Write back Cache with write allocate policy

Question

Consider a computer with the following features:

90% of all memory accesses are found in the cache (hit ratio = 0.9);
The block size is 2 words and the whole block is read on any miss;
The CPU sends references to the cache at the rate of 10⁷ words per second;
25% of the above references are writes (writes = 25%, reads = 75%);
The bus can support 10⁷ words per second, read or writes (total bus bandwidth = 10⁷);
The bus reads or writes a single word at a time;
Assume at any one time, 30% of the block frames in the cache have been modified;
The cache uses write allocate on a write miss.

Calculate the percentage of the bus bandwidth used on the average if cache is WRITE BACK:

Comments

Is it 25%???

Answer 1

WRITE BACK = When cached data is modified, it is just marked using dirty bit. The original data is updated when the cached data is deallocated.

and

Write Allocate – If a write miss occur, load the block into cache and then update. Write back policy generally uses this.

Consider the scenarios for using  bandwidth

1. Read hit: While Read hit, block is present in cache, so no usage of bandwidth
2. Read miss: We must write back the block is dirty, then load new block. Probability of read miss= (1-0.9)*0.75; Replacing the dirty block =0.3* one block =0.3*2 words
3. Write hit= As block is not updated in main memory immediately, no bandwidth use
4. Write miss= 10⁷ * 0.1 * 0.25 * [2 * 0.3 + 2]

Percentage of the bus bandwidth used = sum of these/total =0.26 * 10 ⁷ /10 ⁷ =0.26

Comments

When cache miss has occured during read why are we multiplying with 0.3??? If read miss is there we have to read 2 blocks hence bnadwidth used irrespective of modified or not. So why 0.3*2???

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it is given that on a miss,whole block is read..so during read miss,time for 2 words to be brought into main memory should be taaken into account.

time for 2 words is 0.2 us

so for read miss,time should be 0.75*( 0.1*(0.2 us + 0.3 * 0.1) as in case of modified frames,one word is updated to main memory.

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Akriti we do not need to calculate time here, we just have to calculate average number of memory access in one second.

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ooh okay..alright..but is my approach right??as in case of miss,2 words will be updated and for 30% modified data words,1 word will be updated in main memory.
and same goes for write references...pls tell me why for Write ,we are multiplying with 10^7...??

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@shiva I think when read miss occurs 0.3*2(for modified block) + 2( for bringing block in cache) and when write miss is there 2( for write allocate) + 2( for actual write)

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in case of modifies block,one word is updated or whole block is updated??

and in case of Write also,we need to take 30% modification into account

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@akriti I think it is given that cache block frame is dirty 30% of time so we should be updating 2 words(cache block size). Most probably should be 2 words to be updated.(not sure)

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see this -https://gateoverflow.in/35154/write-back-and-write-through

yes,whole block is to be updated in main memory when 30% are updated.

but can u pls tell me why are we multiplying with 10⁷ in only write..not read

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Find average memory access and multiply by 10^7 bcoz it is number of memory refrences sent by CPU. Now you will get x*10^7 memory access per second.

x will be in decimal(0.something) and it is given bandwidth is 10^7(divide 10^7*x by 10^7) so x*100% is the answer

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@shiva @akriti plz verify this,

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@rahul,for Write also,u need to consider 30% modification.

and i am not understanding how to calculate the bandwidth here

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well..you can check this link  http://www.inf.ed.ac.uk/teaching/courses/car/Notes/2015-16/solution04.pdf

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@Akriti thanks for the link☺

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welcome..:)