GATE CSE 2014 | Set 3 | Question: 44

Question

The memory access time is $1$ nanosecond for a read operation with a hit in cache, $5$ nanoseconds for a read operation with a miss in cache, $2$ nanoseconds for a write operation with a hit in cache and $10$ nanoseconds for a write operation with a miss in cache. Execution of a sequence of instructions involves $100$ instruction fetch operations, $60$ memory operand read operations and $40$ memory operand write operations. The cache hit-ratio is $0.9$. The average memory access time (in nanoseconds) in executing the sequence of instructions is ______.

Comments

Here directly the time required to access the data when it is hit and miss is already give for the read and write operations for this question i think no need to check simultaneous or hierarchal

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How many members marked answer as 336 ns 🤡??

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Instruction Fetch (100 times): The CPU needs to read the next machine code instruction from memory. These are read operations.

Memory Operand Read (60 times): Instructions like Load R1 , [X] need to read data from memory. These are also read operations.

Memory Operand Write (40 times): Instructions like STORE [X] , R1 need to write data to memory. These are write operations.

Total accesses = \( 100 + 60 + 40 = 200 \).

Total Read accesses = \( 100 + 60 = 160 \).

Total Write accesses = \( 40 \).

Read Operation 

Hit time = \( 1 \) ns

Miss time = \( 5 \) ns

Hit ratio = 0.9, Miss ratio = 0.1

\[

T_{read} = (0.9 \times 1) + (0.1 \times 5)

\]

\[

T_{read} = 0.9 + 0.5 = \mathbf{1.4 \text{ ns}}

\]

Write Operation 

Hit time = \( 2 \) ns

Miss time = \( 10 \) ns

Hit ratio = 0.9, Miss ratio = 0.1

\[

T_{write} = (0.9 \times 2) + (0.1 \times 10)

\]

\[

T_{write} = 1.8 + 1.0 = \mathbf{2.8 \text{ ns}}

\]

Average Memory Access Time 

The CPU does 160 reads and 40 writes out of 200 total operations. 

\[

AMAT_{total} = \left( \frac{\text{Number of Reads}}{\text{Total Accesses}} \times T{read} \right) + \left( \frac{\text{Number of Writes}}{\text{Total Accesses}} \times T_{write} \right)

\]

\[

AMAT_{total}  = \left( \frac{160}{200} \times 1.4 \right) + \left( \frac{40}{200} \times 2.8 \right)

\]

\[

AMAT_{total}  = (0.8 \times 1.4) + (0.2 \times 2.8)

\]

\[

AMAT_{total}  = 1.12 + 0.56 = \mathbf{1.68 \text{ ns}}

\]

 

Answer 1

The question is to find the time taken for,

$\frac{100 \space \text{fetch operations and $60$ operand read operations and $40$ memory operand write operations}}{\text{total number of instructions}}$.

Total number of instructions $=100+60+40 =200$

Time taken for $100$ fetch operations(fetch = read) $= 100*((0.9*1)+(0.1*5))$

$1$ corresponds to time taken for read when there is cache hit $= 140 \,\text{ns}$

$0.9$ is cache hit rate

Time taken for $60$ read operations,

$= 60*((0.9*1)+(0.1*5))$
$= 84\,\text{ns}$

Time taken for $40$ write operations

$= 40*((0.9*2)+(0.1*10))$
$= 112\,\text{ns}$

Here, $2$ and $10$ are the times taken for write when there is cache hit and no cache hit respectively.

So,the total time taken for $200$ operations is,

$= 140+84+112$
$= 336\,\text{ns}$

Average time taken $=$ time taken per operation

$=\dfrac{336}{200}$

$= 1.68\,\text{ns}$

Comments

The question mentions 100 instruction fetch operations, 60 memory operand read operations and 40 memory operand write operations
so total = 100 + 60 + 40 = 200 instructions

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@Sudeep 1

Thank you for the reply brother. Please have a look at this example problem from Hamacher. 

Check the highlighted statement. Read/Write operations are included in the instructions. They are not treated as seperate instructions.

Every instruction executed will perform a memory operation for instruction fetch. Additionally there may be other read/write memory operations based on the instruction (this percent will be specified in the question)

Would love to hear your thoughts. 

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bro if 100 were total then they should mention that "out of which 60 are read and 40 are write" and we can also think that during fetch cycle operand read is done...

Answer 2

Fetch is also a memory read operation.

Avg access time $= \dfrac{160 (0.9 \times 1+0.1 \times 5)+ 40(0.9 \times 2+0.1 \times 10)}{200} \\=\dfrac{160\times 1.4+40 \times 2.8}{200} \\=\dfrac{336}{200} \\=1.68$

Comments

@ iamhusayn   and @sushmita @bharti

why do we divide 336 by 200? Here it is not mentioned to find out Average Access Time per instruction. So answer should be 3.36 instead of 1.68 .

Average Memory Access Time = Time  for 1 memory access. 

Total number of instructions= 100+60+40 =200

Here for 200 memory accesses total time is 336 ns .

so for 1 memory access time is 336 ns / 200 = 1.68 ns

Hence Average mem Access Time = 1.68 ns

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i got each and every thing but here the no. of instructions are 100 only and according to me out of 100 fetched instructions 60 are read and 40 are for write operation..? but why is it wrong..?

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It is not given 100 fetch, out of which 60 are read and 40 are write, then it would have been 100. It says " involves 100 instruction fetch operations, 60 memory operand read operations and 40 memory operand write operations" which means fetch instruction and Read write instructions are separate and should be added

 

Answer 3

TavgR = 0.9 * 1 + 0.1 * 5 = 1.4 //TavgR here is avg time needed to one read operation

TavgW=0.9 *2 + 0.1*10 =2.8 //TavgW here is avg time needed to one write operation

Tavg = Total time needed to read/write operations / total no.of operations

Here Read operations are=100(fetch) + 60(memory read) = 160

Here Write operations are=40

So= (160 * 1.4 + 40 * 2.8 ) / 200

    =336/200=1.68ns

Comments

@Bikram sir,

yes sir, now I got it .thank you for explaining me.

Earlier, i was considering the total time 336 ns as 1 memory access time. i had wrongly interpreted the question .Thank you, sir for clarifying this

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its clearly written that 100 fetch which means 100 instructions, why are u taking 200 instructions??

200 is total of memory references.

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@ sushmita

Fetching means loading ,rt?

And in microprocessor load is an instruction.

I think that is why we are calculating w.r.t.200 instructions.

Answer 4

……………………....………….….………..……..........................….……..…

Comments

Nice explanation

Answer 5

Answer is 1.68ns

Comments

Best explanation!

 

Answer 6

Memory access time means howlong it takes for an element in memory to be transferred to or from the cpu....so its the relation between the cpu and memory(in theory....now this memory can be either cache or main memory(in practical)...so when its given memory access time when cache hit then its means the time related to cache and cpu...and when its given memory access time when cache miss then its means its the time related with cpu and higher level memory (not cache)...

Comments

Can we do it like -

Total time for 200 instructions =

.9 (100*1 + 60*1 + 40*2) + .1 (100*5 + 60*5 + 40*10)

= 336ns

Time for 1 instruction = 336/200

= 1.68ns

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Yes you can do like this …….