

5

3

5

122

24

1 43

2

5

3

5

122

24

1 43

2

5

3

5

122

24

1 43

2

5

3


Maximum Flow on a Path

453424 ttt

25242312 tttt

342313 ttt

1312 ttZMax

5

122

24

1 43

2

5

3

453424 xxx

25242312 xxxx

342313 xxx

4545

3434

2525

2424

2323

1313

1212

x2tx5tx3tx1tx2tx2tx4t


The Shortest Route ProblemThe shortest route between two points

l ij : The length of the directed arc ij. l ij is a parameter, not a decision variable. It could be the length in term of distance or in terms of time or cost ( the same as c ij ) For those nodes which we are sure that we go from i to j we only have one directed arc from i to j.

For those node which we are not sure that we go from i to j or from j to i, we have two directed arcs, one from i to j, the other from j to i. We may have symmetric or asymmetric network.

In a symmetric network lij = lji ij In a asymmetric network this condition does not hold


Example

6

3

4

2

5

7

2

6

5

6

4

8

7

2

2

12

2

2


Decision Variables and Formulationxij : The decision variable for the directed arc from node i to nod j.

xij = 1 if arc ij is on the shortest route

xij = 0 if arc ij is not on the shortest route

xij - xji = 0 i N \ O and D

xoj =1

xiD = 1

Min Z = lij xij 6

3

4

2

5

7

2

6

5

6

4

8

7

2

2

12

2

2


Example

6

3

4

2

5

7

2

6

5

6

4

8

7

2

2

12

2

2

+ x13 + x14+ x12= 1- x57 - x67 = -1+ x34 + x35 - x43 - x13 = 0+ x42 + x43 + x45 + x46 - x14 - x24 - x34 = 0

….…..

Min Z = + 5x12 + 4x13 + 3x14 + 2x24 + 6x26 + 2x34 + 3x35

+ 2x43 + 2x42 + 5x45 + 4x46 + 3x56 + 2x57 + 3x65 + 2x67


The ShR Problem : Binary Decision Variables

8

3

4

5

7

10

4

3

5

6

4

5

3

2

2

1

2

6

9

11

6

2

4

34

6

6

OD

2

3

6

53

2

3

2

2

1 43

2

Find the shortest route of these two networks.

But for red bi-directional edges you are not allowed to define two decision variables.

Only one.

Solve the small problem.

Only using 5 variables


Do not worry about the length of the arcs, we do not need to write the objective functions.

Note that we do not know whether we may go from node 2 to 3 or from 3 to 2.

Now we want to formulate this problem as a shortest route.


143

2


To formulate the problem as a shortest route, you probably want to define a pair of decision variables for arcs 2-3 and 3-2, and then for example write the constraint on node 2 as followsX23+X24 = X12 + X32

But you are not allowed to define two variables for 2-3 and 3-2. You should formulated the problem only using the following variables

X23 is a binary variable corresponding to the NON-DIRECTED edge between nodes 2 and 3.

It is equal to 1 if arc 2-3 or 3-2 is on the shortest route and it is 0 otherwise.



As usual we can have the following variables

X12 X13 X24 X34

each corresponding to a directed arc

The solution of (X12 = 1 , X23 = 1, X34 = 1 all other variables are 0) means that the shortest route is 1-2-3-4.

The solution of (X13 = 1 , X32 = 1, X24 = 1 all other variables are 0) ) means that the shortest route is 1-3-2-4.

The solution of (X13 = 1 , X34= 1 all other variables are 0) ) means that the shortest route is 1-3-4.

The ShR Problem : Smaller Number of Binary Variables


Now you should formulate the shortest route problem using defining only one variable for each edge.

Your formulation should be general, not only for this example.

How many variables do you need to formulate this problem?

The ShR Problem : Smaller Number of Binary Variables


How Many Binary Variables for this ShR problem

8

3

4

5

7

10

4

3

5

6

4

5

3

2

2

1

2

6

9

11

6

2

4

34

6

6

OD

2

3

6

53

2

3

2

2


When are “non-integer” solutions okay?· Solution is naturally divisible

· Solution represents a rate

· Solution only for planning purposes

When is rounding okay?

Integer Programming


1 2 3 4 5

1

2

3

4

5

x1

x2· Rounded solution may not be feasible

· Rounded solution may not be close to optimal

· There can be many rounded solutions

Example: Consider a problem with 30 variables that are non-integer in the LP solution. How many possible rounded solutions are there?

The Challenge of Rounding


1 2 3 4 5

1

2

3

4

5

x1

x2

1 2 3 4 5

1

2

3

4

5

x1

x2

How Integer Programs are solved


56789

1011

GTotal

=SUMPRODUCT(C4:F4,C6:F6)

Total Outflow=SUMPRODUCT(C4:F4,C9:F9)=SUMPRODUCT(C4:F4,C10:F10)=SUMPRODUCT(C4:F4,C11:F11)

1234567891011

A B C D E F G H IMercer Development Capital Budgeting

Project 1 Project 2 Project 3 Project 4Undertake? 1 1 0 1

TotalNPV ($million) 35 18 24 16 69

Total Outflow AvailableYear 1 10 8 6 12 30 <= 30Year 2 8 5 4 0 13 <= 15Year 3 8 0 6 0 8 <= 12

Cash Outflow Required ($million)

Spreadsheet Solution to Example #1


Suppose the Washington State legislature is trying to decide on locations at which to base search-and-rescue teams. The teams are expensive, and hence they would like as few as possible. However, since response time is critical, they would like every county to either have a team located in that county, or in an adjacent county. Where should the teams be located?

