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Solution for chapter 6
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2.1
1. 4 (1 byte = 8 bit ; 16*8/32 = 4)
2. I,J
3. A[I][J]
2.2
1. Tag = address/16
Index = add%16
address tag index H/M
3 0 3 M
180 11 4 M
43 2 11 M
2 0 2 M
191 11 15 M
88 5 8 M
190 11 14 M
14 0 14 M
181 11 5 M
44 2 12 M
186 11 10 M
253 15 13 M
2.4
1. 8 (offset 32 byte = 8 word )
2. 32
3. 1+( 22/(8*32)) = 1.086
4.
M/M Block Address : Address DIV/32
Tag : M.M Block Address DIV/32
Index: M.M Block Address MOD/32
Address M.M Block
Address
tag index Replace? Hit?
0 0 0 0 N N
4 0 0 0 N Y
16 0 0 0 N Y
132 4 0 4 N N
232 7 0 7 N N
160 5 0 5 N N
1024 32 1 0 Y N
30 0 0 0 N Y
140 4 0 4 N Y
3100 96 3 0 Y N
180 5 0 5 N Y
2180 68 2 4 Y N
3
5. 5 5/12 = 0.417
6.
2.6
1. P1 : 1/0.66n = 1.52GHz P2 = 1.11GHz
2. AMAT for P1 = 0.66 + (8%*70) = 6.26ns 6.26n/0.66n =9.48 cycles
AMAT for P2 = 5.1ns 5.1n/0.9 = 5.67 cycles
3. CPI for P1 = 1+1.36*0.08*70/0.66 = 12.54
CPI for P2 = 1+1.36*0.06*70/0.9 = 7.36
P2 faster
4. AMAT = 0.66+0.08*5.62+0.08*0.95*70 = 6.43 ns
Cycles = 6.43/0.66 = 9.74 worst
5. CPI = 1+(1.36*0.08*5.62/0.66)+(1.36*0.08*0.95*70/0.66) = 12.89
6.
P1 AMAT 0.66 ns + 0.08 * 70 ns = 6.26 ns
P2 AMAT 0.90 ns + 0.06 * (5.62 ns + 0.95 * 70 ns) = 5.23 ns
For P1 to match P2s performance:
5.23 = 0.66 ns + MR * 70 ns
MR = 6.5%