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

H = 0.7333333

Step-by-step 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 * H+(1-H) * (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 = (T_c -T_e)/(T_m) \\ \\ H = 1+ ((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 \\ \\ (T_c - T_e )/(T_m) = (-4 ns)/(150) \\ \\ (T_c - T_e )/(T_m) = -0.02666667`

From (1)

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