[CO] Memory Organization

Question

3 level memory has the following specifications:-

Level	AccessTime/Word	Block Size in words	Hit Ratio
1	20 ns	-	.7
2	100 ns	2	.9
3	200 ns	4	1

If the referenced block is not in L1,then transfer from L2 to L1,If not in L2,then transfer from L3 to L2 to L1.How long will it take to access a block?

Answer 1

heiarchical acces method is used why ?? "If the referenced block is not in L1,then transfer from L2 to L1,If not in L2,then transfer from L3 to L2 to L1".

$T avg= H1(L1)+(1-H1)(H2)(L1+L2)+(1-H1)(1-H2)H3(L1+L2+L3) =0.7(20)+0.3*0.9(200+20)+0.3*0.1*1(800+200+20) =14+59.4+30.6 =104nsec$

l2 access time is 200 nsec
L3 access time is 800 nsec
since they told block is transfered.

Comments

Focus_Gate,

Why do you say hierarchical method is used?If this is hierarchical ,then first we access L1 with time T1,Then if a miss then we access L2 with time (T2+T1) and then transfer word to L1 also making it to total of (T2+T1+T3).Please clarify?

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because in the question they said " if there is miss in L1 than the block will be transferred from L2 to L1 " this shows they are asking for hierarchical access and the time here (T1+T2) means that in T2 time  it access the block of l2 level memory and in T1 time it transferthe block from  L2 to L1.

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Yes that is correct and also your answer,but by hierarchical means we access them in order like L1 then L2 then Main meory.So if you are accessing like this then at L2 it must be (T1+T2) to read in L2 and further we need to add T1 to transfer it back to L2?So for L2 it must be (T1+T2+T1) if you say it is hierarchical .But i think it should be simultaneous as they have not told that we first need to access l1 and then l2.As per your answer we aer accessing all of them simultaneously ,its just we transferring data back to L1 in case of miss and we keep up adding the other times.Please correct me if i understanding it wrong

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NO ,IT WILL NOT BE LIKE THAT WHEN THERE IS MISS THAN WE HAVE ONLY (T2+T1) TIME TO TRANSFER BLOBK FROM L2 TO L1.
I can tell u how to recognize between hierarchical access and simultaneous access method .
in the above question they told " when there is miss the block is transferred form one level memory to other level memory " means hierarchical access. if nothing is given like this than use simply simultaneously access method.

now the thing which u are telling that " simultaneously ,its just we transferring data back to L1 in case of miss and we keep up adding the other times" this case never happen in simultaneous as if miss occur no need to add both the times ie (T1+T2) in this case the equation will be like this 

$T access= H1(T1) + (1-H1)(H2)(T2) +(1-H1)(1-H2)(H3)(L3)$

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So as the answer you have posted. The times T1,T2,T3 are the times to access the corresponding memory words.Do we also need to take care of time it takes to copy the data from Main memory to L2 to L1?

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Can someone help in above comment please?

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Yes in case of hierarchical acess method we all all those times.

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As per https://gateoverflow.in/2352/gate2010-48. If we have different block size then the requester block size is used.So if there is a miss in cache then shall we access only 1 Word(Block size) in Level 2 ?And if there is a miss in level2 then we should access 2 word block from main memory instead of 4 blocks.

Please clear confusion here.

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@rahul  sharma
yes you are right  block size of requester will be used but  while fetching  only  and before that we need to read the entire block  for required  data.
suppose the required data is  the last word of block then what would be the case when we will read only data a/c to  requester 's  capacity . so we will  read the whole  block and the required  data will be fetched a/c to requester 's capacity   and that will be chosen by seeing  the offset  of the data .

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@ Rahul sharma 

your confusion is 

"Yes that is correct and also your answer,but by hierarchical means we access them in order like L1 then L2 then Main meory.So if you are accessing like this then at L2 it must be (T1+T2) to read in L2 and further we need to add T1 to transfer it back to L2?So for L2 it must be (T1+T2+T1) if you say it is hierarchical"

 now see,  in heirarical access  a/c to this question  3 cases possible:

1. when data is present  in level 1 

then  we will got to know  the availability (hit/miss)  by registers( like TLB) and if there is accessing time of that register also given  then we will consider that time too . otherwise( like here ) we will simply  ignore that accessing  time of register  . 

so  the Tavg  = Hit ratio* (access  time of the block ) 

2. when data is not present in level 1:

then we got to know but hit/miss in level 2 by registers (negligible time ) so the total time  time will be  the reading time of l2 block and yes now the required data will be fetched in level 1 a/c to the capacity of level 1 as we can see level 1 can store only 1 word  of data.   this time will also be consider in total time .

now total  time =(1-H1 ).H2. (T2 time for reading  the block + T1 time for accessing the data  from T1 finally)

Cz in cache we will access the required  data from level 1 only . 

case 3: when  data is not in level 2

simply like case 2 

Tavg= (1- H1)(1-H2)(T3+T2+T1)

hope it will clear you . :)