1. On approximating a geometric prize-collecting traveling salesman problem with time windows Reuven Bar-Yehuda Guy Even Shimon (Moni) Shahar

2. 7:00-5:00 7:00-6:00 12:00-8:00 6:00-7:00 leave home at 5:00 get back at 20:00 10:00-11:00 16:00-17:00 17:00-18:00 0.5 hour rest 1 hour 2 hours 1.5 hours 2 hours 4 hours 2 hours Motivation – nurse visiting patients

3. Prize-collecting TSP with time windows Definition: –Sites in a metric space (e.g. the plane). –A time-interval for each site (release-time, deadline). –Moving agent with speed in [0,1]. –Goal: max #sites the agent visits on-time. –Extra: service-time per site.

4. Known results: scheduling with locations Feasibility is NPC even for points on a line [Tsitsiklis92]. Polynomial algorithm for the case where all intervals are [0,t_i] (using dynamic programming) [Khanna] Min makespan (completion time of last job): –1.5-approx for points on a line with release times, processing times, and no deadlines [KNI98]. –2-approx for points on a line, no deadlines, multiple agents (vehicles) [KN01]. –PTAS for trees with O(1) leaves, single & multiple agents [AS02].

5. Our results Logarithmic approximation for points on a line. Optimal algorithm for points in any metric space, for special case: no round-trips within time windows. Heuristic for any metric space. Achieves an approximation ratio that depends on a “density” measure.

6. Points on a line Speed in [0,1], so dist(A,B) = amount of time required to travel from A to B. Construct the following 2D view of an instance: X t speed = 1/slope  Slope in 90 0 +[-45 0,+45 0 ] Simplify tour: discrete speed in {-1,0,1}.

7. Point on a line (cont.) Now we rotate the view by 45 0, and obtain a weakly x-monotone tour {0 0,45 0,90 0 }. t-X t+X  2-D generalization of the “max monotone subsequence” problem

8. An 8-approx for unit intervals Construct a 1/  2 -square grid. Each interval intersects exactly 2 grid lines. 1 1 1 1 1 1 1 11 1 1 2 1 Direct the grid up & right. Assign edge weights (#intersecting intervals). Find a longest path on the obtained DAG. An instance (slanted intervals in the plane).

9. Analysis - definitions horizontal slice is the region defined by two consecutive horizontal lines. vertical slice is the region defined by two consecutive vertical lines. C i dominates C j if: Right(C i )  Left(C j ) && Top (C i )  Bott(C j ) Domination ensures that concatenation of sub-tours is feasible. CiCi CjCj

10. Analysis (cont.) THM: The longest path in the graph intersects no less than OPT/8 intervals. Each interval intersects 2 grid- segments  weight(path) : wrong by at most a factor of 2. Segment counted twice

11. Analysis (cont.) “decompose” OPT into alternating horizontal and vertical blocks. (each block within a slice). Sub-tours through red (blue) blocks can be concatenated.  If there is a grid-path that is 4-approx in each block, pick red or blue blocks. This gives an 8-approx. Optimal path

12. Analysis (cont.) Select either left-top path or bottom-right path in the DAG. Every interval in block must intersect the perimeter  OPT intersects at most 2 times more intervals than the better path.  Error in weight of path is at most by a factor of 2. Overall: 8-approximation OPT in a block

13. Remarks For [1,2) intervals this is a 16-apx algorithm. This can be extended to a 16log(I)-approx. alg. for arbitrary intervals, where I = I max /I min. The size of the DAG is weakly polynomial. It becomes strongly poly if we consider only grid lines that intersect some interval.

14. An O(log n) approximation If I = I max /I min is is super-poly, then O(log I)> O(log n). We need a different approach to divide the intervals into “disjoint-sets”. We use an interval tree for this partition.

15. Recursive bisection Claim:  separating vertical line (at most half the intervals lie strictly on each side).

16. Recursive bisection (cont.) bisect recursively  log(n) “combs” Level 1 Level 2 2nd comb A comb defines subset of intervals that intersect exactly one comb-tooth.  comb C i such that: C i  OPT contains at least OPT/log n intervals.

17. O(log(n)) Approximation Partition the intervals into (log n) combs. For each comb, 12-approximate an optimal tour.  O(log(n)) approximation.

18. Approximation for comb Form a grid. Construct a DAG. Find the longest path. An interval crosses exactly 1 vertical line. An interval can cross many horz lines (but a grid-path crosses at most 2).

19. Approx ratio = 12 Decompose OPT into alternating sub-tours: –horizontal sub-tours inside a slice –vertical sub-tours between two comb teeth –Note the slices are non-uniform, but the proof still holds.

20. No round-trips within time windows DEF: for all pairs (v i,v j ): dist(v i,v j ) + dist(v j,v i ) > interval (v i ). Service time: dist(v i,v j ) + dist(v j,v i ) + serv(v j ) > interval(v i ). IDEA: can’t zig-zag between sites.

21. Dynamic program for no round-trips within time windows Intuition: examine two tours A,B such that: Both visit k sites Both end at site v i. tour A ends at time t A. tour B ends at time t B. t A < t B  A+best-aug(A)  B+best-aug(B) Proof:  aug(B) is also aug(A). Prize-collecting is additive because aug(B) cannot visit sites visited by A.

22. Dynamic program for no round-trips within time windows (cont.) The algorithm works in layers, where in each layer, it keeps a list of states. A state in layer i is a pair (v,t) which signifies: a path that ends in v at time t after visiting i sites. Layer 0: (origin, t=0)

23. Dynamic program for no round-trips within time windows (cont.) Layer i+1 (induction step): For each state (v i,t) in Layer i go from (v i,t) to all v j. Consider the state (v j,t+dist(v i,C) After deadline? Don’t add to layer i+1. On time? is it earlier than previous states that end in v j ? If so replace. Before release time? Use release time instead of arrival time.

24. Analysis of the dynamic program Correctness: lexicographic order all optimal tours by considering vector of service times. observation: algorithm computes an optimal tour that is minimal in this order. Running time: each layer has at most one interesting point per interval. each point produces at most n candidates for next layers. number of layers is at most n.  The running time is O(n 3 ).

25. Generalized no-round trip The density of an instance is:  (  ) = max uv {I(u) / (L uv + L vu ) } –Density bounds number of zig-zags. –The approximation ratio of the dynamic program is:   (  )  +1. –This follows from the fact that between every two visits of the same site, at least 1/  (  ) of its length passed. Non-unit profits: the problem is knapsack hard, and an approximation scheme of (1+  ) is possible.

26. Remarks This algorithm works also for asymmetric distances. It shows that difficulty is due to “close” intervals.

27. Further research Improve approximation ratio in 1-D. Nothing known for 2-D.

28. The end