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

"5ns with a miss in cache" statement given in the question, it says if it is a miss (cache+M.M =5ns). so we can safely assume read access time of main memory is 4ns.

---

I'm confused with these qsn terms.

someone please clarify.

here access time is given, So isn't like in that time hit is already included??

we should take h.t or t

---

Instruction Fetch is included in the Read Operation from the memory . 

Total Read Operation : 100+60 = 160 

Total Write Operation: 40 

Read : Hit time = 1ns , Miss time =  5ns 

Write :Hit time = 2ns , Miss time = 10ns 

Cache Hit : 90% .

Now do the basic calculation :) Do it . 

Answer 1

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

---

Yes you can do like this …….

Answer 2

Answer is 1.68ns