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Solution Manual For Computer Organization and Design MIPS Edition The Hardware Software Interface 6th Edition By David Patterson, John Hennessy (All Chapters, 100% Original Verified, A+ Grade)14,29 €
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Solution Manual For Computer Organization and Design MIPS Edition The Hardware Software Interface 6th Edition By David Patterson, John Hennessy (All Chapters, 100% Original Verified, A+ Grade)
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Computer Organization and Design MIPS Edition The
Institución
Computer Organization And Design MIPS Edition The
Solution Manual For Computer Organization and Design MIPS Edition The Hardware Software Interface 6th Edition By David Patterson, John Hennessy (All Chapters, 100% Original Verified, A+ Grade)
Solution Manual For Computer Organization and Design MIPS Edition The Hardware Software Interface 6th...
Chapter 1 Solutions S-3To protect the rights of the author(s) and publisher we inform you that this PDF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and typesetter MPS. It is not allowed to publish this proof online or in print. Th is proof copy is the copyright property of the publisher and is confi dential until formal publication.
1.1 Personal computer: Computer that emphasizes delivery of good performance to a single user at low cost and usually executes third-party soft ware.
Server: Computer used for large workloads and usually accessed via a network.
Embedded computer: Computer designed to run one application or one set of related applications and integrated into a single system.
1.2 a. Performance via Pipelining
b. Dependability via Redundancy
c. Performance via Prediction
d. Make the Common Case Fast
e. Hierarchy of Memories
f. Performance via Parallelism
g. Use Abstraction to Simplify Design
1.3 Th e program is compiled into an assembly language program, which is itself assembled into a machine language program.
1.4 a. 1280 × 1024 pixels = 1,310,720 pixels => 1,310,720 × 3 = 3,932,160 bytes/
frame.
b. 3,932,160 bytes × (8 bits/byte) /100E6 bits/second = 0.31 seconds
1.5 Desktop
Processor Year TechMax. Clock Speed (GHz)Integer IPC/
core CoresMax. DRAM Bandwidth (GB/s)SP Floating Point (Gfl op/s) MiB
Westmere i7-6202010 32 3.33 4 2 17.1 107 4
Ivy Bridge i7-3770K2013 22 3.90 6 4 25.6 250 8
Broadwell i7-6700K2015 14 4.20 8 4 34.1 269 8
Kaby Lake i7-7700K2017 14 4.50 8 4 38.4 288 8
Coffee Lake i7-9700K2019 14 4.90 8 8 42.7 627 12
Imp./year 20% 4% 7% 15% 10% 19% 12%
Doubles every 4 years 18 years 10 years 5 years 7 years 4 years 6 years
solution 1123.indd 3solution 1123.indd 3 22-09-2021 20:52:2222-09-2021 20:52:22Computer Organization and Design MIPS Edition The Hardware Software Interface, 6e David Patterson, John
Hennessy
Solution Manual all Chapters S-4 Chapter 1 SolutionsTo protect the rights of the author(s) and publisher we inform you that this PDF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and typesetter MPS. It is not allowed to publish this proof online or in print. Th is proof copy is the copyright property of the publisher and is confi dential until formal publication.
1.6 a. performance of P1 (instructions/sec) = 3 × 109/1.5 = 2 × 109
performance of P2 (instructions/sec) = 2.5 × 109/1.0 = 2.5 × 109
performance of P3 (instructions/sec) = 4 × 109/2.2 = 1.8 × 109
b. cycles(P1) = 10 × 3 × 109 = 30 × 109 s
cycles(P2) = 10 × 2.5 × 109 = 25 × 109 s
cycles(P3) = 10 × 4 × 109 = 40 × 109 s
c. No. instructions(P1) = 30 × 109/1.5 = 20 × 109
No. instructions(P2) = 25 × 109/1 = 25 × 109
No. instructions(P3) = 40 × 109/2.2 = 18.18 × 109
CPInew = CPIold × 1.2, then CPI(P1) = 1.8, CPI(P2) = 1.2, CPI(P3) = 2.6
f = No. instr. × CPI/time, then
f(P1) = 20 × 109 × 1.8/7 = 5.14 GHz
f(P2) = 25 × 109 × 1.2/7 = 4.28 GHz
f(P3) = 18.18 × 109 × 2.6/7 = 6.75 GHz
1.7 a. Class A: 105 instr. Class B: 2 × 105 instr. Class C: 5 × 105 instr. Class D: 2 × 105 instr.
Time = No. instr. × CPI/clock rate
Total time P1 = (105 + 2 × 105 × 2 + 5 × 105 × 3 + 2 × 105 × 3)/(2.5 × 109) = 10.4 × 10−4 s
Total time P2 = (105 × 2 + 2 × 105 × 2 + 5 × 105 × 2 + 2 × 105 × 2)/(3 × 109) = 6.66 × 10−4 s
CPI(P1) = 10.4 × 10−4 × 2.5 × 109/106 = 2.6
CPI(P2) = 6.66 × 10−4 × 3 × 109/106 = 2.0
b. clock cycles(P1) = 105 × 1+ 2 × 105 × 2 + 5 × 105 × 3 + 2 × 105 × 3 = 26 × 105
clock cycles(P2) = 105 × 2+ 2 × 105 × 2 + 5 × 105 × 2 + 2 × 105 × 2 = 20 × 105
1.8 a. CPI = Texec × f/No. instr.
