consider the following

Question

Consider a single-level cache with an access time of 2.5 ns, a line size of 64 bytes, and a hit ratio of H 0.95. Main memory uses a block transfer capability that has a first word (4 bytes) access time of 50 ns and an access time of 5 ns for each word thereafter.

a. What is the access time when there is a cache miss? Assume that the cache waits until the line has been fetched from main memory and then re-executes for a hit.

b. Suppose that increasing the line size to 128 bytes increases the H to 0.97. Does this reduce the average memory access time?

Comments

This is the official solution given.

@Deepakk Poonia (Dee) @Arjun Could you please once check, which one is correct? There's always confusion regarding average memory access time. 

---

@Arjun why you didn't reply.. your solution contradicting with original one...??? At least leave a reply and @Deepakk Poonia (Dee) you too https://gateoverflow.in/user/Deepakk+Poonia+%28Dee%29

---

Because I'm busy implementing a code to prevent unauthorised user tagging here. Some people here think that others are jobless.

Answer 1

a. Hierarchical access is mentioned in question. 

$T_{avg}=  time_{cache}+ \left(1-hit_{cache}\right)\left[time_{memory}\right]$

$=2.5+(1-.95)[125]$

as line size is 64 bytes there are 16 words in a cache line fetch and first word is of four bytes having access time 50 ns and remaining fifteen have access time 75 ns(i.e.,125 ns )

$=2.5 + 6.25$

$=8.75$

---

(b)

now line size increases ie 128 bytes as now there are 32 words .

first word access is 50 ns than after that 5 for each 31 words

$=50+155$

$=205$(block access time) 

now hit ratio is .97

$T_{avg}=2.5+(1-.97)[205]$

$=2.5 + 6.15$

$=8.65$

Comments

Ram Swaroop The access time will be 157 ns whereas the average access time will be 36.2.

 

---

How can you explain

---

T(access)= 6 + 40 + (7*15) + 6 = 157

As line size is 64 bytes there are 8 words in a cache line. The first word is of 8 bytes having access time 50 ns and the remaining 7 have access time 15  ns(i.e.,105 ns )

T(avg) = 0.8(6) + (0.2)(157) = 36.2