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1 Material to Cover  relationship between different types of models  incorrect to round real to integer variables  logical relationship: site selection.

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Presentation on theme: "1 Material to Cover  relationship between different types of models  incorrect to round real to integer variables  logical relationship: site selection."— Presentation transcript:

1 1 Material to Cover  relationship between different types of models  incorrect to round real to integer variables  logical relationship: site selection  weak and strong formulation: uncapacitated facility location problem  set covering problems: airline crew scheduling  generalized piecewise linear approximation

2 2  max/min z = c 1 x 1 + c 2 x 2 + … + c n x n  s.t.  a i1 x 1 + a i2 x 2 + … + a in x n b i, i = 1,…, m  0  x j  u j, j = 1,…, n  x j integer for some or all j =1,…, n {    } Linear Integer Programming - IP

3 3  mixed IP (MIP): some x j  Z #1, some x j   #2  pure IP: all x j  Z  binary decision variable: x j = 1 or 0 (e.g., a variable for a yes-no decision)  binary IP (BIP): all x j being binary Linear Integer Programming - IP #1 Z: the set of integers; Z + : the set of positive integers #2  : the set of real numbers ;  + : the set of positive real numbers

4 4 Motivation of Studying IP  integer variables in some context  e.g., machine, manpower  logical relationship  incorrect to round continuous variables

5 5 Incorrect to Round Continuous Variables                     optimal LP solution X1X1 X2X2 optimal IP solution iso-cost line    usually all right to round in real life problems with large x i

6 6 Example to Motivate IP  site selection: three designs A, B, C on sites 1, 2, 3, 4  total amount for investment: $100 M  how to invest? OptionA1A1A2A2A3A3A4A4B1B1B2B2B3B3B4B4C1C1C2C2C3C3C4C4 Net Income ($M) Investment ($M)

7 7 Example to Motivate IP  I = {A, B, C, D}, J = {1, 2, 3, 4}  y ij = 1 iff design i used at site j, i  I, j  J  max z =  i  I  p ij y ij  s.t.  i  I  j  J a ij y ij  100  y ij  {0, 1}, i  I, j  J  optimal solution: y A1 = y A3 = y B3 = y B4 = y C1 = 1; z * = 40

8 8 Example to Motivate IP  boss says NO!  at most one design at a site  a building at site 2 (required)  at most two designs at the three sites  design A considered for sites 1, 2, and 3 only if being used at site 4  how to model?

9 9 Example to Motivate IP  at most one design at a site and a building at site 2 (required)  y A1 + y B1 + y C1  1, y A2 + y B2 + y C2 = 1,  y A3 + y B3 + y C3  1, y A4 + y B4 + y C4  1  design A considered for sites 1, 2, and 3 only if being used at site 4  y A1 + y A2 + y A3  3y A4

10 10 Example to Motivate IP  at most two designs at the three sites  w i = 1, if design i is used, = 0, o.w., i = A, B, C  w A + w B + w C  2  y i1 +y i2 +y i3 +y i4  4w i, i = A, B, C  optimal solution: y A1 = y A4 = y B2 = y B3 = 1; others = 0; z * = 37

11 11 Logical Constraints for Variables  n situations how to model (i) at most k of them hold, (ii) at least k of them hold, and (iii) exactly k hold  y j binary variables for j = 1 to n; y j = 1 if j holds, and = 0 otherwise  mutually exclusive y j : y 1 + y 2 + … + y n  1  at most k of y j = 1: y 1 + y 2 + … + y n  k  at least k of y j = 1: y 1 + y 2 + … + y n  k  exactly k of y j = 1: y 1 + y 2 + … + y n = k

12 12 Logical Constraints for Expressions  either-or constraints  either f 1 (x 1, …, x n )  b 1 or f 2 (x 1, …, x n )  b 2 or both  IP formulation: let y be a binary variable  M: a large positive number, practically “  ”  two constraints: f 1 (x 1, …, x n )  b 1 +My and f 2 (x 1, …, x n )  b 2 +M(1-y)  only one of these two being picked by the optimization procedure

