Scheduling Conflicting Jobs Problems and Techniques Guy Kortsarz

Scheduling Conflicting Jobs: Problems and Techniques Guy Kortsarz, Rutgers University, Camden

Scheduling dependent jobs Jobs compete on resources Create a graph. Each job is a vertex. Two vertices are adjacent if dependent Two possibilities: Single unit jobs Jobs that require more than one unit of processing Two conflicting jobs can not be executed at the same time unit

Relation of single units jobs to graph coloring • • Given a graph G, find a mapping : V N so that if (v)= (u) then uv E Objective, usually, is to minimize the number of colors used. Classic applications: Timetabling, frequency allocation 3

Multicoloring bipartite graphs 3 4 2 2 1 4

Multicoloring bipartite graphs 3 4 2 2 1 4

Formal definition of the multicoloring problem • Input: Graph G=(V, E), with lengths x(v) on the vertices • Output: An assignment : V 2 N such that adjacent vertices do not receive overlapping times. (v) (u) implies that uv E • Goal: minimize the maximum integer: • Minimize Maxu V { f (u) } • f (u)- The largest color of u.

Multicoloring is often easy 6 -clique More generally, chordal graphs, even perfect, are no harder to multicolor than to color. Maybe one reason why multicolorings are not more common. 7

Some other objectives than makespan (= number of colors) • • Sum of completion times of jobs – For each vertex, count the last time unit assigned, and sum these values up Weighted sum of completion times – Vertices additionally have importance value attached Total lateness – Assumes deadline for each task Sum of flow times – Assumes release time of each job

“Sum of completion times” in Graph coloring? • • Recall we color with numbers, : V N Sum of a coloring = Sum of the values assigned to the vertices Sum coloring problem: – Find a coloring that minimizes the chromatic sum, This measure is more favorable to the users (as a whole), while the makespan is desired by the system (machines) 9

The Sum Multicoloring problem Ø Each vertex v V requires x(v) 1 distinct colors. Ø A proper coloring of G is a function Ψ: V → 2 N, such that adjacent vertices are assigned distinct sets of numbers (colors). Let fΨ(v) denote the maximum color assigned to v by Ψ. Ø The sum multicoloring (SMC) problem: find a multicoloring Ψ that minimizes v V fΨ(v). • The preemptive model (p. SMC): each vertex can get any set of colors. • The no-preemption model (np. SMC): assign to each vertex a contiguous set of colors. 10

Is “Sum coloring” any different from ordinary coloring? YES! 12 1 2 3 11 11

Ok, isn’t at least the sum of an ordinary coloring “good enough”? Factor k for kchromatic graphs. Any 3 -coloring has sum of 2 n. . . while a certain 4 -coloring has sum of n+6 12

Applications • Wire minimization in VLSI design; The conflict graph is an interval graph • Train scheduling; Permutation graphs • Minimizing distance in storage allocation; Interval graphs • Session scheduling in path arrays • Biprocessor scheduling; Line graphs • Data migration; Line graphs • Dynamic storage allocation; Interval graph • Open shop scheduling; Line graphs of bipartite graphs

Some interesting results in sum coloring; Upper bounds • • • NP-hard on planar graphs, but approximation scheme (PTAS) exists. Halldorsson, K. Bipartite graphs: 27/26 approximation. The following problem is polynomial: Color the vertices by 3 colors, so that the sum is minimized but the third color-class may not be an independent set. M. Malafiejski , K. Giaro, R. Janczewski and M. Kubale Approximable within factor 4, with an oracle for Max Independent Set It was well known that produces O(log n) colors Bar-Noy, Bellare, Halldorsson, Schachnai, Shapira

Some hardness results • APX-hard on bipartite graphs. [Bar-Noy, K. 98] • Sum coloring is as hard as Graph Coloring in general, i. e. n 1 -o(1) -approximation hardness [Feige-Killian 96, Bar-Noy+ 96] • Therefore, SMC is at least as hard. • APX-hard on interval graphs [Gonen 01] • Open shop scheduling APX-hard; [Hoogoveen, Shurman and Woeginger 98] • Hardness not well understood

Very Few Exact Results • Sum multicoloring non preemptively of trees is in P. [Halldórsson, K, Proskurowski, Salman, Shachnai and Telle] • Preemptive sum multicoloring of Paths is in P. An ICALP paper (!) [Kovács] • Preemptive coloring of trees is NPC (!) [Marx] • See the collection of papers maintained by Daniel Marx for more results: http: //www 2. informatik. huberlin. de/~dmarx/sum. php

