GATE CSE 2020 | Question: 21

Question

A direct mapped cache memory of $1$ MB has a block size of $256$ bytes. The cache has an access time of $3$ ns and a hit rate of $94 \%$. During a cache miss, it takes $2$0 ns to bring the first word of a block from the main memory, while each subsequent word takes $5$ ns. The word size is $64$ bits. The average memory access time in ns (round off to $1$ decimal place) is______.

Comments

Why sequencial access is not considered here?

Please tell us how to treat this?

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As per GATE official key both $13.3$(Simultaneous access) and $13.5$(Hierarchical access) are answers.

Source: https://web.archive.org/web/20200618164834/http://www.gate.iitd.ac.in/CSKey.php

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Thanks Sagar

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@Arjun Sir, please update the answer here as given in the GATE 2020 official answer key as provided by @Shiva Sagar Rao with 13.3 to 13.3 OR 13.5 to 13.5

It’s also present here link: cs_2020.pdf (iitk.ac.in)

Answer 1

Block size is 256 Bytes,word size is 64 bits or 8 bytes. So Block size in words is 8 words.

Number of words per block=32

Time to fetch a word from main-memory to cache is: $20+31 \times 5=175$ns because first word takes 20ns and rest each subsequent words take 5ns each.

So average Memory acces time is

$0.94(3)+0.06(3+175)=13.5$ ns

Comments

@arjun sir, can we challenge  the question by writing  above explanation given by aditya jain?

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I want to but we need good explanation. I don't know your explanation is enough or not. We should not be desperate to just prove a point. I am waiting for a teacher to give proper explanation. If only arjun sir help us. We just want to know whether is there exist any weight on 13.3 side.

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formula for Average access time = h1*t1+(1-h1)(t1+t2) = t1+(1-h1)t2

 

Answer 2

Considering simultaneous access

Cache m/m = 1 MB

Block = 256 B

Word = 64 b = 8 B

1 Block = 256/8 = 32 words

$T_{access}$ = 3 ns

$T_{mem}$ = 20 ns (first word) + 5 ns (each of the rest 31 words)

T = 0.94 * (3) + 0.06 * (20 + 5 * 31) = 13.3 (ans)

Comments

I think here Hierarchical access.

Answer is 13.5 ns

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Check out the official key: http://gate.iitd.ac.in/CSKey.php

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This is the link for official solution key: http://gate.iitd.ac.in/gate2020/CSKey.php

Answer 3

Block size = $256$ B

Word size = $64$ bits = $8$ B

Number of words in a block= $256/8$ = $32$ words i.e whenever a miss occurs a block will be fetched from the memory containing $32$ words in which first word will take $1$ ns and rest $31$ words will take $5$ ns each.

Average m/m access time = $0.94 * 3 + 0.06 * [1*20 + 31 * 5 + 3] = 13.5 ns$

Answer 4

.94x3+.06x(20+5x31+3) = 13.5