Question

1) Consider a system with 2 level caches. Access times of Level 1 cache, Level 2 cache, and main memory are 2 ns, 10 ns, and 200 ns, respectively. The hit rates of Level 1 and Level 2 caches are 0.8 and 0.9, respectively. What is the average access time of the system ignoring the search time within the cache?
a) 8.6 ns
b) 6.2 ns
c) 7.4 ms
d) 6.4 ns
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Answer 1

Answer:

8.6ns

We assuming the formula of avarage access time.that is

average memory access time = hit time0 + miss rate0 * (hit time1 + miss rate1 * miss penalty1 )

Answer 2

Answer:

Concept:

The cache is an addition to and an extension of the primary memory of a computer. Internal random-access memories (RAMs) known as main memory and cache both make use of transistor circuits based on semiconductors. Only the information or programme codes that are used the most frequently from the main memory are copied into the cache. The cache's smaller volume reduces the amount of time needed to locate data there and deliver it to the CPU for processing.

Given:

Consider a system with 2 level caches. Level 1 cache access timings are 2 ns, Level 2 cache access times are 10 ns, and main memory access times are 200 ns. Level 1 and Level 2 caches have hit rates of 0.8 and 0.9, respectively. What is the system's typical access time, taking into account only the time spent searching the cache?

a) 8.6 ns

b) 6.2 ns

c) 7.4 ms

d) 6.4 ns

Find:

find the right response to the provided question's inquiry.

Answer:

The correct answer is option (b). 7.4 ns

Given:

$T_{1} = 2 ns\\T_{2} = 10 ns\\T_{3} = 200 ns \\\\H_{1} = 0.8,\\H_{2} = 0.9$

Average access time $= T_{avg}$

$T_{avg} =H_{1}T_{1} +(1-H_{1})H_{2}(T_{1}+T_{2})+(1-H_{1})(1-H_{2})(T_{1}+T_{2}+T_{3})$

Because a cache ignores search time,

$T_{avg}=H_{1}T_{1}+(1-H_{1})H_{2}(T_{2})+(1-H_{1})(1-H_{2})(T_{3})$

$T_{avg}=0.8*2+0.2*0.9*10+0.2*0.1*200\\T_{avg}=7.4ns$

An SI unit of time known as a nanosecond (ns) is one billionth of a second, or 1/1 000 000 000 of a second, or 10-9 seconds. The basic unit for one-sixtieth of a minute is combined with the prefix nano- in the phrase. 1000 picoseconds or 1/1000 microseconds make up a nanosecond.

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