CS232 Discussion 9: Model Solutions
1. Here is a series of address references given as word addresses: 2, 3, 11, 16, 21, 13, 64, 48,
19, 11, 3, 22, 4, 27, 6, and 11. Assuming a direct-mapped cache with 16 one-word blocks
that is initially empty, label each reference in the list as a hit or a miss and show the final
contents of the cache.
Solution:
2 miss, 3 miss, 11 miss, 16 miss, 21 miss, 13 miss, 64 miss, 48 miss, 19 miss, 11 hit, 3 miss,
22 miss, 4 miss, 27 miss, 6 miss, 11 miss.
Index Cache Contents
0 16 64 48
1
2 2
3 3 19 3
4 4
5 21
6 22 6
7
8
9
10
11 11 27 11
12
13 13
14
15
2. Assume that 33% of the instructions in a program are data accesses. The cache hit ratio
is 97% and the hit time is one cycle, but the miss penalty is 20 cycles.
AM AT = Hit time + (Miss rate ×Miss penalty)
= 1 cycle + (3% ×20 cycles)
= 1.6 cycles
Let Ibe the total number of instructions.
Memory Stall Cycles = Memory Accesses ×Miss rate ×Miss penalty
= 0.33I×0.03 ×20
= 0.198I
If the cache was perfect and never missed, the AMAT would be one cycle. But even with
just a 3% miss rate, the AMAT here increases 1.6 times!
1
