Class-constrained Resource Allocation Problems Tami Tamir Thesis advisor: Hadas Shachnai

A Resource Allocation Problem: Devices: Each device has a limited amount of some resource. - A disk with a limited broadcast capacity. - A processor with specific rate. - A machine with a limited amount of material. Clients: that present requests for resources. - Transmission of some video-program file. - Processing of a job that needs s processing units. - Production of a product that needs s units of the material.

A set of N knapsacks, K 1,...,K N. Each knapsack, K j, has limited volume, v j. A set, I, of n items, u 1,...,u n. Each item, u I, has a size, s(u), and a value, p(u). The objective: A legal placement of some U I with maximal value. Legal = For all j, the total size of items placed in K j is at most v j. The Multiple Knapsack Problem (MK) devices clients

An Example (MK) For all u, p(u) = s(u) v= Optimal value = 16 n=

A Resource Allocation Problem: Devices: Each device has a limited amount of some resource, - A disk with a limited broadcast capacity, - A processor with a specific rate, Clients: that present requests for resources, - Transmission of a video-program file. - Processing of a job that needs u processing units. A Class-constrained and a limited number of settings. and a limited storage capacity. and a limited number of configurations. and need specific settings. the video program file Tom & Jerry of type red that needs s processing units.

A set of N knapsacks, K 1,...,K N. Each knapsack, K j, has limited volume, v j, A set, I, of n items, u 1,...,u n. Each item, u I, has a size, s(u), and a value, p(u). The objective: A legal placement of some U I with maximal value. Legal = For all j, the total size of items placed in K j is at most v j, The Multiple Knapsack Problem (MK) The Class-constrained Multiple Knapsack Problem (CCMK) and a limited number, c j, of compartments. a value, p(u), and a type (color), c(u) {1,…,M}. and the number of different colors included in K j is at most c j.

An Example (CCMK) For all u, p(u) = s(u) v=8, c= Optimal value = 15 n=6, M=

Related Work IP formulation of CCMK ( Dawande and Kalagnanam, 1998 ) A PTAS for MK ( Chekuri and Khanna, 2000 ) A dual PTAS for BP ( Hochbaum and Shmoys, 1987, Epstein and Sgall, 1999 ) Cardinality-constrained packing problems - when the number of items is limited (no types) ( Kellerer and Pferschy, 1999; Caprara, Kellerer, Pferschy, and Pisinger, 2000 ) Data placement ( Golubchik, Khanna, Khuller, Thurimella, and Zhu, 2000 ).

Related Work (Cont) Multi-dimensional packing problems: May seem related, but there is a crucial difference: In class-constrained packing - Each bin has the 2-dim size v j c j - Each item has the 2-dim size s(u): s(u) 1 if u is the first item of its color in K j s(u) 0 otherwise s(u) =

A Special Case: Unit-sized Items When each clients request is for one unit of some resource, then in the resulting packing problem each item, u I, has size s(u)= = 24 v = 8 c = 2 An important application: Data Placement for Multimedia-on-demand Systems.

Multimedia-on-Demand Systems A centralized server. A large database of video programs transmitted to clients. Storage devices (disks). The system has to play multiple streams of many different videos simultaneously, based on customer demand. Read only, basically predictable requests. Rigid real-time scheduling requirements. Fixed transfer rate. The Challenge: Balance the load on the disks.

The I/O Subsystem A database of M video-program files {f 1, f 2, …, f M } N storage devices (shared disks) {d 1, d 2,..., d N } Each disk, d j, has: c j = Storage Capacity: the number of program files that can be stored on d j. v j = Load Capacity:the maximum number of concurrent multimedia streams that can be transmitted by d j.

The I/O Subsystem (Cont) The number of broadcasts allocated to each file must be determined by its popularity p i = the popularity of the file f i

Maximizing the Throughput How many broadcasts should be allocated to each of these copies? Given the file popularities, how many copies of each file should be kept? Where should we store each copy?

The Static Assignment A Static Assignment Algorithm allocates the resources to the files. The output of a static assignment consists of two M N matrices: The Assignment matrix, A a {0, 1} – matrix, A ij = 1 iff a copy of f i is stored on d j. The Broadcast matrix, B B ij {0, 1,..., v j }. B ij is the number of broadcasts allocated to f i on d j.

The Static Assignment (Cont) A legal static assignment must satisfy the following conditions: 1. A ij = 0 B ij = 0 [A disk can broadcast a video program only if it holds the video program file]. 2. j, i B ij < v j [the sum of allocated broadcasts does not exceed the broadcast capacity of the disk]. 3. j, i A ij < c j [the number of copies stored on a disk does not exceed its storage capacity].

