GATE CSE 2017 Set 2 | Question: 29

Question

In a two-level cache system, the access times of $L_1$ and $L_2$ caches are $1$ and $8$ clock cycles, respectively. The miss penalty from the $L_2$ cache to main memory is $18$ clock cycles. The miss rate of $L_1$ cache is twice that of $L_2$. The average memory access time (AMAT) of this cache system is $2$ cycles. The miss rates of $L_1$ and $L_2$ respectively are

• A. $0.111$ and $0.056$
• B. $0.056$ and $0.111$
• C. $0.0892$ and $0.1784$
• D. $0.1784$ and $0.0892$

Comments

Can you please type more clean.

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@BASANT KUMAR

You are partially right.

The equation for AMAT will be as follows:

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

AMAT = (1-2x)*(1) + (2x)(1-x)(1+8) + (2x)(x)(1+8+18)

=1-2x+18x-18x$^{2}$+54x$^{2}$

=1+16x+36x$^{2}$

Using the information given in question, we have EMAT = 2 cycles.

36x$^{2}$ + 16x + 1 = 2

x = $\frac{-16 \pm \sqrt400 }{72}$

Ignoring the negative root we get,

x = (-16 + 20)/72 =1/18 = 0.055

2x = 0.111

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This is the formula for Hierarchal access right? How to determine whether it is Hie. Or Simul?

AMAT = HitTIme of L1 + Miss Rate of L1(Hit time of L2 + Miss Rate of L2*Miss Penalty of L2)

Is this formula correct?

Answer 1

Access time of L1 cache=1 clock cycle

Access time of L2 cache =8 clock cycles

miss penalty of L2 cache to main memory=18 clock cycle

miss rate of L1 cache= 2 * miss rate of L2 cache

AMAT=2 cycles

Avg. Access time of L1 cache= access time of L1 cache+ miss rate of L1 cache=1+2 L2

Avg access time of L2 cache=access time of L2 cache+miss rate of L2 cache* miss penalty of L2 cache=8+18 L2

Acc to given equation  

1+2 L2 +8+18 L2=2

 L2= 0.35

L1=0.70

Answer 2

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