Seek time (given) = 9ms
RPM = 5400 rpm rotation in 1 min [60 sec]
So, 1 rotation will be =60/5400 =0.01ms [rotation speed]
Rotation latency= 1/2 * 0.01 ms=0.025ms
# To access a file,
Total time includes =seek time + rot. Latency +transfer time
TO calc. transfer time, find transfer rate
Transfer rate = bytes on track /rotation speed
So, transfer rate = 60*100/6ms =1000 B/ms
Transfer time= total bytes to be transferred/ transfer rate
so, Transfer time =100*150/1000 = 15ms
Given as each sector requires seek tim + rot. latency
= 9ms+0.025ms =9.025ms
Total 7500 sector takes = 7500*9.025ms =67687.5 ms
To read entire file, total time =67687.5 + 100(transfer time)
= 67787.5 ms
Seek time (given) = 6ms
RPM = 5400 rpm rotation in 1 min [60 sec]
So, 1 rotation will be =60/7200 =0.0083ms [rotation speed]
Rotation latency= 1/2 * 0.083 ms=0.0415ms
# To access a file,
Total time includes =seek time + rot. Latency +transfer time
TO calc. transfer time, find transfer rate
Transfer rate = bytes on track /rotation speed
So, transfer rate = 60*67787.5 /6ms =4067 B/ms
Transfer time= total bytes to be transferred/ transfer rate
So, Transfer time =100*150/4067 = 3.688ms
Given as each sector requires seek time + rot. Latency
= 9ms+4067ms =4076ms
Total 7500 sector takes = 7500*4076ms =3057 ms
To read entire file, total time =3057 + 100(transfer time)
= 3157ms
Effective Access Time (EAT)
EAT = (1 – p) x memory access x cost value
+ p (absolute fastest system overhead
+ [fastest system out]
+ Fastest system in
+ restart overhead)
Let’s say memory access (absolute fastest system
Table accesses) = effective cost 3157
Overhead at both ends say 3157 instructions ≈
10000nsecs
Absolute fastest system in = 15 msecs = 4067000 nsecs
50% of time effective access time = 8 000 000 nsecs
So EAT = (1 – p) x 20 + p(10000 + 8 000 000)
≈ 20 + 8 000 000p nsecs
~8000020nsecs
Let as consider the Absolute fastest system value=9000ms
Total memory of accessing time=8000020nsecs
Half way of the absolute fastest system on cost=67787.5Ms
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