Definitions The broadcast potential of the system is V = The broadcast potential of f i is B i = The broadcast request of f i, denoted by W i, is the desired number of broadcasts of f i. i, p i V < W i < p i V and

The Goals 1. Fairness: An assignment is -fair if for every file f i, B i > W i allocated required [0, 1] 2. Utilization: The utilization of an assignment is U = i min(W i, B i ) required allocated U is the total number of broadcasts that may be required and provided. A Perfect Assignment is both 1-fair and with utilization V ( i, W i =B i ).

f 1 : Pokemon p 1 = 0.8 f 2 : Telletubies p 2 = 0.1 f 3 : The Flinstones p 3 = 0.1 Maximizing Fairness and Utilization B 1 = 60, B 2 = 20, B 3 = 20 3/4 fair (60/80 for f 1 ) ; U = = 80 Most Fair! V=100 W 1 = 80, W 2 = W 3 = 10 d 1 : c 1 = 1 v 1 =60 d 2 : c 2 = 2 v 2 =40 B 11 = 60 B 22 =B 23 =20

f 1 : Pokemon p 1 = 0.8 f 2 : Telletubies p 2 = 0.1 f 3 : The Flinstones p 3 = 0.1 Maximizing Fairness and Utilization Most Utilized! V=100 W 1 = 80, W 2 = W 3 = 10 d 1 : c 1 = 1 v 1 =60 d 2 : c 2 = 2 v 2 =40 B 11 = 60 B 12 =B 22 =20 B 1 = 80, B 2 = 20, B 3 = 0 0 fair, U = = 90

Our Results The fair-allocation problem and the utilization problem are strongly NP-hard. For some instances a perfect assignment always exists and can be found in polynomial time. How a perfect assignment can be approximated (the approximation ratio depends on the uniformity of the disks). The utilization problem can be solved optimally in polynomial time by adding a single storage unit to each disk (a dual approximation algorithm).

The Moving Window Algorithm To fill the knapsack K j, scan the list of items with a moving window that covers items of c j colors, until for the first time the window contains at least v j items. The knapsacks are sorted such that v 1 /c 1 v 2 /c 2 v N /c N. The sets of items are sorted such that |I 1 | |I 2 | |I M |. 3, 4, 6, 6, 8, 8 v j =17 c j =

The Moving Window Algorithm We use the moving window algorithm to show: 1.When the knapsacks are homogenous (with identical v/c ratio) and c j N+M-1, a perfect assignment always exists, and can be found efficiently. 2.If we are allowed to add a single compartment to each knapsack, then we can achieve the optimal utilization.

Back to General Instances…. where items may have arbitrary sizes and arbitrary profits v=8, c= n=6, M=

The Class-constrained Bin- Packing Problem (CCBP) A set of Identical bins, each having volume v and c compartments. A set, I, of items. Each item, u, has a size, s(u) < v, and a color, c(u) 1,...,M. The goal: A legal placement of all the items in I in a minimal number of bins. In resource allocation terms, CCBP = complete all the requests using as few (identical) devices as possible.

In resource allocation (examples): 1.A request to print a file of 600 pages, both yellow and pink papers are fine. 2.A request to watch a video-program-file, any episode from Seinfeld will do. Generalized Class-constrained Packing (GCCP) Each item, u I, has a size, s(u), a value, p(u), and a set, c(u) {1,...,M}, of colors. It is legal to place u in any knapsack that allocates a compartment to some color in c(u).

Our Results for General Instances A dual PTAS for CCBP, (packs any instance, I, into m OPT(I) bins of size (1+ )v ). A PTAS for CCMK, (packs a set U I such that P(U) (1- )OPT(I)). An FPTAS for the 0-1 class-constrained knapsack. - optimal when all the items have the same value. APX-hardness proofs for GCCP (already for the case where all items have the same size and the same value). - For the PTASs we assume that M is a constant. - The FPTAS is suitable for M= (n). - In the APX-hardness proof M= (n).

Scheduling With Machine Allotment and Parallelism Constraints Each job, J j, has : –a processing requirement, t j, –an allotment parameter, a j, –a parallelism parameter, p j. The execution of J j can be shared by a j machines. At most p j machines can process it simultaneously. Traditional scheduling problems: –No parallelism (p j =1), –Non-preemptive (a j = 1) or preemptive (a j =m). j, a j p j

Our Results for the Scheduling Problem Strongly NP-hard even for easy instances: –identical machines, and j, a j = p j = c –identical jobs, and j a j 3m/2 (reductions from 3-partition) A max j (1+1/p j )- approximation algorithm. (based on the moving window algorithm) A PTAS for the case j, a j p j is a fixed constant.

Results Relevant to Classic Scheduling Problems (p j =1). Known Results: An optimal algorithm for j, a j = m. Hardness proof for j, a j = 1. Our Results: An optimal algorithm for identical machines and j, a j 3. Hardness proof for uniform machines and j, a j =3. An optimal algorithm for uniform machines and j, a j 5.

On-line Class-constrained Packing The input is given as a sequence,, of items each having size 1 and color in {1,…,M}. We need to pack the items upon arrival, without any knowledge on the subsequent items. On-line CCBP: use as few bins to pack all the items in. On-line CCMK: pack as many items from (arriving items may be rejected).

Deterministic Algorithm for CCBP. For a sequence,, of length n, let k = M/c -1 andh = n/v -1. S c,v (k,h) = the set of input sequences with k,h. Lower Bound: For any deterministic algorithm, r d (S c,v (k,h)) 1+ (k+1- kc+1/v )/(h=1). (asymptotically, r d 2) This bound is achieved by first-fit. Conclusion: An a-priori knowledge of n, M does not help.