Cache memory(made easy)

Question

During a program execution out of 1000 memory references there are 250 and 120 misses in L1 (Level1) and L2(Level2) caches respectively. Hit times for L1 and L2 cache are 24 and 40 cycles respectively. If there are 2.5 memory references per instruction, how many average stall cycles per instruction? (Assume L2 to memory miss penalty is 250 cycles)?

a)50

b)100

c)150

d)200

Answer 1

Average stall/instr. = Misses in L1/instruction * Hit time of L2 + Misses in L2/instruction * Misspenalty in L2

  no of instruction = 1000/2.5 = 400.

so, Average stall/instruction = 250/400 * 40 + 120/400 * 250 = 100.

option B is answer.

Comments

@sayantan can u tell me why dont we consider hit time of L1.I mean it is memory stall so ideally hit time in L1 also need to be consider na.....plz if u can provide any source/info for derivation of this formula :

Average stall/instr. = Misses in L1/instruction * Hit time of L2 + Misses in L2/instruction * Misspenalty in L2

i

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ya i got explanation 4 that because if word found in cache there wont be any stall because CPI is still 1 assuming hit time in L1 is 1cycle.....thanks for ur help :)

Answer 2

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 ⇒ 10002.5 instructions for 1000 references = 400 instructions
∴ Average number of stalls per instruction
=(250400×40)+(120400×250)
= 25 + 75 = 100 cycles