# mm class 8

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Education

Published on June 26, 2007

Author: BAWare

Source: authorstream.com

Service Models:  Service Models Example Server has 50 movies, 100 min. each Request rate: 1 movie/min Max. capacity: 20 streams Random Access Model Case 1: after 20 movies, no more memory left. 21st movie waits for 80 minutes, 22nd movie waits for 81 minutes … Case 2: after 20 movies, more memory can be allocated. 21st movie has to wait (initial latency) till one round of the previous 20 movies each has been served. EPPV Model: At any time 20 movies are served, movies are initiated every 5 minutes Streams are distributed uniformly during these 20 minutes A Disk Striping Model:  A Disk Striping Model Assume compressed videos stored on m disks with stripe units of size d – each round gets 1 stripe unit Period P for a video: # rounds to get successive streams q = max. number of videos serviced in a round / disk If total memory=R Optimal d given by: If R=2GB, m=50, display rate=1.5Mbps d=0.9Mb, q=20, round=600ms Period for a video of 100 min = 1 min. round duration buffer/video is 2d tseek= avg. time to reach a track Vertical Striping:  Vertical Striping A situation 4 supervideos V1 .. V4 V1, V2: 3 stripe units per disk V3, V4: 2 stripe units per disk 2 stripe units/round/disk Round 1..3: get data for V1 and V2 from disk 0 Round 4, 5: get data for V3 and V4 from disk 0 get data for V1 and V2 from disk 1 Round 6: get data for V1 and V2 from disk 1 get data for V3 and V4 from disk 1 4 streams from a disk!!! A solution Partition videos into sets that reduce the difference between the lengths of videos of any two sets as much as possible Scheduling for Vertical Striping:  Scheduling for Vertical Striping The Problem: Get w stripe units for every super video Vi at intervals of Pi from disk 0, when a maximum of q streams can be retrieved from a disk in one round Schedule non-preemptible tasks T1…Tr with periods P1…Pr and computation time w on q processors How complex is the problem? What can be done? Result 1: if n.w  gcd(P1…Pn) then the tasks are schedulable on a single processor, where n=r/q If tasks T1..T4 have periods 4,4,6,6 and w=1, they can be scheduled at 0,2,1,3 respectively – a stronger result is needed Example:  Example Tasks T1..T4 have periods 4,4,6,6 and w=1 g=2 n=4, so n andgt; g Make tasks T1…T4 with periods 2, 2, 3, 3 Initial subsets S1=T1, T2 S2=T3, T4 Next iteration: From S1: {(T1, 0) (T2, 1)} From S2: {(T3, 0) (T4, 1)} Step 5: {(T1, 0) (T2, 2) (T3, 1) (T4, 3)}

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