MadeEasy Test Series: CO & Architecture - Cache Memory

Question

 

Comments

Is it 3.6?

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Yes. Can you explain?

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Out of 1000 memory accesses , 40 are miss in L1 cache . And out of those 40 , 10 misses happens in L2 cache.

So AMAT = 1 clock cycle + 40/1000[15 + 10/40 x 200]

 = 1+ 0.04x15 + 0.04 x 0.25 x 200 = 3.6 clock cycles

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how you got this formula?

Answer 1

AMAT = h1*t1 + (1-h1)*h2*(t1+t2) + (1-h1)(1-h2)*(t1+t2+t3)

=>$\frac{960}{1000}*1 + \frac{40}{1000}*\frac{30}{40}*(1+15) + \frac{40}{1000}*\frac{10}{40}*(1+15+200)$

=>3.6 clock cycles

Comments

just a suggestion,

use simplified formula it will take less time-

$T1+(1-H1)T2+(1-H1)(1-H2)T3$

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Thanks for this.