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1 Solutions
Solution 1.
1.1.1 Computer used to run large problems and usually accessed via a network:
(3) servers
1.1.2 10 15 or 2^50 bytes: (7) petabyte
1.1.3 A class of computers composed of hundred to thousand processors and tera-
bytes of memory and having the highest performance and cost: (5) supercomputers
1.1.4 Today’s science fiction application that probably will be available in near
future: (1) virtual worlds
1.1.5 A kind of memory called random access memory: (12) RAM
1.1.6 Part of a computer called central processor unit: (13) CPU
1.1.7 Thousands of processors forming a large cluster: (8) data centers
1.1.8 Microprocessors containing several processors in the same chip: (10) multi-
core processors
1.1.9 Desktop computer without a screen or keyboard usually accessed via a net-
work: (4) low-end servers
1.1.10 A computer used to running one predetermined application or collection
of software: (9) embedded computers
1.1.11 Special language used to describe hardware components: (11) VHDL
1.1.12 Personal computer delivering good performance to single users at low cost:
(2) desktop computers
1.1.13 Program that translates statements in high-level language to assembly
language: (15) compiler
S2 Chapter 1 Solutions
1.1.14 Program that translates symbolic instructions to binary instructions: (21) assembler
1.1.15 High-level language for business data processing: (25) Cobol
1.1.16 Binary language that the processor can understand: (19) machine language
1.1.17 Commands that the processors understand: (17) instruction
1.1.18 High-level language for scientific computation: (26) Fortran
1.1.19 Symbolic representation of machine instructions: (18) assembly language
1.1.20 Interface between user’s program and hardware providing a variety of services and supervision functions: (14) operating system
1.1.21 Software/programs developed by the users: (24) application software
1.1.22 Binary digit (value 0 or 1): (16) bit
1.1.23 Software layer between the application software and the hardware that includes the operating system and the compilers: (23) system software
1.1.24 High-level language used to write application and system software: (20) C
1.1.25 Portable language composed of words and algebraic expressions that must be translated into assembly language before run in a computer: (22) high-level language
1.1.26 10 12 or 2^40 bytes: (6) terabyte
Solution 1. 1.2.1 8 bits × 3 colors = 24 bits/pixel = 3 bytes/pixel.
a. Confi guration 1: 640 × 480 pixels = 179,200 pixels => 179,200 × 3 = 537,600 bytes/frame Confi guration 2: 1280 × 1024 pixels = 1,310,720 pixels => 1,310,720 × 3 = 3,932, bytes/frame b. Confi guration 1: 1024 × 768 pixels = 786,432 pixels => 786,432 × 3 = 2,359, bytes/frame Confi guration 2: 2560 × 1600 pixels = 4,096,000 pixels => 4,096,000 × 3 = 12,288, bytes/frame
AQ 1
S4 Chapter 1 Solutions
1.3.2 No. cycles = time × clock rate time = (No. Instr × CPI)/clock rate, then No. instructions = No. cycles/CPI
a. 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 No. instructions(P1) = 30 × 10 9 /1.5 = 20 × 109 No. instructions(P2) = 25 × 10 9 /1 = 25 × 109 No. instructions(P3) = 40 × 10 9 /2.2 = 18.18 × 109 b. cycles(P1) = 10 × 2 × 109 = 20 × 109 s cycles(P2) = 10 × 3 × 109 = 30 × 109 s cycles(P3) = 10 × 4 × 109 = 40 × 109 s No. instructions(P1) = 20 × 10 9 /1.2 = 16.66 × 109 No. instructions(P2) = 30 × 10 9 /0.8 = 37.5 × 109 No. instructions(P3) = 40 × 10 9 /2 = 20 × 109
1.3.3 timenew = time (^) old × 0.7 = 7 s
a. CPI (^) new = CPIold × 1.2, then CPI(P1) = 1.8, CPI(P2) = 1.2, CPI(P3) = 2. f = No. Instr × CPI/time, then f(P1) = 20 × 109 × 1.8 / 7 = 5.14 GHz f(P2) = 25 × 10 9 × 1.2 / 7 = 4.28 GHz f(P1) = 18.18 × 109 × 2.6 / 7 = 6.75 GHz b. CPI (^) new = CPIold × 1.2, then CPI(P1) = 1.44, CPI(P2) = 0.96, CPI(P3) = 2. f = No. Instr × CPI/time, then f(P1) = 16.66 × 109 × 1.44/7 = 3.42 GHz f(P2) = 37.5 × 10 9 × 0.96/7 = 5.14 GHz f(P1) = 20 × 10 9 × 2.4/7 = 6.85 GHz
1.3.4 IPC = 1/CPI = No. instr/(time × clock rate)
a. IPC(P1) = 0. IPC(P2) = 1. IPC(P3) = 2. b. IPC(P1) = 2 IPC(P2) = 1. IPC(P3) = 0.