Tools and techniques “The craftsman is known by his tools” 17

Techniques Geometric series and randomization Grouping by length Universal colorings Delaying long vertices Reducing to Independent Set, via SC Rounding and scaling Reducing sum to makespann LP techniques

A simple guessing game • • • Player A decides on a number x. Player B tries a sequence x 1, x 2, . . . , of guesses until it finds xi that Player A says satisfies xi ≥ x. The value of the game is the performance ratio 19

A simple deterministic strategy • Guess 1, 2, 4, 8, 16, . . . • Performance ratio of 4: – The last number is at most 2 x – The next-to-last number is less than x – The previous numbers are a geometric series, at most x. • Bad instance: – x=2 k+1, then =(2 k+2 -1)/(2 k+1) 4 • This is also best possible. . . deterministic.

Lower bound on a deterministic algorithm • We show that no deterministic approximation can have ratio better than 4 - • Let >0 be a small enough constant • xi= i j i-1 xj; • xi+1=µi xi

Assumptions • The start point assumption: x 1 max{4, 1/ } Otherwise, wait for the guess to be large enough • We may assume that i 4. Otherwise the adversary waits a bit and then stops. Declares x=xi+1. • The ratio is (4 xi+xi)/(xi+1)=5 xi/(xi+1) 4 The last inequality follows as x 1 4

Strategy for an adversary • Stopping condition: If at some iteration µi (1 - /8)(1+ i) the adversary stops and says x=xi+1 • Else: let r=x 2/x 1 and =8 ln (3/r)/ • If for +2 times the stopping condition does not apply then the adversary stops • Answers: your last guess xq equals x

Analysis • • If at one of the +2 iterations µi (1 - /8)(1+ i) The ratio is: (1/ i +1+µi) xi/(xi+1) 4 - The last inequality is proved using: 1) small enough 2) i 4 3) x 1 1/ 2 4) x 2+1/x 2 for every x

If +2 consecutive failures • For every i: i+1=µi i/( i+1) (1 - /8) i • Thus +2<(1 - /8) <1/3 • Thus: 3 xq< j q-1 xj • The ratio is at least: (xq+3 xq)/xq=4

A randomized strategy • Defeat the worst-case instance, by randomizing the initial guess. • Work in the log scale – Choose an arithmetic series with a fixed multiple; for instance, e. – Randomize the starting point, R[0, 1) – Define guess xi = ei+ 0 +1 +2 +3 log length +4

Analysis of randomized strategy • Write x=et, or t = ln x and let i=i be such that i-1+ ≤ t ≤ i+. • As ranges from 0 to 1, i+ -t also ranges uniformly from 0 to 1. i-2 • i-1 t i Thus, 27

Analysis, cont. • The expected amount spent by the algorithm is then • Hence, a performance ratio of 2. 71 28

Applications of strategy • Geometric series has been used on different measures for different problems – Lengths of vertices [BBHST’ 98] Non-preemptive SMC of bipartite graphs – Maximum k-colorable subgraph [HKS’ 01] Sum coloring interval graphs – LP values of vertices [GHKS, WAOA’ 04] SMC of line graphs (pre and non-pre) and interval graphs (non-pre) • All, in some sense, are lower bounds on the optimal solution

Grouping by length • The challenge of multicoloring is that vertices can have widely divergent lengths • Handling separately vertices with roughly similar lengths often simplifies the problem • It is also natural to expect that vertices of like length go together

Grouping by length • Divide the real line into segments: V 1 V 2 V 3 V 4 V 5 length • • Vertices of lengths that fall within the same induce a subproblem that is solved separately. Subsolutions are pasted together in order to produce final solution

How to group VSMALL VBIG length • Cost of whole (given that we break)= Cost of coloring VSMALL + | VBIG | * #colors used on VSMALL + Cost of coloring VBIG • Avoid x x xx SMALL x x BIG

The Markov Inequality: Several Break Points • The Markov inequality: Given a collection of n positive numbers with average µ : • |{ ai | ai 2µ }| n/2 • |{ ai | ai 3µ }| n/3 • What happens if we consider all break points r, r+1, r+2, ……. , s? Can we get a point where there are less than 1/i numbers that are at least i times the average?

