CS351 - Sec 123 Quiz 1

CS351 - Sec 123 Quiz 1 - CS351Quiz1 Quiz1Section1,...

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CS 351 – Quiz 1 Quiz 1 - Section 1,  September 28, 2009 Question: s = 15 msec r = 5 msec btt = 0.80 msec ebt = 0.83 msec n = 100,000 (records in the file including deleted ones) Blocking factor = 4 R (Record Size) = 300 bytes B (Block size) = 2400bytes a.  T = ? b.  T = ? c.  T = ? d.  T = ? (Assume that %25 of the records are deleted) e.  T = ? (after reorganization) Solution: a. # of blocks = # of records / bf b = 100,000 / 4 = 25,000 T F   = (b / 2) * ebt     T F   =  (25,000 / 2) * 0.83 T F   = 10,375 msec b. T =  T F   = 10,375 msec c. T = (s + r + btt) + 2r T I  = (15 + 5 + 0.80) + 2*5  T = 30.8 msec d. T = (b before  * ebt) + (n active  / bf) *ebt  n = 100,000 * 0.75 n = 75,000 T = (25,000 * 0.83) + (75,000 / 4) * 0.83 T = 36,312.5 msec e. T = (b after  / 2) * ebt  b = 25,000 * 0.75
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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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CS351 - Sec 123 Quiz 1 - CS351Quiz1 Quiz1Section1,...

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