Presentation is loading. Please wait.

Presentation is loading. Please wait.

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

Similar presentations


Presentation on theme: "Integer Programming. 2 Ardavan Asef-Vaziri June-2013Integer Programming Computationally speaking, we can partition problems into two categories. Easy."— Presentation transcript:

1 Integer Programming

2 2 Ardavan 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 n k. 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 n 2. Where n is the number of variables in the problem. Difficult problems are those problems with their solution time proportional to k n, or in our case 2 n. Computational Complexity

3 3 Ardavan Asef-Vaziri June-2013Integer Programming Now suppose we have an easy problem with n variables. Therefore, the solution time is proportional to n 2. 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 n21 n2 (10000) (100)(10000)(n+x) 2 Computational Complexity

4 4 Ardavan Asef-Vaziri June-2013Integer Programming 1 n 2 (10) 10 (n+x) 2 1 (n+x) 2 = (10) 10 n 2 (n+x) 2 / n 2 = (10) 10 (n+x / n ) 2 = (10) 10 n+x / n = (10) 5 = 100000 n+x = 100,000n The number of variables in the new problem is 100,000 times greater that the number of variables in the old problem. Computational Complexity

5 5 Ardavan Asef-Vaziri June-2013Integer Programming Now supposed we have a hard problem with n variables. Therefore, the solution time is proportional to 2 n. 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 2 n (10) 10 2 (n+x) Computational Complexity

6 6 Ardavan Asef-Vaziri June-2013Integer Programming 1 2 n (10) 10 2 (n+x) (1) 2 (n+x) = (10) 10 2 n 2 (n+x) / 2 n = (10) 10 2 x = (10) 10 log 2 x = log (10) 10 x log 2 = 10 log (10) x (.301) = 10(1) x = 33 The number of variables in the new problem is 33 variables greater that the number of variables in the old problem. Computational Complexity

7 7 Ardavan Asef-Vaziri June-2013Integer Programming Working with Binary Variables

8 8 Ardavan Asef-Vaziri June-2013Integer Programming 1- Exactly one of the two projects is selected X1+X2 =1 2- At least one of the two projects is selected X1+X2 ≥1 3-At most one of the two projects is selected X1+X2 ≤1 4- None of the projects should be selected X1+X2 =0 5- Both projects must be selected X1+X2 =2 6- none, or one or both of projects are selected 7- If project 1 is selected then project 2 must be selected X2 ≥ X1 8- If project 1 is selected then project 2 could not be selected X1+X2 ≤1 !!! Binary Variables

9 9 Ardavan Asef-Vaziri June-2013Integer Programming More Practice

10 10 Ardavan Asef-Vaziri June-2013Integer Programming Either project 1&2 or projects 3&4&5 are selected X1=X2 X3=X4=X5 X1+X3=1 Suppose 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

11 11 Ardavan Asef-Vaziri June-2013Integer Programming 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 hrs Product 1 Product 2 Product 3 Plant 1 3 4 2 30 Plant 2 4 6 2 40 Unit profit 5 7 3 Sales 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.

12 12 Ardavan Asef-Vaziri June-2013Integer Programming Required hrsAvailable hrs Product 1 Product 2 Product 3 Plant 1 3 4 2 30 Plant 2 4 6 2 40 Unit profit 5 7 3 Sales potential 7 5 9 One constraint out of two X1+X2+X3 ≤ 2 Y1+Y2 ≤ 1 Yi Plant i produces, i=1,2

13 13 Ardavan Asef-Vaziri June-2013Integer Programming Consider a linear program with the following set of constraints: 12x 1 + 24x 2 + 18x 3 ≤ 2400 15x 1 + 32x 2 + 12x 3 ≤ 1800 20x 1 + 15x 2 + 20x 3 ≤ 2000 18x 1 + 21x 2 + 15x 3 ≥ 1600 Suppose that meeting 3 out of 4 of these constraints is “good enough”. Meeting a Subset of Constraints

14 14 Ardavan Asef-Vaziri June-2013Integer Programming 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 y 1 + y 2  100 y 1 +2 y 3  160 y 2 + y 3  50 y 1 + y 2 + y 3  170 In general we want to satisfy k constraints out of n constraints Still Fun

15 15 Ardavan Asef-Vaziri June-2013Integer Programming A Constraint With k Possible Values y 1 + y 2 = 10 or 20 or 100 Mastering Formulation of Binary Variables