The basic breakpoint lemma Consider the collection of breakpoints r, r+1, …. . , s There exists r j s so that: |{ ai | ai j µ }| n/(j ln(s/r))

Breakpoint lemma [Halldorsson, K ’ 98] • Can break into size groups BIG and SMALL such that – The largest item in SMALL is negligible in comparison with the average item in BIG • Thus, if the number of colors used is proportional to the largest item, then the cost of breaking up is minor – Holds for constant-colorable graphs Breakup overhead = | VBIG | * #colors used on VSMALL

Breakpoint Lemma • For any q, there is a sequence of breakpoints bi satisfying such that the total breakpoint overhead is at most b 1 b 2 b 3 b 4 36

PTAS for planar graphs of roughly equal length vertices • • Let [a, b] be the range of vertex lengths Round the lengths of the jobs to a multiple of a factor overhead • Scale the lengths by a factor [ -2 , -1 b/a] No overhead for nonpreemptive problems • Baker decomposition into k-outerplanar Gk and outerplanar and small GO. • Solve optimally by dp, but truncate the coloring after (1/ )b/a colors. Cost ≤ OPT • Finish by 4 -coloring GO and remaining vertices Cost of O(n/k)* c (1/ )b/a ≤ n/c when k >> -2 (b/a).

Input graph G Round and scale, by factor of a Baker’s decomposition Solve optimally by DP Color with minimal amount of colors Retain the (1/ )b/a smallest colors Combine Output a coloring 38

Preemptive multicoloring of planar graphs • A general tool for O(1) makspan coloring: preemptive scaling. We are able to reduce job lengths paying only (1+ ) • Claim 1: Say that all x(v) are divisible by q. Then then if x(v)/q c ln n for every v then: (I) q (I/q) (1+ ) (I)

Proof • Take the optimum for I • Include every independent set with probability (1+ ’)/q in a solution for I/q • With constant probability the makespan is not larger than (1+ ’)(1+ ’’) (I)/q • By the Chernoff bound and the union bound every vertex gets at least x(v)/q scheduling units

A general lemma on reducing jobs lengths • Assume that a graph is constant colorable • It is possible to reduce all job lengths to O(log n) with only (1+ ) penalty • The number of preemption can be reduce to O(log n) • This holds true even if tasks are initially of size exponential in n (!)

Proof • Split jobs to (roughly) log n most significant bits and the rest • The large parts of numbers are all divisible by some large q • Reduce these numbers to log n bits (small loss). The O(log n) is derived • Color firs by round robin the small parts nonpreemptively • Small delay because constant colors and small numbers • Then take the solution to the O(log n) instance • The coloring for x’/q is repeated q times

Planar decomposition • A planar graph G can be decomposed into two graphs G’ and G’’ so that: • G’ has treewidth k 2 • G’’ has treewidth 2, and color requirement sum O(S/k 2) • If we have an approximation for p-SMC for graphs with treewidth k, we can use to get c’ and c’’ for G’ and G’’ • To get the combine coloring: Each k color classes of c’ put one of c’’

Universal family of colorings • Works for graphs that have constant treewidth • It is a family of colorings so that for every instance there exists a coloring in the family that approximates the SMC by (1+ ) factor • The key property: the number of different colorings of v in the family is polynomial in n • This allows finding the best solution via DP if treewidth constant • The existence is proven by modifying OPT

A somewhat surprising fact • The colorings in our family depend only on a specific k-coloring of the graph, on n and on p • In particular, it does not depend on the actual connections in the graph, nor on the distribution of color requirements • Hence the name universal

The universal family • Split the colors of every vertex in powers of (1+ ) • Segment i: colors (1+ )i to (1+ )i+1 • In every segment, treat the coloring as a makespann instance • Thus in every segment O(log n) colors • Make the number of segments in which a vertex is colored, constant

How to bound # of segments • For every vertex v, its colors that are smaller than x(v) or larger than (2/ )x(v) are removed from OPT • Instead they are replace by round robins that are performed every (roughly) 1/ rounds • The key: the number of “non-standard” segments for every v is O(log 1+ ( 2 ) )

The number of preemption for v In every segment v has O(log n) preemptions • # colors per segment clog n. Choose c’ log n of them for v • The number of possibilities for v is O((2 clog n) f( ) ) hence polynomial in n • The round robin, executed only every 1/ rounds and adds O( opt)

Delay of large jobs • This technique is illustrated via its Application on two classical problems: • 1) Data migration • 2) Open shop scheduling

Data migration pe pe pe • • pe pe Storage Area Networks = length of data transfer along edge e At most one active transfer per storage device Minimize (weighted) sum of completion times of the storage devices

m 1 3 Open Shop Scheduling m 2 4 m 4 2 wt=40 Jobs m 1 2 m 3 5 m 4 3 m 1 m 2 wt=30 Out In m 1 4 m 2 1 wt=15 m 3 6 m 2 2 m 3 2 m 4 2 wt=100 The Shop m 3 m 4