1.3.
a. Time (^) new/Timeold = 7/10 = 0.7. So fnew = fold /0.7 = 2.5 GHz/0.7 = 3.57 GHz. b. Timenew/Timeold = 5/8 = 0.625. So fnew = fold/0.625 = 4.8 GHz.
AQ 3
Chapter 1 Solutions S
1.3.
a. Timenew /Timeold = 9/10 = 0.9. Then Instructionsnew = Instructions (^) old × 0.9 = 30 × 109 × 0.9 = 27 × 10 9. b. Time (^) new /Timeold = 7/8 = 0.875. Then Instructionsnew = Instructions (^) old × 0.875 = 26.25 × 109.
Solution 1.
1.4.
Class A: 10^5 instr. Class B: 2 × 10 5 instr. Class C: 5 × 10 5 instr. Class D: 2 × 105 instr.
Time = No. instr × CPI/clock rate
a. Total time P1 = (10^5 + 2 × 10 5 × 2 + 5 × 10 5 × 3 + 2 × 105 × 3)/(2.5 × 109 ) = 10.4 × 10 −^4 s Total time P2 = (10^5 × 2 + 2 × 105 × 2 + 5 × 105 × 2 + 2 × 105 × 2)/(3 × 10 9 ) = 6.66 × 10 −^4 s b. Total time P1^ =^ (10^5 ×^2 +^2 ×^105 ×^ 1.5^ +^5 ×^10 5 ×^2 +^2 ×^10 5 )/(2.5^ ×^109 )^ =^ 6.8^ ×^10 −^4 s Total time P2 = (10^5 + 2 × 10 5 × 2 + 5 × 10 5 + 2 × 10 5 )/(3 × 10 9 ) = 4 × 10 −^4 s
1.4.2 CPI = time × clock rate/No. instr
a. CPI (P1) = 10.4 × 10 −^4 × 2.5 × 109 /10^6 = 2. CPI (P2) = 6.66 × 10 −^4 × 3 × 109 /10^6 = 2. b. CPI (P1)^ =^ 6.8^ ×^10 −^4 ×^ 2.5^ ×^10 9 /10^6 =^ 1. CPI (P2) = 4 × 10 −^4 × 3 × 10 9 /10^6 = 1.
1.4.
a. clock cycles (P1) = 10 5 × 1 + 2 × 10 5 × 2 + 5 × 105 × 3 + 2 × 105 × 3 = 26 × 105 clock cycles (P2) = 105 × 2 + 2 × 10 5 × 2 + 5 × 105 × 2 + 2 × 105 × 2 = 20 × 105 b. clock cycles (P1) = 17 × 105 clock cycles (P2) = 12 × 105
1.4.
a. (650 × 1 + 100 × 5 + 600 × 5 + 50 × 2) × 0.5 × 10–9^ = 2,125 ns b. (750 × 1 + 250 × 5 + 500 × 5 + 500 × 2) × 0.5 × 10–9^ = 2,750 ns
1.4.5 CPI = time × clock rate/No. instr
a. CPI = 2,125 × 10–9^ × 2 × 10^9 /1,400 = 3. b. CPI = 2,750 × 10–9^ × 2 × 10^9 /2,000 = 2.
Chapter 1 Solutions S
Solution 1.
1.6.1 CPI = Texec × f/No. Instr
Compiler A CPI Compiler B CPI a. 1.8 1. b. 1.1 1.
1.6.2 f (^) A/fB = (No. Instr(A) ´ CPI(A))/(No. Instr(B) ´ CPI(B))
a. fA/fB = 1 b. fA/fB = 0.
1.6.
Speedup vs. Compiler A Speedup vs. Compiler B a. T (^) new /TA = 0.36 Tnew/T (^) B = 0. b. T (^) new /TA = 0.6 Tnew/T (^) B = 0.
1.6.
P1 Peak P2 Peak a. 4 × 10^9 Inst/s^ 2 × 10^9 Inst/s b. 4 × 10^9 Inst/s^ 3 × 10^9 Inst/s
1.6.5 Speedup, P1 versus P2:
a. T 1 /T 2 = 1. b. T 1 /T 2 = 1.
1.6.
a. 4.37 GHz b. 6 GHz
Solution 1.
1.7.
Geometric mean clock rate ratio = (1.28 × 1.56 × 2.64 × 3.03 × 10.00 × 1.80 × 0.74) 1/7^ = 2.
