search
Log In
9 votes
1.1k views
A system has cache main memory and disk for virtual memory. If referenced word in cache $30$ ns to access it. If it is not in cache $80$ ns to load it in cache and reference is started again. If the word not in memory then $22$ms to bring from disk to memory and $80$ ns from memory to cache and start again. Cache hit ratio is $0.8$ memory hit ratio is $0.9$
in CO and Architecture
edited by
1.1k views
0
What this question is asking?
1
This qstn is asking about Average access Time .

3 Answers

3 votes
 
Best answer
In this question nothing is mention about search time inside memory and disk, so we safely ignore search time .

so that average access time is :

cache access time + MM access time + Disk access time

$$\begin{align*} & 0.8 * 30 \text{ ns } \\ &+ 0.18 * \left ( 30 + 80 \right ) \text{ ns } \\ &+ 0.02 * \left ( 30 + 80 + 22000000 \right ) \text{ ns } \\ &=\bf 440046 \text{ ns } \\ \end{align*}$$

in each above case , cache  and MM and disk we consider search time to find that referenced word = 0  .

edited by
0
From where you have used 0.18.....
0
it is (1-0.8)*0.9 = 0.18
8 votes

Total time = Cache Access + Memory Access + Disk Access

= Cache time + Cache Miss * (Transfer from Memory to cache + 2nd Cache reference ) + Cache Miss * Memory Miss ( Fetch Disk + Transfer from Disk to Memory )
= 30 + 0.20*(80+30) + 0.20*0.10(80 + 22,000,000)
=30 + 22 + 440001.6
= 440053.60 ns
$\approx 440 \mu s$
 

0
80 + 22000 + 30 rt?
1
Sir if der is miss in Cache, miss in memory so we will go upto disk and then transfer that file to maine memory. now at this time condition is like cache miss and memory hit.

 

Actually second reference of cache already counted in 0.20*(80+30) .. is it not ?
1
yes, I missed that :)
0
why dont we consider hit ratios ....??
0
0.8 * (30) + 0.2 (0.9 * (80+30) + 0.1 (80+30+22000000) )=440026ns

ans should be like this, isn't it?

i'm not getting logic here....can u explain further?
1
you can expand that 0.8 * 30 + 0.2 * 30 = 30 - we can rewrite like this also. But given answer is wrong due to "ms".
0
what about it ratio for cache
1 vote
I did something Like this

 

0.8*30 + 0.2*0.9*(80+30) + 0.2 *0.1*(80+30 + (22000000))=440046 ns.
Answer:

Related questions

0 votes
0 answers
2
92 views
What events happen when we say that a cache at level i is accessed? (I am able to use the cache formulas as given in textbooks and also most of the times, I arrive at the correct answer, but I want to fully understand the basic details.) My understanding is the following ... in the access duration? Do we include any extra time for STORING the data in the level (i-1)th cache as is done here ?
asked Sep 29, 2018 in CO and Architecture Harsh Kumar 92 views
0 votes
0 answers
3
226 views
If the question is given, L1 cache access time is 150ns/word and one block of L1 cache contains 5words. Should I multiply the access time by 5 or leave it at 150ns as memory is word addressable?
asked Aug 6, 2016 in CO and Architecture prasitamukherjee 226 views
0 votes
1 answer
4
444 views
Consider a two level memory hierarchy, L1 (cache) has an accessing time of 5 ns and main memory has an accessing time of 100 ns. Writing or updating contents takes 20 ns and 200 ns for L1 and main memory respectively. Assume L1 gives misses 20% of the time with 60% of the instructions are read only instructions. What is the average access time for system (in ns) if it uses WRITETHROUGH technique?
asked Nov 25, 2018 in CO and Architecture Shivangi Parashar 2 444 views
...