13 13 Logical Constraints for Expressions  m constraints, at least k out of m being true  f 1 (x 1, …, x n )  b 1, …, f m (x 1, …, x n )  b m  modeling procedure  m binary variables y i, one for each constraint  f 1 (x 1, …, x n )  b 1 +M(1-y 1 ), …, f m (x 1, …, x n )  b m +M(1-y m )  y 1 + … + y m  k

14 14 An Example of Logical Constraints for Expressions  single processor for three jobs, of processing times 3 hr, 5 hr, and 7 hr, respectively  objective: minimizing the total completion time of the three jobs  how to formulate it as an integer program?  note. The IP is for the illustration of formulation. The problem has very simple solution.

15 15 Definitions of Parameters and Variables  s i : the processing start time of job i  c i : the completion time of job i  p i : the processing time of job i (i.e., p 1 = 3, p 2 = 5, p 3 = 7)  C:total completion time

16 16 How About This?  if job 1 before job 2, and before job 3, C = 26  if job 1 before job 3, and before job 2, C = 28  if job 2 before job 1, and before job 3, C = 28  if job 2 before job 3, and before job 1, C = 32  if job 3 before job 1, and before job 2, C = 32  if job 3 before job 2, and before job 1, C = 34

17 17 How About This?  s 1  s 2  s 3, C = 26  if (y 12 =1 & y 23 =1), C = 26  s 1  s 3  s 2, C = 28  if (y 13 =1 & y 32 =1), C = 26  s 2  s 1  s 3, C = 28  if (y 21 =1 & y 13 =1), C = 28  s 2  s 3  s 1, C = 32  if (y 23 =1 & y 31 =1), C = 32  s 3  s 1  s 2, C = 32  if (y 31 =1 & y 12 =1), C = 32  s 3  s 2  s 1, C = 34  if (y 32 =1 & y 21 =1), C = 34  then setting conditions on y ij …, which obviously not working

18 18 The Formulation  min c 1 + c 2 + c 3,  s.t. c 1 -s 1 = 3;c 2 -s 2 = 5;c 3 -s 3 = 7;  one of c 1  s 2 and c 2  s 1 holds;  one of c 1  s 3 and c 3  s 1 holds;  one of c 2  s 3 and c 3  s 2 holds;  c i  0, s i  0, i = 1, 2, 3.

19 19 The Formulation  min c 1 + c 2 + c 3,  s.t. c 1 -s 1 = 3;c 2 -s 2 = 5;c 3 -s 3 = 7;  c 1  s 2 +My 12 ; c 2  s 1 +My 21 ; y 12 +y 21 = 1;  c 1  s 3 +My 13 ; c 3  s 1 +My 31 ; y 13 +y 31 = 1;  c 2  s 3 +My 23 ; c 3  s 2 +My 32 ; y 23 +y 32 = 1;  c i  0, s i  0, i = 1, 2, 3;  y ij  {0, 1}, i  j, and i, j = 1, 2, 3 y ij = 1 if job j is before job i.

20 20 Equivalence Between BIP and General PIP  BIP  PIP  BIP  PIP  a PIP of bounded integer variables  BIP  max5x 1 + 2x 2  s.t.2x 1 + x 2  15  x 1  0, x 2  Z +  conversion  0  x 2  15  x 2 = y 1 + 2y 2 + 4y 3 + 8y 4, y i  binary

21 21 Fixed-Charge Problem  costs for having a facility at site j, j = 1 to n  set up cost k j  variable cost c j per unit of capacity  capacity of the whole system  C  minimum cost site selection for the capacity constraint

22 22 Fixed-Charge Problem n j=1 min  f j (x j ) where f j (x j ) = { k j + c j x j, if x j > 0 0, if x j = 0 k j = set-up cost, c j = per unit cost IP formulation: min n j =1  ( c j x j + k j y j ) s.t.s.t. x j  My j, j = 1, …, n;  j x j  C; y j  {0,1}, j = 1, …, n ; x j  0, j = 1, …, n ;

