A file has r = 20,000 STUDENT records of fixed-length. Each record has the follo

Question

A file has r = 20,000 STUDENT records of fixed-length. Each record has the following fields:

NAME (30 bytes) , SSN (9 bytes) , ADDRESS (40 bytes) , PHONE (9 bytes) , BIRTHDATE (8 bytes) , SEX (1 byte) , MAJORDEPTCODE (4 bytes) , MINORDEPTCODE (4 bytes) , CLASSCODE (4 bytes) , DEGREEPROGRAM (3bytes) , An additional byte is used as a deletion marker.

The file is stored on the disk whose parameters are given as :

block size B = 512 bytes

interblock gap size G = 128 bytes

number of lbocks per track = 20

number of tracks per surface = 400

A disk pack consists of 15 double-sided disks.

(a) Calculate the record size R in bytes... (MY ANSWER I got was 113 bytes)

(b) Calculate the blocking factor bfr and the number of file blocks b assuming an unspanned organization. (ANSWER I got was bfr = 4 records, and b = 5,000 blocks)

(c) Calculate the average time it takes to find a record by doing a linear search on the file if

(i) the file blocks are stored contiguously and double buffering is used, and

(ii) the file blocks are not stored contiguously.

(d) Assume the file is ordered by SSN; calculate the time it takes to search for a record given its SSN calue by doing a binary search.

Explanation / Answer

(a) R = (30 + 9 + 40 + 9 + 8 + 1 + 4 + 4 + 4 + 3) + 1 = 113 bytes

(b) bfr = floor(B / R) = floor(512 / 113) = 4 records per block

b = ceiling(r / bfr) = ceiling(20000 / 4) = 5000 blocks

(c) For linear search we search on average half the file blocks= 5000/2= 2500 blocks.

i. If the blocks are stored consecutively, and double buffering is used, the time to read

2500 consecutive blocks

(k is the number of cylinders required for 2500 blocks)

= ks+rd+(2500*(B/btr))= 5*30+12.5+(2500*(512/409.6))

= 3287.5 msec = 3.2875 sec

(a less accurate estimate is = s+rd+(2500*btt)= 30+12.5+2500*1= 2542.5 msec)

ii. If the blocks are scattered over the disk, a seek is needed for each block, so the time

is: 2500 * (s + rd + btt) = 2500 * (30 + 12.5 + 1) = 108750 msec = 108.75 sec

(d) For binary search, the time to search for a record is estimated as:

ceiling(log 2 b) * (s +rd + btt) = ceiling(log 2 5000) * (30 + 12.5 + 1) = 13 * 43.5 = 565.5 msec = 0.5655 sec

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