16 16 Ardavan Asef-Vaziri June-2013Integer Programming 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 DC1DC2DC3DC4DC5 Market 115181317 Market 2152261414 Market 32818788 Market 4100120889 Market 53020203040 Each DC can satisfy the demand of one, two, three, four, or five market A Location Allocation Problem

17 17 Ardavan Asef-Vaziri June-2013Integer Programming 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

18 18 Ardavan Asef-Vaziri June-2013Integer Programming 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 OutflowRequired ($million)Cash Available Project1 oject2 oject3 oject4($million) Year 110861230 Year 2854015 Year 3806012 NPV35182416 Example #1 (Capital Budgeting)

19 19 Ardavan Asef-Vaziri June-2013Integer Programming Maximum Flow Problem : D.V. and OF 5 1 2 2 2 4 1 43 2 5 3

20 20 Ardavan Asef-Vaziri June-2013Integer Programming Maximum Flow Problem : Arc Capacity Constraints 5 1 2 2 2 4 1 43 2 5 3

21 21 Ardavan Asef-Vaziri June-2013Integer Programming Maximum Flow Problem : Flow Balance Constraint 5 1 2 2 2 4 1 43 2 5 3  t ij =  t ji  i  N \ O and D

22 22 Ardavan Asef-Vaziri June-2013Integer Programming Maximum Flow Problem : Flow Balance Constraint 5 1 2 2 2 4 1 43 2 5 3 2 1 2 2 2 3 1 43 2 5 3 1 1 2 1 2 4 1 43 2 5 3 2 1 2 1 1 4 1 43 2 5 3

23 23 Ardavan Asef-Vaziri June-2013Integer Programming Maximum Flow Problem with Restricted Number of Arcs x ij : The decision variable for the directed arc from node i to nod j. x ij = 1 if arc ij is on the flow path x ij = 0 if arc ij is not on the flow path  x ij  4

24 24 Ardavan Asef-Vaziri June-2013Integer Programming Maximum Flow Problem with Restricted Number of Arcs 5 1 2 2 2 4 1 43 2 5 3 1 1 2 1 2 4 1 43 2 5 3

25 25 Ardavan Asef-Vaziri June-2013Integer Programming  Divisibility 1.5, 500.3, 111.11  Certainty c j, a ij, b i  Linearity No x 1 x 2, x 1 2, 1/x 1, sqrt (x 1 ) a ij x j a ij x j + a ik x k  Nonnegativity The relationship between flow and arc variables t 23 could be greater than 0 while x 23 is 0,

26 26 Ardavan Asef-Vaziri June-2013Integer Programming Relationship between Flow and Arcs 5 1 1 2 2 4 1 43 2 5 3 5 1 2 2 2 4 1 43 2 5 3

27 27 Ardavan Asef-Vaziri June-2013Integer Programming On a Path 5 1 2 2 2 4 1 43 2 5 3 5 1 2 2 2 4 1 43 2 5 3 5 1 2 2 2 4 1 43 2 5 3 5 1 2 2 2 4 1 43 2 5 3

28 28 Ardavan Asef-Vaziri June-2013Integer Programming Maximum Flow on a Path 5 1 2 2 2 4 1 43 2 5 3

29 29 Ardavan Asef-Vaziri June-2013Integer Programming The Shortest Route Problem The 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 l ij = l ji  ij In a asymmetric network this condition does not hold

30 30 Ardavan Asef-Vaziri June-2013Integer Programming Example 6 3 4 2 5 7 2 6 5 6 4 8 7 2 2 1 2 2 2

31 31 Ardavan Asef-Vaziri June-2013Integer Programming Decision Variables and Formulation x ij : The decision variable for the directed arc from node i to nod j. x ij = 1 if arc ij is on the shortest route x ij = 0 if arc ij is not on the shortest route  x ij -  x ji = 0  i  N \ O and D  x oj =1  x iD = 1 Min Z =  l ij x ij 6 3 4 2 5 7 2 6 5 6 4 8 7 2 2 1 2 2 2

