ME-CO write through

Question

@arjun sir please see this
________________________________________________________________________________________
if picture is not clear
consider a 2 level memory hierarchy L1(cache) has an accessing time of 20ns and MM has accessing time 40nsec. writing or updating contents into their memory takes 40nsec.. and 50nsec for L1 and L2 respectively.assume L1 gives misses 65% of the time. the average writing time if it uses write through technique____
________________________________________________________________________________________
 main memory is updated in any case so shudnt it be 50ns only?
i dont think we shud care abt hit ration in such case

Answer 1

This question is asking about Average writing time for Write-Through technique ...In write through Technique, cache write / update and memory write/ update is done paralley means both have done at the same time  ( parallely)  .

Now, H = hit rate in cache
T_c = cache access time
T_m = memory access time

H= 35%
Tc= 20  and   Tm = 40
writing / updating time into memory = 50

so formula is
T _{avg write time} = H * T _{memory update time} + ( 1-H) * ( T _{memory update time} )

= 0.35 * 50 + 0.65 * (  50 )
= 17.5 + 32.5
= 50 

When cache miss occurs , in Write Through it use Write No-Allocate policy, in Write No-Allocate Missed main memory block is updated while residing in the Main memory itself and NOT brought to the cache.

That's why we need to consider Memory update time only when cache miss.

 In Write - Through Technique , every write operation to the CACHE is repeated to the main memory AT THE SAME TIME. 



Write-Through Technique use Write No Allocate Policy . Where the missed Main memory block is updated while residing in the main memory and not brought to the cache. 
so for this problem formula used is :

T_{Average writing time} = H * T _{mem update time} + ( 1 - H ) * ( T _{mem update time} ) .

// In this question  all the data given is not required .

And we don't need extra search time as it is clearly mention in the question that " consider a 2 level memory hierarchy " . Means if miss in cache then surely found in main memory .

So the question implicitly assumes that all the blocks required are contained within the main memory,  we NO need to check whether a block exists or not because, if it doesn't exist then we have to go to secondary storage but that is not mentioned in the question , in question only 2 level memory hierarchy is given .

Comments

@Anusha and @uddipto

Yes, it is  50 ns .

T_{Avg writing Time}  = H * T _{mem update time} + ( 1 - H ) * ( T _{mem update time} ) 

only memory update time is considered as it is write through .

see my explanation and Read the book i mention, though from explanation it is clear why 50 is correct answer.

Write-Through Technique use Write No Allocate Policy . Where the missed Main memory block is updated while residing in the main memory and not brought to the cache.  Hope it is clear now.

---

got that

---

@anusha and @aboveallplayer

Update my answer ..

Uddipto is correct it is 50 ns though question seems ambiguous to me due to all the given data is not needed ..please check my answer ..

Answer 2

L₁ cache miss chances = 65%

L₁ cache hit chances = 100-5 = 35%

T_avgW = 0.35*T_MemUpdate + 0.65*(T_MemUpdate) = 0.35*50 + 0.65*(50) = 50 nsec

Since only 2 levels of memory is given no need to consider check time in main memory

Comments

but in case of cache hit we need to find block in cache also.. we have to take cache acess time also...??

---

In write through cache access and memory update is done parallely