Question

When there is miss in cache. CPU access main memory to transfer a block of 64B to cache. Main memory is byte addressable with access time of 100 nsec. If CPU takes 10 nsec to send the transfer request (which includes main memory address and control signals) to main memory and memory system transfers content to cache with a rate of 1 GB/sec, then total miss penalty time is ______nsec.

Answer 1

64 Byte block access time from main memory = 64 * 100 nsec
= 6400 nsec
[because 1 byte can be accessed in 1 main memory access]
$\text{Byte transfer time to cache =}\frac{1sec}{1G}=\text{1 nsec}$
64 Byte transfer time to cache = 64nsec
Total time = Request time + block access time + block transfer from main memory time to cache
= 10 + 6400 + 64
= 6474 nsec