23 23 A More Realistic Fixed-Charge Problem  telecommunication network  source nodes S = {1, 3, 7}; destination nodes D = {2, 4, 5, 8}; transshipment node T = {6}  solid links: existing; dotted links: planned; total A = {1, 2, …, 17}  each link: (capacity, cost)  planned links A’ = {1, 2, …, 5}; fixed costs f 1 = 8; f 2 = 6; f 3 = 9, f 4 = 7, f 5 = 7  min cost construction to satisfy the demands & flows

24 24 A More Realistic Fixed-Charge Problem  decision variables  x k : the amount of flow in link k  y k : the binary variable of constructing link k  A'  parameters  b i : the demand of node i (source = -demand)  u k : the upper bound of flow of link k  f k :the fixed cost coefficient  c k :the variable cost coefficient

25 25 A More Realistic Fixed-Charge Problem  min z =  k  A’ f k y k +  k  A c k x k  s.t.  total in-flow – total out-flow = b i, conservation of flow  node i  x k  u k y k, capacity constraint  (proposed) arc k  A'  x k  u k, capacity constraint  (existing) arc k  A  y k = 0 or 1, binary variable  (proposed) arc k  A'  x k  0,  arc k  A  A'

26 26 Facility Location Problem  distributing goods to n customers possibly through m warehouses  warehouse i  fixed cost f i  variable cost v i per unit capacity  maximum capacity u i  shipment cost c ij per unit from warehouse i to customer j C1C1... Cn W1W1... Wm

27 27 Facility Location Problem  distributing goods to n customers possibly through m warehouses  data  d j : demand for customer j  u i : maximum capacity at warehouse i  f i : fixed cost to build warehouse i  v i : variable cost/unit of capacity of warehouse i  c ij : variable cost/unit of goods sent from warehouse i to customer j  decision variables  y i : build a warehouse at site i? (1 = yes, 0 = no)  z i : capacity (supply) of warehouse i  x ij : shipment from warehouse i to customer j

28 28 Facility Location Problem

29 29 Strong versus Weak Formulation – An Illustration through the Facility Location Problem  uncapacitated version of the facility location problem  intuitively optimal to have each customer satisfied by one warehouse  simplified the formulation  re-definition  c ij : shipment cost to customer j, possibly including the variable cost of operating warehouse i for demand d j  x ij : proportion of demand j satisfied by warehouse i

30 30 Weak Formulation

31 31 Strong Formulation

32 32 Comparison of the Strong and Weak Formulations  strong: more constraints  x ij  y i, i = 1, …, m; j = 1, …, n  weak: less constraints   i x ij  ny i, j = 1, …, n  which one is better?  strong: more precise constraints and possibly shorter computation time

33 33 Covering Problems and Partitioning Problems  S:a set of m items  S j :a subset of S that includes one or more of the items, j = 1, …, n  c j :the cost of selecting subset j  selecting the minimum cost collection of subsets S j to include elements of S  set covering: fine as long as including all items of S  set partitioning: each element of S is included exactly once

34 34 Airline Crew Scheduling (Set Covering Problem)  service network  group legs into tours according to constraints LAX SEA CHI DEN DFW

35 35 Airline Crew Scheduling (Set Covering Problem)  a tour: feasible assignment for a crew, starting & ending at DFW  a leg: a flight scheduled between two cities  covering 11 legs by 3 crews on 12 possible tours  minimizing the total tour cost

36 36 Airline Crew Scheduling (Set Covering Problem)  optimal solution: “Dead heading” on first leg Min2x2x 1 +3x3x 2 +4x4x 3 + … +8x8x 11 +9x9x 12

37 37 The Days-Off Scheduling Problem  (5,7)-cycle: 5 working days + 2 consecutive days off  7 days-off patterns  parameters  r i = number of employees required on day i  c j = weekly cost of pattern j per employee  decisions  x j = # of employees using days-off pattern j

38 38 The Days-Off Scheduling Problem  min z =  c j x j  s.t. (  x j ) – x i – x i-1  r i, i = 1,…7  x j  0 and integer, j = 1,…,7; x 0 = x 7 7 j=1 7 j=1  solve problem to get x * j  minimum cost workforce W =  x * j 7 j=1

