Column generation to solve planning problems

Transcription

1 Column generation to solve planning problems ALGORITMe Han Hoogeveen 1

2 Continuous Knapsack problem We are given n items with integral weight a j ; integral value c j. B is a given integer. Goal: Find a subset of the items with maximum value and total weight B. Assume that it is possible to select an item partially. 2

3 Linear Programming formulation Use variables x j to model how much of item j is selected (x j = 1: take item j completely). max n j=1 c jx j n j=1 a jx j B. (1) 0 x j 1 j = 1,..., n. (2) 3

4 Shadow price. Constraint i; right-hand side value b. Decrease (increase) b by ɛ Change in outcome value (ɛ) Shadow price of this constraint is (ɛ) ɛ Comparable to the left (right) derivative of the outcome value with respect to b. The shadow price signals the unit price you want to pay for a little more. The shadow price is a by-product of the solution. 4

5 Reduced cost. New activity (item) 0; variable x 0. Add x 0 to the problem (with c 0 and a 0 ). Compute net gain (gain minus investment ) of putting x 0 ɛ. Reduced cost x 0 is the net gain per unit. Adding item 0 will improve the LP-solution only if its reduced cost is positive (in case of maximization). 5

6 Column generation. Take any LP-problem (maximization) with large number of variables. Only a few variables will get positive value. Remainder of the variables is lumber. Remedy Solve the LP-problem for a small subset of the variables The remaining variables are ignored (they get value 0). Is it worthwhile to add a variable? Check reduced cost. Smart implementation: pricing problem Maximize the reduced cost over all possible variables. 6

7 The Beunhaas problem. Beun de Haas is an independent entrepreneur. Clients contact him for small jobs. For each job j is given: the reward (c j ); the time it takes (a j ); the days on which they can be done (characterized by binary parameters h jt ). Planning period: days 1,..., T. Beun has Q t time on day t (t = 1,..., T ). Goal. Choose and plan the work to earn as much as possible. 7

8 Standard ILP formulation Use binary variables y jt to signal if job j is done on day t. max n j=1 T t=1 c jy jt T t=1 y jt 1 j (3) n j=1 a jy jt Q t t (4) y jt {0, 1} j, t (5) Preprocessing: y jt = 0 if h jt = 0, for all j, t. 8

9 Advanced ILP formulation Formulation with day plans. A day plan for day t is a set of jobs that Beun can do on day t. Dayplan (j, t) for day t is characterized by parameters g ijt (i = 1,..., n): g ijt = 1 if job i is part of this day plan (and 0 otherwise). The reward of day plan (j, t) is equal to C jt S t is the set of feasible day plans for day t (t = 1,..., T ). Use a binary variable x jt for each day plan (j, t). x jt = 1 if day plan j is the chosen day plan for day t (and 0 otherwise). 9

10 Advanced ILP formulation (2) max T t=1 j S t C jt x jt T t=1 g ijt x jt 1 i (6) j S t x jt 1 t (7) j S t x jt {0, 1} j, t (8) The feasibility of the day plans ensures Beun does not have to work too long per day. Availability of chosen jobs. 10

11 What can we do with it? Downside of this formulation: There are very many day plans possible Not all day plans are known We do not want to enumerate them Most of them are useless anyway. Heuristic approach. Select a number of presumably useful day plans. Solve the ILP formulation for this subset 11

12 How to find this subset? Strategy: Simplify (relax) the problem Solve this relaxation Put the discovered day plans in the subset Solve the original problem Use here the LP-relaxation; solve it using column generation. 12

13 LP relaxation max T t=1 j S t C jt x jt T t=1 g ijt x jt 1 i (9) j S t x jt 1 t (10) j S t x jt 0 j, t (11) Replace x jt {0, 1} by 0 x jt 1 x jt 1 is redundant. 13

14 Reduced cost Consider a day plan (j, t) Characterize the day plan by paramaters g i g i = 1 if job i is included (and 0 otherwise). Reduced cost = net gain: Direct gain of choosing the day plan: n i=1 g i c i Cost of choosing the day plan: you have to pay for the change in right-hand side values 14

15 LP relaxation max T t=1 j S t C jt x jt T t=1 g ijt x jt 1 i π i (12) j S t x jt 1 t λ t (13) j S t x jt 0 j, t (14) 15

16 Pricing problem The reduced cost of this day plan for day t amounts to n i=1 g ic i n i=1 g iπ i λ t = n i=1 g i(c i π i ) λ t If n i=1 g i (c i π i ) λ t > 0, then improvement possible. Assume day t given: λ t is constant (ignore). 16

17 Pricing problem (2) Maximize n i=1 g i (c i π i ) such that (g 1,..., g n ) corresponds to a feasible day plan. Put job i in day plan (g i = 1): Gain c i π i ; Extra work a i in day t. Requirements Beun has Q t time available on day t Job i can be chosen only if it is available on day t. 17

18 Pricing problem for day t Consider all jobs available on day t Select the most valuable set with total processing time Q t Knapsack problem: weight p j value c j π j. Size of knapsack: Q t. 18

19 Improving the solution Use more columns in the ILP: After solving pricing problem for day t: solve the pricing problem again by removing chosen jobs. Store these in a column pool and add them to the ILP. Use branch-and-bound Compute an upper bound from the LPrelaxation Use a branching rule that can be combined with the column generation 19

20 Extensions Beun gets some co-workers Processing times vary per day Travel times between jobs Stochastic processing times 20

21 Travel times The ILP-formulation remains the same Travel times are incorporated in the day plans A day plan becomes an ordered set of jobs Travelling repairman problem No simple algorithm to solve it It is not necessary to solve the pricing problem to optimality Generate good day plans by applying Local Search 21

22 Stochastic processing times The ILP-formulation remains the same Feasibility of a day plan becomes more difficult Beun has Q t time available on day t Use a chance constraint: the probability that Beun needs more than Q t time should be less than a given value. Interesting combination: Stochastic processing times Stochastic travel times 22