#1 Testbook Mock Test (COA - Cache Memory)

Question

Suppose that in $500$ memory references there are $50$ misses in the first level cache and $20$ misses in the second level cache. Assume miss penalty from the $L2$ cache to memory in $100$ cycles. The hit time of $L2$ cache is $20$ cycles. The hit time of the $L1$ cache is $10$ cycles. If there are $2.5$ memory references per instruction, how many average stall cycles per instructions.

• A. $10$
• B. $5$
• C. $15$
• D. $\text{None of these}$

Comments

Given data is for $500$ memory references.

find for $2.5$ memory references.

i.e $0.25$ miss rate for L$_1$ and $0.1$ for L$_2$.

$\therefore$ Stalls $=0.25\times20+0.1\times100=15 $ cycles/instruction

Answer 1

Average Number of stalls per instruction = (# misses per instruction in L1   × Hit time in L2) + (# misses per instruction in L2 × Miss penalty of L2)
2.5 memory references per instruction ⇒ 500/2.5 instructions for 500 references = 200 instructions
∴ Average number of stalls per instruction
=(50/200)×20+(20/200)×100
= 5 + 10 = 15 cycles.