Compiler A CPI = 1.1
Compiler B CPI = 1.25
solution 1123.indd 4 solution 1123.indd 4 22-09-2021 20:52:22 22-09-2021 20:52:22 Chapter 1 Solutions S-5To protect the rights of the author(s) and publisher we inform you that this PDF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and typesetter MPS. It is not allowed to publish this proof online or in print. Th is proof copy is the copyright property of the publisher and is confi dential until formal publication.
b. fB/fA = (No. instr.(B) × CPI(B))/(No. instr.(A) × CPI(A)) = 1.37
c. TA/Tnew = 1.67
TB/Tnew = 2.27
1.9 1.9.1 C = 2 × DP/(V
2 × F)
Pentium 4: C = 3.2E–8F
Core i5 Ivy Bridge: C = 2.9E–8F
1.9.2 Pentium 4: 10/100 = 10%
Core i5 Ivy Bridge: 30/70 = 42.9%
1.9.3 (Snew + Dnew)/(Sold + Dold) = 0.90
Dnew = C × Vnew 2 × F
Sold = Vold × I
Snew = Vnew × I
Th erefore:
Vnew = [Dnew/(C × F)]1/2
Dnew = 0.90 × (Sold + Dold) − Snew
Snew = Vnew × (Sold/Vold)
Pentium 4:S
new = Vnew × (10/1.25) = Vnew × 8
Dnew = 0.90 × 100 − Vnew × 8 = 90 − Vnew × 8
Vnew = [(90 − Vnew × 8)/(3.2E8 × 3.6E9)]1/2
Vnew = 0.85 V
Core i5:S
new = Vnew × (30/0.9) = Vnew × 33.3
Dnew = 0.90 × 70 − Vnew × 33.3 = 63 − Vnew × 33.3
Vnew = [(63 − Vnew × 33.3)/(2.9E8 × 3.4E9)]1/2
Vnew = 0.64 V
solution 1123.indd 5 solution 1123.indd 5 22-09-2021 20:52:22 22-09-2021 20:52:22 S-6 Chapter 1 SolutionsTo protect the rights of the author(s) and publisher we inform you that this PDF is an uncorrected proof for internal business use only by the author(s), editor(s), reviewer(s), Elsevier and typesetter MPS. It is not allowed to publish this proof online or in print. Th is proof copy is the copyright property of the publisher and is confi dential until formal publication.
1.10 1.10.1
p # arith inst. # L/S inst. # branch inst. cycles ex. time speedup
1 2.56E9 1.28E9 2.56E8 1.92E10 9.60 1.00
2 1.83E9 9.14E8 2.56E8 1.41E10 7.04 1.364 9.14E8 4.57E8 2.56E8 7.68E9 3.84 2.508 4.57E8 2.29E8 2.56E8 4.48E9 2.24 4.29
1.10.2
p ex. time
1 41.02 29.34 14.68 7.33
1.10.3 3
1.11 1.11.1 die area15cm = wafer area/dies per wafer = π × 7. = 2.10 cm2
yield15cm = 1/(1 + (0.020 × 2.10/2))2 = 0.9593
die area20cm = wafer area/dies per wafer = π × 102/100 = 3.14 cm2
yield20cm = 1/(1 + (0.031 × 3.14/2))2 = 0.9093
1.11.2 cost/die15cm = 12/(84 × 0.9593) = 0.1489
cost/die20cm = 15/(100 × 0.9093) = 0.1650
1.11.3 die area15cm = wafer area/dies per wafer = π × 7.52/(84 × 1.1) = 1.91 cm2
yield15cm = 1/(1 + (0.020 × 1.15 × 1.91/2))2 = 0.9575
die area20cm = wafer area/dies per wafer = π × 102/(100 × 1.1) = 2.86 cm2
yield20cm = 1/(1 + (0.03 × 1.15 × 2.86/2))2 = 0.9082
1.11.4 defects per area0.92 = (1–y .5)/(y .5 × die_area/2) = (1 − 0.92.5)/
(0.92.5 × 2/2) = 0.043 defects/cm2
defects per area0.95 = (1–y .5)/(y .5 × die_area/2) = (1 − 0.95.5)/
(0.95.5 × 2/2) = 0.026 defects/cm2
1.12 1.12.1 CPI = clock rate × CPU time/instr. count
clock rate = 1/cycle time = 3 GHz
CPI(bzip2) = 3 × 109 × 750/(2389 × 109)= 0.94
solution 1123.indd 6 solution 1123.indd 6 22-09-2021 20:52:22 22-09-2021 20:52:22
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