Geometric mean power ratio = (1.24 × 1.20 × 2.06 × 2.88 × 2.59 × 1.37 × 0.92)1/7^ = 1.
S8 Chapter 1 Solutions
1.7. Largest clock rate ratio = 2000 MHz/200 MHz = 10 (Pentium Pro to Pentium 4 Willamette) Largest power ratio = 29.1 W/10.1 W = 2.88 (Pentium to Pentium Pro)
1.7. Clock rate: 2.667 × 109 /12.5 × 106 = 213. Power: 95 W/3.3 W = 28.
1.7.4 C = P/V 2 × clock rate 80286: C = 0.0105 × 10 −^6 80386: C = 0.01025 × 10 −^6 80486: C = 0.00784 × 10 −^6 Pentium: C = 0.00612 × 10 −^6 Pentium Pro: C = 0.0133 × 10 −^6 Pentium 4 Willamette: C = 0.0122 × 10 −^6 Pentium 4 Prescott: C = 0.00183 × 10 −^6 Core 2: C = 0.0294 × 10 −^6
1.7.5 3.3/1.75 = 1.78 (Pentium Pro to Pentium 4 Willamette)
1.7. Pentium to Pentium Pro: 3.3/5 = 0. Pentium Pro to Pentium 4 Willamette: 1.75/3.3 = 0. Pentium 4 Willamette to Pentium 4 Prescott: 1.25/1.75 = 0. Pentium 4 Prescott to Core 2: 1.1/1.25 = 0. Geometric mean = 0.
Solution 1. 1.8.1 Power = V 2 × clock rate × C. Power 2 = 0.9 Power 1
a. C 2 /C 1 = 0.9 × 1.75^2 × 1.5 × 10^9 /(1.2^2 × 2 × 10^9 ) = 1. b. C 2 /C 1 = 0.9 × 1.1^2 × 3 × 10^9 /(0.8 2 × 4 × 10^9 ) = 1.
1.8.2 Power 2 /Power 1 = V 22 × clock rate 2 /(V 12 × clock rate 1 )
a. Power 2 /Power 1 = 0.62 => Reduction of 38% b. Power 2 /Power 1 = 0.7 => Reduction of 30%
S10 Chapter 1 Solutions
1.9.
a. Power (^) st /Power (^) dyn = 10/50 = 0. b. Power (^) st /Power (^) dyn = 60/90 = 0.
1.9.4 Powerst/Powerdyn = 0.6 => Powerst = 0.6 × Powerdyn
a. Power^ st = 0.6 × 35 W = 21 W b. Power^ st = 0.6 × 30 W = 18 W
1.9.
1.2 V 1.0 V 0.8 V a. P^ st = 12.5 W Pdyn = 62.5 W
Pst = 10 W Pdyn = 50 W
Pst = 5.8 W Pdyn = 29.2 W b. Pst = 24.8 W Pdyn = 37.2 W
Pst = 20 W Pdyn = 30 W
Pst = 12 W Pdyn = 18 W
1.9.
a. 29. b. 23.
Solution 1.
a. Processors Instructions per Processor Total Instructions 1 4096 4096 2 2048 4096 4 1024 4096 8 512 4096
b. Processors Instructions per Processor Total Instructions 1 4096 4096 2 2048 4096 4 1024 4096 8 512 4096
a. Processors Execution Time (μs)
1.11.2 Cost per die = cost per wafer/(dies per wafer × yield)
Then defect per area = (2/die area)(y−1/2^ − 1) 1.12.1 CPI = clock rate × CPU time/instr count
- Chapter 1 Solutions S
- 1.10.
- 1 4.
- 2 2.
- 4 1.
- 8 1.
- 1 4. b. Processors Execution Time (μs)
- 2 2.
- 4 1.
- 8 0.
- 1.10.
- 1 5. a. Processors Execution Time (μs)
- 2 3.
- 4 1.
- 8 1.
- 1 5. b. Processors Execution Time (μs)
- 2 3.
- 4 1.
- 8 1.
- 1.10.
- 1 4. a. Cores Execution Time (s) @ 3 GHz
- 2 2.
- 4 1.
- 8 1.
- 1 3. b. Cores Execution Time (s) @ 3 GHz
- 2 2.
- 4 1.
- 8 0. - Chapter 1 Solutions S
- Yield = 1/(1 + (defect per area × die area)/2)
- a. Yield = 0.
- b. Yield = 0.
- a. Cost per die = 0.
- b. Cost per die = 0.
- 1.11.
- a. Dies per wafer = 1.1 × 84 =
- Defects per area = 1.15 × 0.02 = 0.023 defects/cm
- Die area = wafer area/Dies per wafer = 176.7/92 = 1.92 cm
- Yield = 0.
