1. a. According to your textbook (figure 1.11, page 52) the low-end access time
ID: 3812684 • Letter: 1
Question
1. a. According to your textbook (figure 1.11, page 52) the low-end access time for SSDs is 25,000 ns, and magnetic disk is 5,000,000 ns. You create a hybrid SSD/hard disk that caches hard drive reads 98% of the time on SSD storage. If all data is stored on the hard drive and the SSD is only used for caching, what is the effective access time? (in ns)
b. A second hybrid SSD/hard disk is designed on a different principle. Data usage is logged and frequently accessed data is moved in the background onto the SSD portion where it is addressed directly, and less used data is moved onto the hard drive. Access times for both drives are identical, but only 95% of the time our reads done from the SSD.
What is the effective access time?
c. A regular memory lookup requires a page retrieval followed by a memory call. If a TLB hit rate is 85%, what is the effective access time?
d. If you had no cache implemented, every memory retrieval would require 2 memory calls at 80 ns each. What is the speed up factor of the TLB call? (old access time/new access time)
Explanation / Answer
a.
The low-end access time for SSDs =25,000ns
The low-end access time for magnetic disk = 5,000,000 ns
A hybrid SSD that caches hard drive is created and it is used for caching, the effective access time is 5 nanoseconds
b.
The memory access time is 100nano sec.
c.
A regular memory look up requires a page retrieval followed by a memory call. If a TLB hit rate is 85%, the effective access time is given as follows:
(5+100) *(0.85) +(5+100+100) *(0.15) =120 nanoseconds
d.
Now suppose the hit ratio increase to 98%
Then the new Effective Access time = (100+5) *0.98 + (100+100+5) *(0.02) =107 nanoseconds
The speed up factor of the TLB call is: old access time/ new access time=120/107=1.121
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.