6-2
1615: 000000 1 000000 0 100010 0 0 0 0001 110 00000010101 21: 100001 0 100001 0 100011 0 0 0 0010 000 00000000000
6-3 00000000 00000000 0 00 00000000 0
00000111 00000011 0 01 00010101 0 11111111 11111111 1 11 11111110 1
CHAPTER 7 SOLUTIONS 7-1
T
eff
= .95 × 100ns + .05 × 800ns = 135ns
7-2
Tag: 8 bits Set: 4 bits Word: 7 bits
7-3 page offset
2050 = 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 virtual 4096 = 1 0 0 0 0 0 0 0 0 0 0 1 0 physical
25 = 0 0 0 0 0 0 0 0 1 1 0 0 1 physical 4121 = 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 1 virtual
7-4
a 0, 2, 4, 5, 3, 11, 10 b 0, 2, 4, 5, 3, 11, 2, 10
7-5
a N - M N b M - F M
c T1×N - M N + T2 × M - F M + T3 × F
7-6 C 10
7-7
Iteration 1: 5 misses and 60 hits Iterations 2 - 10: 2 misses and 63 hits per iteration
Hit ratio = hitsmisses = 60 + 9 x 63 65 x 10 = 96.56
7-8
Tag Set
Byte 8 bits
4 bits 7 bits
7-9
B 5; page faults: 1, 8, 7, 2, 3
7-10
False
7-11 a
Tag Slot
Byte 7 bits
8 bits 4 bits
b
Hit ratio = 34 + 9 x 38 10 x 38 = 376 380 = 98.94 T
eff
= [4 misses x 210nsmiss + 34 + 9 x 38 hits x 10 nshit] 380 accesses = 12.1 ns
7-12 a
00000000000
b
This address is in Page 1 which is not present, so there is no virtual address.
7-13
24
7-14 D 31
7-15 T
eff
= .9 × T1 + .9 × 1 - .9 × T2 + [1 - .9 × 1 - .9] × T3 CHAPTER 8 SOLUTIONS
8-1
1 Read 0, 1, 2, 3 on first rotation. 2 Write 0, 1, 2, 3; read 4, 5, 6, 7 on second rotation
3 Write 4, 5, 6, 7 on third rotation. Total time = 3.0 rotations x 16 msrotation = 48 ms.
8-2 a
16 surfaces x 2048 trackssurface x 256 sectorstrack x 512 bytessector = 2
32
bytes
b
7200 revmin x 160 minsec x 1 trackrev x 256 sectorstrack x 512 bytessector = 15.7 MBsec
c
Avg seek time = 7ms. Avg rotational delay = [17200 minrev x 60 secmin]2 = .00417s = 4.17ms
Sector readwrite time = 18 x [17200 minrev x 60 secmin] = .00104s = 1.04ms Avg transfer time = [7ms + 4.17ms + 1.04ms read] + [7ms + 4.17ms + 1.04ms write] = 24.42ms
8-3
10
6
wordss per bus × 1.5 × 10
6
sword per disk = 20 disksbus
CHAPTER 9 SOLUTIONS 9-1
1 1 1 0 1 0 0 1 0 1 1 1 -- -- -- -- -- -- -- -- -- -- -- --
12 11 10 9 C8 7 6 5 C4 3 C2 C1
9-2 0 0 0 0 0 0 0 0 0 1 0 1 1 1 1 1 0 1 0 1 0
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- 21 20 19 18 17C16 15 14 13 12 11 10 9 C8 7 6 5 C4 3 C2 C1
9-3
We want a Hamming distance of 2 between valid numbers, in base 10. In base 2, we looked for evenodd parity, which can be generalized to mod2sum of digits in number = 0 for even parity. We
want mod10sum of digits in number = 0 for this problem. So, mod_104+4+5+2+6+5+4 = 0 is a valid telephone number 445-2654 whereas mod_104+4+5+3+5+2+3 = 6 is not 445-3523.
Since only 110 of the numbers will come out as mod_10 = 0, only 10 of the possible numbers can be assigned.
9-4
For each of the 10
7
valid phone numbers, there is an uncorrupted version, and 9×7 ways to corrupt each of the 7 original digits, plus 9r ways to corrupt each of the r check digits.
The following relationship must hold: 10
7
9×7 + 9r + 1 = 10
7 + r
which simplifies to 64 + 9r = 10
r
for which r = 2 is the smallest value that satisfies the relation.
9-5 a
C 1 00010 D 0 00011
E 1 00100 F 0 00101
C 0 00110 Chk 0 00110
b The letter ‘M’ 9-6 B 43
CHAPTER 10 SOLUTIONS 10-1 1 + P