Graph formulation of Open Shop w 1 = 0 0 v 1 v 2 v 3 v 4 Machines v 6 v 7 v 8 Jobs 30 100 15 pe=3 4 v 5 2 w 5= 40 • Assign interval (Ce-pe, Ce] to each edge e; intervals of adjacent edges non-overlapping; minimizing v V wv. Cv (Cv = maxv e Ce ) • minimizing e E we. Ce operation completion sum is the SMC problem in bipartite line graphs

History Data Migration • Coffman+ ’ 85 • Kim ’ 03 • GHKS 05 2 9 5. 06 makespan v wv C v v wv Cv, rv Open shop scheduling • • Chakrabarti+ ’ 96 Hoogeveen+ ’ 98 Queyranne+ ’ 02 GHKS 05 5. 78 fixed# APX-hard fixed# 5. 83 5. 06

Problems with greediness 1 1 1 Q Long job delays short ones 1 1 2 2 1 2 Q Maximality can cause long delays 54

Linear programming relaxation • From [Hall+ 96, Kim ´ 03] • Let Ev = the set of edges incident on vertex v • Let p(X) = pe = sum of lengths of edges in set X e X (LP) minimize wv C v v V subject to: Cv Ce, for each edge e and incident vertex v Cv p(Ev), for each vertex v pe Ce ½ [p(Sv)2 + p(Sv 2)], for any Sv Ev e Sv Ce 0, for each edge e • Solvable in poly-time using ellipsoid method [Queyranne] 55

Capacity inequality e 1 e 2 e 3 e 4 e 5 Sv v • When the edges are scheduled in some sequence, without a break: Denote pi = pei p 1 p 2 p 3 p 4 p 5 C 1 C 2 C 3 C 4 C 5 i pi Ci = p 1 p 1 + (p 1+p 2)p 2 +. . +(p 1+p 2+ +pt)pt e Sv = i j pi pj = ½ [( i j pi)2 + i j pi 2] = ½ [p(Sv)2 + p(Sv 2)]

Delay technique An edge is. . . • active for pe steps, once it becomes active. • finished, after being active • delayed, if an adjacent edge is active • waiting, until it has waited We steps • waiting/active/finished, if no active nbor Priority rule: If two adjacent edges can become active, give priority to the smaller LP value

The first delay function We = 1 Ce* is the value of Ce in LP sol. • Theorem: With this weight function and 1 chosen optimally, each vertex v completes within 5. 83 Cv* steps • 5. 83 ratio Same as Queyranne & Sviridenko!

The second delay function Edges w/ LP-value Ce* Se(u) u e Se(v) v We = 2 max (p(Se(u)), p(Se(v)) • Also 5. 83 -ratio on per-vertex basis • 5. 0553 -ratio by trying both weight functions, and using the one that gives a better result

The last edge incident on u that the algorithm completes Proof outline: Ratio 6 Edges e w/ Ae < Aeu Du u eu Dv v • The algorithm’s completion time of u: Au = Aeu = [Time eu is waiting] = Weu = Ceu* Cu* + [eu is active] By LP C u* } + [eu delayed by Du] inequality + [eu delayed by Dv] = p(Dv)

Bounding size by LP values • Let Xt = {e Ev : Ce* t} = edges incident on v with LP-value at most t • [Hall+’ 96] p(Xt) 2 t Proof follows from main LP-inequality. • Tight example: 2 t edges, pe = 1. All Ce*= t 1 1 1

Edge in D with Proof outline: Ratio 6 largest (cont)LP-value. v Du u eu Dv e’ v • Time that eu is delayed by neighbors of v : p(Dv) 2 Ce’* = 2 We’ If e’ is waiting, then edges incident on v are inactive We’ [(eu waiting) + (eu delayed by Dv)] 2 C u* • Algorithm’s completion time of u is then Au = Cu* + p(Dv) 2 Cu* + 2 We’ 6 Cu*

Simulating the algorithm We=pe 3 3 4 1 Waiting Delayed Active Finished 2 1 2 3 63

Simulating the algorithm We=pe 3 3 4 1 Waiting Delayed Active Finished 2 1 2 3 4 5 64

Simulating the algorithm We=pe 3 3 4 1 Waiting Delayed Active Finished 2 1 2 3 4 5 6 65

A stronger property of the LP • Recall Xt = {e Ev : Ce* t} Know p(Xt) 2 t • Suppose p(Xt/2) =t, what is then p(Xt) = t • New: p(Xt) t + sqrt(t 2 – 2 (t – t/c)) where = p(Xt/c) • E. g. if p(Xt/2) =t, then p(Xt) = t ! • Or, if p(Xt) =2 t, then p(Xt/2) = 0 !

