In a two-level cache system, the access times of L1 and

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In a two-level cache system, the access times of L₁ and L₂ caches are 1 and 8 clock cycles, respectively. The miss penalty from the L₂ cache to main memory is 18 clock cycles. The miss rate of L₁ cache is twice that of L₂. The average memory access time (AMAT) of this cache system is 2 cycles. The miss rates of L₁ and L₂ respectively are :

1. 0.111 and 0.056
2. 0.056 and 0.111
3. 0.0892 and 0.1784
4. 0.1784 and 0.0892

Correct Option: A

As given that :
The access time L₁ = 1 clock cycles
or (Hit time)
The access time L₂ = 8 clock cycles
or (Hit time)
Miss penality L₂ = 18 clock cycles
(T_Avg = 2 cycles)
Miss Rate (L₁ = 2X)
Miss Rate (L₂ = X)
As we know that,
T_Average = Hit time L₁ + (Miss Rate L₁ × Miss Palenality L₁) ...(1)
Miss Palenality L₁ = Hit time L₂ + (Miss Rate L₂ × Miss Palenality L₂) ....(2)
Put the given values in equation (1) and (2).
2 = 1 + 2 × (Miss Palenality 4)
1 = 2 × (L₁)

X =		1		.....(3)
		2L₁

(Miss Palenality L₁) = 8 + X. (18)

From eq. (3) put (L₁) value to eq. (4)

L₁ =		1
		2X

	1	= 8 + 18X
	2X

	1	- 18X		= 8
	2X

1 - 36X² = 16X
36X² + 16X - 1 = 0
Compare to aX² + bX + c = 0
a = 36, b = 16, c = –1

=	- b ± √b² - 4ac
	2a

=	-16 ± √256 + 4 × 36 × 1
	2 × 36

=	-16 ± 2	=	4	(take +ve value only)
	72		72

X = 0.055

Then Miss rate L₁ = 2X = 2 × 0.055 = 0.111
L₂ = X = 0.055
Hence option (a) is correct.