GATE IT 2006 | Question: 42

Question

A cache line is $64$ bytes. The main memory has latency $32$ $ns$ and bandwidth $1$ $GBytes/s$. The time required to fetch the entire cache line from the main memory is:

• A. $32$ $ns$
• B. $64$ $ns$
• C. $96$ $ns$
• D. $128$ $ns$

Comments

 Memory latency is the time (the latency) between initiating a request for a byte or word in memory until it is retrieved by a processor.  OR

Memory latency is also the time between initiating a request for data and the beginning of the actual data transfer.

Reference :-   https://en.wikipedia.org/wiki/Memory_latency#:~:text=In%20computing%2C%20memory%20latency%20is,is%20retrieved%20by%20a%20processor.&text=Memory%20latency%20is%20also%20the,of%20the%20actual%20data%20transfer.

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option C ) 96ns is the answer

Answer 1

Answer is C.

For $1$ sec it is $10^9$ bytes  ( Note - by looking at the options, it can be decided that $1GB$ should be considered as $10^9$, but not as $2^{30}$, and in general, if bandwidth is given, then we consider $1\text{ Giga} = 10^9)$

So, for $64$ bytes?

It is $\dfrac{64\times 1 }{10^9}$ so it is $64\text{ ns}$ but mm latency is $32.$
So, total time required to place cache line is $64+32 = 96\text{ ns}.$
 

Comments

here m geting confused y we cnt assume 1GB= 2^30B if consider by this then getting wrong & please tell me in whci case we used to consider 1GB=2^30B & in which 1GB=10^9

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It's still ambiguous, but even if you take $2^{30}$ answer would be around 92 rt? 

Ref: https://en.wikipedia.org/wiki/Gigabyte

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sorry for ask such equations but plz solve here how 92 plz say 2^-24 sec*10^-9+32ns

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64 / 10⁹ + 32ns = 96 ns

64 / 2³⁰ + 32ns = 91.6 ns

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bw is sending of signal rather than bits so we should take 1GBps = 10^9

 

also its makes sense in this questiona as when we convert sec to nano sec we have to multiply by 10^9, so both will cancel each other

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okay

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if bandwidth is given then consider 10^9,,,,,if capacity is given then consider 2^30

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Hi There,

Shouldn't the answer be 95, as the latency is the time to transfer first word/byte and bandwidth for subsequent transfers.

So considering that,

time to fetch 64 bytes or a cache line from the main memory = 32ns (for first word/byte) + 63ns (for subsequent words/bytes) = 95

Please consider this and reply the corrections if any. Thanks

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Data is always expressed in power of 2 and Bandwidth in power of 10.

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@Rajesh Panwar-MM bandwidth is always in power of 2 and I/O device bandwidth always in power of 10.

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can you please explain what is memory latancy here(32ns)

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Why here we need to add main memory access time? I mean why ans is not 64ns?

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64 ns is for transfering data from Main memory to Cache, whereas 32 nsec is for accessing data from main memory.

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There will be some processing time(latency 32 ns) to fetch the data from main memory.after particular data found, transfer(64ns) from main memory to cache memory.

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For (storing) the data use (power of 2)

For (transfer) the data(speed) use (power of 10)

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As per the definition of memory latency given in carl Hamacher i.e, Memory Latency is the amount of time it takes to transfer a word or byte to or from the memory so basically for 1st byte 32 ns and for the remaining 63 bytes 63 ns will be the time so the total time = 95 ns which is nearly equal to 96ns.

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bro which page of carl hamacher?

can you please tell me?

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@ijnuhb

Ch 8-The memory System: Pg 278 (6^th edition)

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If we stick to the meaning of “Bandwidth” as understood in this question https://gateoverflow.in/302803/gate-cse-2019-question-45. and in this PDF http://opencourse.ncyu.edu.tw/ncyu/file.php/28/tmp/101-1%E8%A8%88%E7%AE%97%E6%A9%9F%E7%B5%84%E7%B9%94/CO-Chapter_05_Part_2_.pdf

And if the total time is taken = 92ns and data transferred = 64 Bytes

Then Bandwidth = 64 bytes / 92 ns = 0.69 GBytes/sec

But given is even more i.e 1 GBytes/sec

What I feel is Bandwidth is covered for everything that’s involved during miss latency. i.e Memory latency time would also count under Bandwidth.

Say t = Time to transfer one block(64Bytes) from MM to Cache

So Bandwidth = Total Data transferred / Total time taken = 64 Bytes/ (32ns+t)  = 1GBytes/sec

                       =  64/(32+t) = 1 bytes/ns

                      t = 32ns

So total time = 32 + t = 64 ns.

Any compliments or comments are highly recommended on this.

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@mani312

Memory latency is the time (the latency) between initiating a request for a byte or word in memory until it is retrieved by a processor.

so i think it is not covered in the bandwidth itself, bandwidth is the rate at which all the subsequent words transferred in this case.

EDIT:

Latency should not be confused with memory bandwidth, which measures the throughput of memory. Latency can be expressed in clock cycles or in time measured in nanoseconds. Over time, memory latencies expressed in clock cycles have been fairly stable, but they have improved in time.

source:

https://en.wikipedia.org/wiki/Memory_latency#:~:text=In%20computing%2C%20memory%20latency%20is,is%20retrieved%20by%20a%20processor.&text=Memory%20latency%20is%20also%20the,of%20the%20actual%20data%20transfer.

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@mani312 How total time taken is 92ns?

Answer 2

Memory latency is the time (the latency) between initiating a request for a byte or word in memory until it is retrieved by a processor.