Question

consider a memory system with a memory access time of 150 ns and a cache access time of 20ns. If the effective access time is 20% greater than the cache access time, what is the hit ratio H g

Answer 1

Answer:

H = 0.7333333

Explanation:

Given that:

The memory access time ($T_m$) = 150 ns

The cache access time $(T_c)$ = 20 ns

Effective access time $(T_e)$  = 20% > $(T_c)$

Then, it implies that:

= $(T_c)$ + 20% of

=$(T_c)$(1+20%)

=$(T_c)$(1+ 0.2)

= 20ns × 1.2

= 24ns

To determine the hit ratio H;

Using the formula:

$T_e = T_c \times H+(1-H) \times (T_c + T_m) \\ \\ T_e = HT_c + T_c + T_m -HT_c -HT_m \\ \\ T_e = T_c +T_m - HT_m \\ \\ T_c -T_e = T_m (H-1) \\ \\ H-1 = \dfrac{T_c -T_e}{T_m} \\ \\ H = 1+ (\dfrac{T_c -T_e}{T_m})--- (1)$

Replacing the values; we have:

$T_c - T_e = 20ns - 24 ns \\ \\ T_c - T_e = -4 ns \\ \\ \dfrac{T_c - T_e }{T_m} = \dfrac{-4 ns}{150} \\ \\ \dfrac{T_c - T_e }{T_m} = -0.02666667$

From (1)

$H = 1+ (-0.2666667) \\ \\ H = 1 - 0.2666667 \\ \\ \mathbf{H = 0.7333333}$