Suppose that a file initially contains r=120,000 records of R=200 bytes each in an unsorted (heap) file. The block size B=2400 bytes, the average seek time s=16 ms, the average rotational latency rd=8.3 ms and the block transfer time btt=0.8 ms. Assume that 1 record is deleted for every 2 records added until the total number of active records is 240,000.

Question

(a) How many block transfers are needed to reorganize the file?

(b) How long does it take to find a record right before reorganization?

(c) How long does it take to find a record right after reorganization?

Answer 1

Let X = # of records deleted
Hence 2X= # of records added.
Total active records
= 240,000 = 120,000 - X + 2X.
Hence, X = 120,000
Records before reorganization (i.e., before deleting any records physically) =
360,000.

(a) No. of blocks for Reorganization
= Blocks Read + Blocks Written.
-200 bytes/record and 2400 bytes/block gives us 12 records per block
-Reading involves 360,000 records; i.e. 360,000/12 = 30K blocks
-Writing involves 240,000 records; i.e., 240000/12 = 20K blocks.
Total blocks transferred during reorganization
= 30K + 20K = 50K blocks.

(b) Time to locate a record before reorganization. On an average we assume that
half the file will be read.
Hence, Time = (b/2)* btt = 15000 * 0.8 ms = 12000 ms.
= 12 sec.

(c) Time to locate a record after reorganization
= (b/2) * btt = 10000 * 0.8 = 8 sec.