32 32 Ardavan Asef-Vaziri June-2013Integer Programming Example 6 3 4 2 5 7 2 6 5 6 4 8 7 2 2 1 2 2 2 + x 13 + x 14 + x 12 = 1 - x 57 - x 67 = -1 + x 34 + x 35 - x 43 - x 13 = 0 + x 42 + x 43 + x 45 + x 46 - x 14 - x 24 - x 34 = 0 …. ….. Min Z = + 5x 12 + 4x 13 + 3x 14 + 2x 24 + 6x 26 + 2x 34 + 3x 35 + 2x 43 + 2x 42 + 5x 45 + 4x 46 + 3x 56 + 2x 57 + 3x 65 + 2x 67

33 33 Ardavan Asef-Vaziri June-2013Integer Programming 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 3 4 6 6 O D 2 3 6 5 3 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

34 34 Ardavan Asef-Vaziri June-2013Integer Programming 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. The ShR Problem : Binary Decision Variables 1 43 2

35 35 Ardavan Asef-Vaziri June-2013Integer Programming 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 follows X 23 +X 24 = X 12 + X 32 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 X 23 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. The ShR Problem : Binary Decision Variables

36 36 Ardavan Asef-Vaziri June-2013Integer Programming As usual we can have the following variables X 12 X 13 X 24 X 34 each corresponding to a directed arc The solution of (X 12 = 1, X 23 = 1, X 34 = 1 all other variables are 0) means that the shortest route is 1-2-3-4. The solution of (X 13 = 1, X 32 = 1, X 24 = 1 all other variables are 0) ) means that the shortest route is 1-3-2-4. The solution of (X 13 = 1, X 34 = 1 all other variables are 0) ) means that the shortest route is 1-3-4. The ShR Problem : Smaller Number of Binary Variables

37 37 Ardavan Asef-Vaziri June-2013Integer Programming 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

38 38 Ardavan Asef-Vaziri June-2013Integer Programming How Many Binary Variables for this ShR problem 8 3 4 5 7 10 4 3 5 6 4 5 3 2 2 1 26 9 11 6 2 4 3 4 6 6 O D 2 3 6 5 3 2 3 2 2

39 39 Ardavan Asef-Vaziri June-2013Integer Programming 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

40 40 Ardavan Asef-Vaziri June-2013Integer Programming  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

41 41 Ardavan Asef-Vaziri June-2013Integer Programming How Integer Programs are solved

42 42 Ardavan Asef-Vaziri June-2013Integer Programming Spreadsheet Solution to Example #1

43 43 Ardavan Asef-Vaziri June-2013Integer Programming 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 7 6 9 10 11 12 8 5 13 14 15 16 17 18 19 20 21 22 23 25 24 26 27 28 2930 31 32 33 34 35 36 37 Counties 1. Clallum 2. Jefferson 3. Grays Harbor 4. Pacific 5. Wahkiakum 6. Kitsap 7. Mason 8. Thurston 9. Whatcom 10. Skagit 11. Snohomish 12. King 13. Pierce 14. Lewis 15. Cowlitz 16. Clark 17. Skamania 18. Okanogan 19. Chelan 20. Douglas 21. Kittitas 22. Grant 23. Yakima 24. Klickitat 25. Benton 26. Ferry 27. Stevens 28. Pend Oreille 29. Lincoln 30. Spokane 31. Adams 32. Whitman 33. Franklin 34. Walla Walla 35. Columbia 36. Garfield 37. Asotin Example #2 (Set Covering Problem)

44 44 Ardavan Asef-Vaziri June-2013Integer Programming Counties 1. Clallum 2. Jefferson 3. Grays Harbor 4. Pacific 5. Wahkiakum 6. Kitsap 7. Mason 8. Thurston 9. Whatcom 10. Skagit 11. Snohomish 12. King 13. Pierce 14. Lewis 15. Cowlitz 16. Clark 17. Skamania 18. Okanogan 19. Chelan 20. Douglas 21. Kittitas 22. Grant 23. Yakima 24. Klickitat 25. Benton 26. Ferry 27. Stevens 28. Pend Oreille 29. Lincoln 30. Spokane 31. Adams 32. Whitman 33. Franklin 34. Walla Walla 35. Columbia 36. Garfield 37. Asotin Spreadsheet Solution to Example #2

45 45 Ardavan Asef-Vaziri June-2013Integer Programming Spreadsheet Solution to Example #2

46 46 Ardavan Asef-Vaziri June-2013Integer Programming Spreadsheet Solution to Example #2 (Formulas)


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

Similar presentations


Ads by Google