1

2

3

4

76

9

10

11

12

8

5

13

14

15

1617

18

1920

21 22

2325

24

26 27 28

29 30

31 32

33

3435

36

37

Counties 1. Clallum 2. Jefferson 3. Grays Harbor 4. Pacific 5. Wahkiakum 6. Kitsap 7. Mason 8. Thurston 9. Whatcom10. Skagit11. Snohomish12. King13. Pierce14. Lewis15. Cowlitz16. Clark17. Skamania18. Okanogan19. Chelan

20. Douglas21. Kittitas22. Grant23. Yakima24. Klickitat25. Benton26. Ferry27. Stevens28. Pend Oreille29. Lincoln30. Spokane31. Adams32. Whitman33. Franklin34. Walla Walla35. Columbia36. Garfield37. Asotin

Example #2 (Set Covering Problem)


1

2

3

4

76

9

10

11

12

8

5

13

14

15

1617

18

1920

2122

2325

24

26 27 28

29 30

31 3 2

33

3435

3 6

37

Counties 1. Clallum 2. Jefferson 3. Grays Harbor 4. Pacific 5. Wahkiakum 6. Kitsap 7. Mason 8. Thurston 9. Whatcom10. Skagit11. Snohomish12. King13. Pierce14. Lewis15. Cowlitz16. Clark17. Skamania18. Okanogan19. Chelan

20. Douglas21. Kittitas22. Grant23. Yakima24. Klickitat25. Benton26. Ferry27. Stevens28. Pend Oreille29. Lincoln30. Spokane31. Adams32. Whitman33. Franklin34. Walla Walla35. Columbia36. Garfield37. Asotin



123456789101112131415161718192021222324

A B C D E F G H I J K L M NSearch & Rescue Location

# Teams # TeamsCounty Team? Nearby County Team? Nearby

1 Clallam 0 1 >= 1 19 Chelan 0 2 >= 12 Jefferson 1 1 >= 1 20 Douglas 0 1 >= 13 Grays Harbor 0 2 >= 1 21 Kittitas 1 1 >= 14 Pacific 0 1 >= 1 22 Grant 0 1 >= 15 Wahkiakum 0 1 >= 1 23 Yakima 0 3 >= 16 Kitsap 0 1 >= 1 24 Klickitat 0 1 >= 17 Mason 0 1 >= 1 25 Benton 0 1 >= 18 Thurston 0 1 >= 1 26 Ferry 0 1 >= 19 Whatcom 0 1 >= 1 27 Stevens 1 1 >= 110 Skagit 1 1 >= 1 28 Pend Oreille 0 1 >= 111 Snohomish 0 1 >= 1 29 Lincoln 0 1 >= 112 King 0 1 >= 1 30 Spokane 0 1 >= 113 Pierce 0 2 >= 1 31 Adams 0 1 >= 114 Lewis 1 2 >= 1 32 Whitman 0 2 >= 115 Cowlitz 0 2 >= 1 33 Franklin 1 1 >= 116 Clark 0 1 >= 1 34 Walla Walla 0 1 >= 117 Skamania 1 2 >= 1 35 Columbia 0 1 >= 118 Okanogan 0 1 >= 1 36 Garfield 1 1 >= 1

37 Asotin 0 1 >= 1Total Teams: 8



3456789

10111213141516171819202122

E# TeamsNearby

=D5+D6=D5+D6+D7+D8+D10+D11=D6+D7+D8+D11+D12+D18=D7+D8+D9+D18=D8+D9+D18+D19=D6+D10+D11+D15+D16+D17=D6+D7+D10+D11+D12+D17=D7+D11+D12+D17+D18=D13+D14+D22=D13+D14+D15+D22+K5=D14+D15+D16+K5=D10+D15+D16+D17+K5+K7=D10+D11+D12+D16+D17+D18+K7+K8=D7+D8+D9+D12+D17+D18+D19+D21+K9=D9+D18+D19+D20+D21=D19+D20+D21=D18+D19+D20+D21+K9+K10=D13+D14+D22+K5+K6+K12+K15

3456789

1011121314151617181920212223

L# TeamsNearby

=D14+D15+D16+D22+K5+K6+K7=D22+K5+K6+K7+K8=D16+D17+K5+K6+K7+K8+K9=K6+K7+K8+K9+K11+K15+K16+K17=D17+D18+D21+K7+K8+K9+K10+K11=D21+K9+K10+K11=K8+K9+K10+K11+K19+K20=D22+K12+K13+K15=K12+K13+K14+K15+K16=K13+K14+K16=D22+K8+K12+K13+K15+K16+K17+K18=K13+K14+K15+K18=K8+K15+K17+K18+K19=K15+K16+K17+K18+K19+K21+K22+K23=K8+K11+K17+K18+K19+K20=K11+K19+K20+K21=K18+K20+K21+K22=K18+K21+K22+K23=K18+K22+K23

24C D

Total Teams: =SUM(D5:D22,K5:K23)

Spreadsheet Solution to Example #2 (Formulas)

Documents

Integer Programming. 2 Ardavan Asef-Vaziri June-2013Integer Programming Computationally speaking, we can partition problems into two categories. Easy