Combining weight functions 6 1= 2=1 h( )=2 max(1, )+2 5 f( )= (4 -2 )+4 4 0 1 2 = p(Se(v) Dv )/Cu* 67

The Effect of Lambda Du u eu Dv v • If = 2 then only “short” edges are responsible for the delay of eu. So, by the stronger LP property p(Dv) 2 Cu* for a ratio of 4

Reducing sum multicoloring to makespann • A general technique to reduce sum problems to makespann problems • Leads to the first constant approximation for SMC non-preemptively interval graphs • Independently invented by Queyranne and Sviridenko (a few months before us). But we gave new applications

The Generic Algorithm ACS(G, q) r uniformly random in (0, 1]. i 0, mi 1 while (G ) do ci max(1, qi+r ) Gi Dual. Thruput(G, ci) Approximate throughput with resource augmentation Color Gi next (with new colors mi, mi+1, …, mi+ ci-1) G G- Gi, mi+1 mi+ ci, , , i i+1 end =d* 70

Approximating Non-preemptive SMC Overview: • Formulate np. SMC as an integer program. Find an optimal fractional solution, f*(v), for all v V. Let Ft be the set of vertices with f*(v) t, then f*(v) satisfies Property 1: max v Ft f*(v) (Ft)/d, for some d 1, where (Ft) is the maximum weight of any clique in Ft. • Partition the time axis into intervals (cm-1, cm], m=1, 2…, M, and let Vm={v V: f*(v) (cm-1, cm] }. • Let A be an approximation algorithm for graph multicoloring that uses (Vm) colors for Vm. 71

Approximating np. SMC (Cont’d) • Assign colors to the vertices in Vm using algorithm A. • Let u be the vertex satisfying f*(u)= maxv Vm f*(v), then, by Property 1, f*(u) (Vm)/d, for some d 1; the largest color assigned to any vertex v by A is fv (Vm) d f*(u) • Suppose that cm /cm-1 , for some 1, then we get a d – approximation for np. SMC. In general, we can argue that d=2. An algorithm of [Gergov 1996] achieves the ratio =3 for multicoloring interval graphs. By optimizing on we have: Theorem: There is a 11. 273 approximation algorithm for np. SMC on interval graphs. 72

Application for sum-coloring • The sum set-cover problem was studied by Feige, Lovats and Tetali. The motivation: Heuristics for speeding Semidefinite solvers • Input: G(A, B, E) as in Set-cover • Output: An ordered subset S={a 1, …, ak} A and a function : B S so that (b, (b)) E • The input is as in set-cover and the output too except that the set-cover is ordered

Objective function • Let Bi={b B | (b)=ai} • Intuitively, Bi the set of ``jobs” covered at time i. • Objective function: Minimize: i i |Bi| • Example: If vertices in A represent independent sets in a graph, we get sum-coloring

Approximation results • Bar-Noy, Bellare, Sachnai and Shapira showed: The algorithm that iteratively picks an maximum independent set gives ratio 4 (if IS can be found in poly time) • The same proof implies: the classic greedy algorithm has ratio 4 approximation for minimum set-cover

Lower bound • Bar-Noy, Halldorsson, K. There is a (rather complex) example where the greedy maximum independent set algorithm has indeed ratio 4 • Feige, Lovats and Tetali: The lower bound for greedy MIS of [BHK] can be adapted to show that unless P=NP no 4 - ratio is possible for sum set-cover for any constant

Better result for sum-coloring • The following was shown by Gandhi, Halldorsson, K. and Shachnai • The reduction from sum to makespann gives an 3. 591 approximation algorithm for sum-coloring assuming a maximum independent set can be found • This ratio can be apllied for all PERFECT graphs

Open problems • For graphs that admit a polynomial time exact algorithm for maximum independent set is there an O(1) for np. SMC on this class? Example: Chordal graphs(? ) • Lower bounds (maybe 4? ? ) for open shop scheduling? The only APX hardness has a very small constant [Hoogeveen, Schurman and Woeginger 98] • Can minimum makespann of PLANAR graphs be approximated within 4/3? Ratio 2 easy: 4 -colorable, Bip graph solvable