Calculating memory stalls while adding second level cache

Question

I am trying to calculate memory stall cycles per instructions when adding the second level cache.

I have the following given values:

Direct Mapped cache with 128 blocks
16 KB cache
2ns Cache access time
1Ghz Clock Rate
1 CPI
80 clock cycles Miss Penalty
5% Miss rate
1.8 Memory Accesses per instruction
16 bit memory address

L2 Cache
4% Miss Rate
6 clock cycles miss penalty

As I understand it, the way to calculate the Memory stall cycles is by using the following formula:

Memory stall cycles = Memory accesses x Miss rate x Miss penalty

Which can be simplified as:

Memory stall cycles = instructions per program x misses per instructions x miss penalty

What I did was to multiply 1.8 x (.05 +.04) x (80 + 6) = 13.932

Would this be correct or am I missing something?

A Farmanbar · Accepted Answer · 2021-11-12 05:21:09Z

2

First of all, I am not sure about the given parameters for miss penalty for L1 and L2 (L1 being 80 cycles and L2 being 6 cycles).

Anyway using the data as it is:

You issue 1 instruction per clock
There are 1.8 memory instructions in an instruction.
There is a 5% that access can miss L1 and another 4% chance that it can miss L2. You would only access the main memory if you miss in both L1 and L2. That would be .04 * .05 = 0.002 = 0.2% This means per memory access, you are likely to access the main memory 0.2% of the time.
Since you have 1.8 memory accesses per instruction, you are likely to access main memory 0.002 * 1.8 = 0.0036 = 0.36% per instruction.
When you encounter a miss in both L1 and L2, you will get stalled for 80 + 6 = 86 cycles (ignoring any optimizations)
Per instruction, you would only encounter .36% main memory accesses. Hence the memory stall cycles per instruction is .0036 * 86 = 0.3096

edited Nov 12, 2021 at 5:21

A Farmanbar

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answered May 8, 2017 at 7:22

Isuru H

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2 Comments

Rick Over a year ago

In #3 you are multiplying the miss rates of L1 and L2 caches to get the main memory access?

Isuru H Over a year ago

Yes. Main memory will only be accessed in case of a miss in both L1 and L2. That is 4%(L2) out of 5%(L1). So the % of main memory accesses is the multiplication of those two.

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