CS351HW1-Fall2009.Sol

CS351HW1-Fall2009.Sol - S a l i m S a r ım u r a t – H W...

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Unformatted text preview: S a l i m S a r ım u r a t – H W # 1 | 1 CS351, Fall 2009, HW1 Solutions S1. Storage Type Speed of Capacity RAM Cost Access 1GB – 4 GB 50ns – 60 ns 0.04 $/MB 100GB – 1 TB 5.55 ms 0.09 $/GB 1.44 MB 100 ms 0.18 $/MB 650 MB – 750 MB 70 ms 0.28 $/GB DVD 4.7GB 110 ms 0.17 $/GB Blu‐Ray Optical Media 25GB 10 ms 0.18 $/GB 1TB 5‐6 ms 1 $/GB Hard Disc Floppy Disc CD‐ROM Magnetic Tapes S2. a) 5400 rpm Period of Revolution b) 7200 rpm = = 1 minutes 5400 1 × 60 5400 Period of Revolution = = seconds = Average Rotational Delay seconds = 0.0083 seconds = 8. 3 milliseconds 0,0 1 seconds = 11. 1 milliseconds = = 11. 1 milliseconds 2 5. 5 milliseconds 1 minutes 7200 1 × 60 7200 Average Rotational Delay = = 8. 3 milliseconds 2 4.16 milliseconds S3. a) Both tracks have the same amount of recording capacity because btt (block transfer time) for every track should be the same on a disk. b) Since the recording capacities are the same and T1 has smaller radius than T2, the recording density (bits/inch) of T1 is greater than T2’s recording density. c) The bits on the inner tracks are stored closer than the bits on the outer track so that the number of bits on each track is the same and the block transfer time for a given block size is the same no matter what track the block is on. So, T1 and T2 have the same amount of recording capacity, but T1’s recording density is higher than T2’s recording density. S a l i m S a r ım u r a t – H W # 1 | 4 S4. Number of blocks 100 blocks 1,000,000 blocks ∞ blocks Ts (total time for sequential processing) Tr (total time for random processing) = 15m sec+ 5m sec+ (100 × 0.84m sec) = 104m sec = s + r + (b × ebt ) = 15m sec+ 5m sec+ (1,000,000 × 0.84m sec) = 100 × (15m sec+ 5m sec+ 0.8m sec ) = 2080m sec = b × (s + r + btt ) = 1,000,000 × (15m sec+ 5m sec+ 0.8m sec ) = 840,020m sec = b × ebt = 20,800,000m sec = s + r + (b × ebt ) = b × (s + r + btt ) Nothing can be ignored. Because we can ignore (s + r ) Because we multiply the number of blocks with (s + r + btt ) For very large number of blocks (=>∞); Tr b × (s + r + btt ) s + r + btt 15m sec+ 5m sec+ 0.8m sec = = = = 24.76 Ts b × ebt ebt 0.84m sec S5. 5.a) Tape Reel Length File Length Unit Length = = = = 2,400 feet × 12 (Unit Length) x (Number of Records) 28,800 inches (Inter Block Gap) + (Record Length) = 0.5 inch + 800bytes 1600bytes / inch = 1 inch File length = 1 inch x 36,000 records = 36,000 inch As 36,600 inches > 28,800 inches, we say that this many records cannot fit on a single tape reel. S a l i m S a r ım u r a t – H W # 1 | 4 5.b) (Tape Reel Length) x (Recording Density) – (Number of Records x Record Size) (28,800 inch) x (1600 bytes/inch) – (36,000 records) x (800 bytes/record) = Space Available to use = 17,280,000 bytes (Space Available) / (Inter Block Gap Size) = Number of Blocks 17,280,000bytes 1600bytes / inch × 0.5inch = 21,600 blocks = (Number of Records) / (Number of Blocks) = Blocking Factor 36,000 21,600 = 1.6 Minimum Blocking Factor is ⎡1.6⎤ = 2 because it should be exact. 5.c) File Length = (36,000records)× (800bytes / records) Number of Blocks = Inter Block Gap Length = Total Length = 1600bytes / inch 36,000 10 36,600 × 0.5 18,000 inches + 1,800 inches = 18,000 inches = 3600 = 1,800 inches = 19,800 inches 5.d) Bkf Total start/stop time 1 Tape Processing Speed File Size = 36,000 x 10ms = 1,600 byte/inch x 200 inch/sec = 36,000 x 800 = 360 sec = 32 x 104 bytes/sec = 288 x 105 bytes Process Time = 288 × 105 + 360 sec 32 × 10 4 = 90 + 360 sec = 450 sec S a l i m S a r ım u r a t – H W # 1 | 4 12 = 360 sec / 12 = 30 sec same S6. Tx = Time for exhaustive reding 100,000 msec = b x ebt b × ebtNew 70,000 msec = 2 7 × ebt ebtNew = 5 ebt = (block transfer) + (Inter Block Gap) = 5x ebtNew = 2 x (block transfer) + (Inter Block Gap) = 7x (block transfer) 2 = is found. So; (Inter Block Gap) 3 Finally, data transfer time (dtt); ⎛ 2x ⎞ ⎛ 4x ⎞ dtt = ⎜ ⎟ × 100 = 40 sec OR dtt = ⎜ ⎟ × 70 = 40 sec ⎝ 2 x + 3x ⎠ ⎝ 4 x + 3x ⎠ same = 288 × 10 5 + 30 sec 32 × 10 4 = 120 sec ...
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This note was uploaded on 12/01/2011 for the course CS 351 taught by Professor Fazlıcan during the Spring '11 term at Bilkent University.

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