Consider a disk drive with the following parameters: seek time = 5 ms, rotational latency = 5 ms, and a transfer rate (after the seek and rotational latency have occured) of 10 MB/s. Recall that 1 ms = 10^(-3) seconds, 1 MB = 10^6 bytes, and 1 MB/s = 1 MB per second.
Answer the following questions:
Total time = seek + latency + transfer.
For 10,000 byte block this is
5ms + 5ms + (10^4 bytes) / (10^7 bytes/sec) =
10ms + 10^(-3)sec = 10ms + 1ms = 11ms
For 20,000 byte block this is
5ms + 5ms + (2*10^4 bytes) / (10^7 bytes/sec) =
10ms + 2*10^(-3)sec = 10ms + 2ms = 12ms
The small block size disk transfers 10,000 bytes in 11ms, which
is
10^4 bytes / (11 * 10^(-3) sec) = 1/11 * 10^7 bytes/sec = 10/11 MB/s.
The large block size disk transfers 20,000 bytes in 12ms, which
is
2*10^4 bytes / (12 * 10^(-3) sec) = 2/12 * 10^7 bytes/sec = 5/3 MB/s.
For a faster small block size disk the time is
5ms + 5/n ms + (10^4 bytes) / (n * 10^7 bytes/sec) =
5ms + 5/n ms + 1/n ms = (5+6/n)ms = (5+6/n) * 10^(-3) sec.
The overall transfer rate is
10^4 bytes / ( (5+6/n) * 10^(-3) sec) = 10/(5+6/n) MB/s.
To match the big disk we need 10/(5+6/n) = 5/3 or 6 = 5+6/n or n=6.