39 39 The Days-Off Scheduling Problem  possibly to be solved as two LP  compact expression  Minimize z = cx  s.t. x  0 and integer

40 40 The Cutting Stock Problem  raw material: rolls of 25 ft  requirements  5-foot: 40 rolls  8-foot: 35 rolls  12-foot: 30 rolls  15-foot: 25 rolls  17-foot: 20 rolls  objective: using minimum # of 25-foot rolls

41 41 General Piecewise Linear Approximations  f j ( x j ), 0  x j  u j  r = number of grid points  (d ij, f ij ) be i th grid point, i = 1…, r

42 42 Linear Transformation for x j  x j =   i d ij  g j (x j ) =   i f ij    i = 1,  i  0, i = 1,…, r  not sufficient to guarantee that the solution is on one of the line segments r i=1 r i=1 r i=1

43 43 Additional Constraints for Piecewise Linear Approximation  condition: at most two positive  i, and positive  i ‘s adjacent  1 ≤ y 1  i ≤ y i -1 + y i, i = 2,…, r–1  r ≤ y r -1 y 1 + y 2 + · · · + y r -1 = 1 y i = 0 or 1, i = 1,..., r–1  not necessary to define  ’s if minimizing a convex function or maximizing a concave function

44 44 Approximation in Minimizing a Convex Objective Function y x1x1 CBAOpoints

45 45 Approximation in Minimizing a Convex Objective Function all right to omit (4) if approximating a convex objective function in minimization or a concave objective function in maximization

46 46 Approximation in Minimizing a Convex Objective Function all right to omit (4) if approximating a convex objective function in minimization or a concave objective function in maximization

47 47 Special Non-linear Objective Functions  machines: A to D  products: P, Q, R  potential sales: P  100, Q  100, R  100 prod mh time PQR available time A B C D

48 48 Special Non-linear Objective Functions  nonlinear unit profit from the products prod sales PQR  How to formulate the problem?

49 49 Special Non-linear Objective Functions  maxZ = f 1 (P) + f 2 (Q) + f 3 (R)  s.t.20P + 10Q + 10R  2400 (mh A)  12P + 28Q + 16R  2400 (mh B)  15P + 6Q + 16R  2400 (mh C)  10P + 15Q  2400 (mh D)  P  100, Q  100, R  100 (marketing)  P  0, Q  0, R  0  How to model f 1 (P), f 2 (Q), f 3 (R)

50 50 Special Non-linear Objective Functions  P i :# of sales of product P in the ith price range  Q i :# of sales of product Q in the ith price range  R i :# of sales of product R in the ith price range  object function: max  Z = 60P 1 +45P P 3 +40Q 1 +60Q Q 3 +20R 1 +70R 2 +20R 3

51 51 Special Non-linear Objective Functions  Z = 60P 1 +45P P 3 +40Q 1 +60Q Q 3 +20R 1 +70R 2 +20R 3 +  for P  0  P 1  30, 0  P 2  30, 0  P 3  40  concave prices, no additional constraints  for Q  0  Q 1  30, 0  Q 2  30, 0  Q 3  40  use second price segment only if Q 1 = 30  use third price segment only if Q 2 = 30  for R  0  R 1  30, 0  R 2  30, 0  R 3  40  use second and third price segments only if R 1 = 30

52 52 Special Non-linear Objective Functions  for Q  y Q2 = 1 if sales in segment 2 are made  = 0 otherwise  y Q3 = 1 if sales in segment 3 are made  = 0 otherwise  30y Q2  Q 1  30, 30y Q3  Q 2  30y Q2, 0  Q 3  40y Q3  for R  y R2 = 1 if sales in segments 2 or 3 are made  = 0 otherwise  30y R2  R 1  30, 0  R 2  30y R2, 0  R 3  40y R2

53 53 Other Examples  Traveling salesman problems  sequence dependent setup times  assembly line balancing


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