Computer Organisation Textbook Questions

Question

Consider the case memory which has 0.8 and 0.9 hit ratio for read and write operation. Whenever there exist here miss either read or write 2 word block is to be moved from main memory to case memory. The case access time is 10 ns per word, main memory access time is 100 ns per word. Let there are 20% references results write operation with write Back technique.(Assume there is 30% data are dirty in cache memory ) .Find the following,

1. first find the average access time for read operation,

2. second find average access time for write operation,

3. average access time overall and throughput.

Comments

What is the amount of dirty bits or percentage of dirty page replacement since only the percentage of reference operations is given??
Also is it write back or write through??
seems like the question is for write through policy

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@none30 sorry I forgot to write info about dirty bits. Please Update your answer 

Answer 1

Edit: Ans is updated

Given , tc=10 ns, tm=100 ns per word i.e., for 2 words Tm=2*100=200 ns, hr=0.8, hw=0.9, % of dirty page replacement = x = 30% = 0.3 and by default hierarchial access
Considering the given question is for

1. Write through policy –
Tr=(hr*tc)+(1-hr)(tc+Tm)=(0.8*10)+(0.2)(10+200)=8+42=50 ns
Tw=100 ns (Write through always goes to memory, so hit rate does not matter for time and only 1 word is written to)
Tavg=0.8*Tr+0.2*Tw = 0.8*50 + 0.2*100 = 40+20 = 60 ns
Throughput = 1/Tavg = 1/60 = 0.016667*(10^9) = 16.67 * (10^6) words per second = 16.67 million words per second

2. Write Back policy –
Tr = (hr*tc)+(1-hr)(tc+Tm+(x*Tm)) = (0.8*10)+(0.2)(10+200+(0.3*200)) = 8+54 = 62 ns
Tw = (hw*tc)+(1-hw)(tc+Tm+(x*Tm)) = (0.9*10)+(0.1)(10+200+(0.3*200)) = 9+27 = 36 ns
Tavg = 0.8*Tr+0.2*Tw = 0.8*62 + 0.2*36 = 49.6+7.2 =  56.8 ns
Throughput = 1/Tavg = 1/56.8 = 0.0176056*(10^9) = 17.6056 * (10^6) words per second = 17.6056 million words per second