- b. Dies per wafer = 1.1 × 100 =
- Defects per area = 1.15 × 0.031 = 0.036 defects/cm
- Die area = wafer area/Dies per wafer = 314.2/110 = 2.86 cm
- Yield = 0.
- 1.11.4 Yield = 1/(1 + (defect per area × die area)/2)
- T1: defects per area = 0.00085 defects/mm 2 = 0.085 defects/cm Replacing values for T1 and T2 we get:
- T2: defects per area = 0.00060 defects/mm 2 = 0.060 defects/cm
- T3: defects per area = 0.00043 defects/mm 2 = 0.043 defects/cm
- T4: defects per area = 0.00026 defects/mm 2 = 0.026 defects/cm
- Solution 1. 1.11.5 no solution provided
- a. CPI(bzip2) = 3 × 10^9 × 750/(2,389 × 10^9 ) = 0. clock rate = 1/cycle time = 3 GHz
- b. CPI(go) = 3 × 10^9 × 700/(1,658 × 10^9 ) = 1.
- a. SPECratio(bzip2) = 9,650/750 = 12. 1.12.2 SPECratio = ref. time/execution time
- b. SPECratio(go) = 10,490/700 = 14.
S14 Chapter 1 Solutions
1.12.
(12.86 × 14.98)1/2^ = 13.
1.12.4 CPU time = No. instr × CPI/clock rate If CPI and clock rate do not change, the CPU time increase is equal to the increase in the of number of instructions, that is, 10%. 1.12.5 CPU time(before) = No. instr × CPI/clock rate CPU time(after) = 1.1 × No. instr × 1.05 × CPI/clock rate CPU times(after)/CPU time(before) = 1.1 × 1.05 = 1.155 Thus, CPU time is increased by 15.5%. 1.12.6 SPECratio = reference time/CPU time SPECratio(after)/SPECratio(before) = CPU time(before)/CPU time(after) = 1/1.1555 = 0.86. Thus, the SPECratio is decreased by 14%.
Solution 1. 1.13.1 CPI = (CPU time × clock rate)/No. instr
a. CPI = 700 × 4 × 10^9 /(0.85 × 2,389 × 10^9 ) = 1. b. CPI = 620 × 4 × 10^9 /(0.85 × 1,658 × 10^9 ) = 1.
1.13.2 Clock rate ratio = 4 GHz/3 GHz = 1.
a. CPI @ 4 GHz = 1.37, CPI @ 3 GHz = 0.94, ratio = 1. b. CPI @ 4 GHz = 1.75, CPI @ 3 GHz = 1.26, ratio = 1.
They are different because although the number of instructions has been reduced by 15%, the CPU time has been reduced by a lower percentage.
1.13.
a. 700/750 = 0.933. CPU time reduction: 6.7% b. 620/700 = 0.886. CPU time reduction: 11.4%
1.13.4 No. instr = CPU time × clock rate/CPI
a. No. instr = 960 × 0.9 × 4 × 10^9 /1.61 = 2,146 × 10^9 b. No. instr = 690 × 0.9 × 4 × 10^9 /1.79 = 1,387 × 10^9
1.13.5 Clock rate = no. instr × CPI/CPU time. Clock ratenew = no. instr × CPI/0.9 × CPU time = 1/0.9 clock rateold = 3.33 GHz
S16 Chapter 1 Solutions
1.14.
a. T(P1) = (5 × 10^5 × 0.75 + 4 × 10^5 × 1 + 10 × 10^5 × 1.5)/(4 × 10^9 ) = 5.86 × 10–4^ s CPI(P1) = 5.86 × 10–4^ × 4 × 10^9 /10 6 = 2. MIPS(P1) = 4 × 10^9 /(2.27 ×10^6 ) = 1.76 × 10^3 T(P2) = (2 × 10^6 × 1.25 + 2 × 10^6 × 0.8 + 1 × 10^6 × 1.25)/(3 × 10^9 ) = 1.78 × 10–3^ s CPI(P2) = 1.78 × 10–3^ × 3 × 10^9 /(5 × 10^6 ) = 1.068 s MIPS(P2) = 3 × 10^9 /(1.068 × 10^6 ) = 2.78 × 10^3 b. T(P1) = (1.5 × 10^6 × 1.5 + 1.5 × 10^6 × 1 + 2 × 10^6 × 2)/(4 × 10^9 ) = 1.93 × 10–3^ s CPI(P1) = 1.93 × 10–3^ × 4 × 10^9 /(5 × 10^6 ) = 1. MIPS(P1) = 4 × 10^9 /(1.54 × 10^6 ) = 2.59 × 10^3 T(P2) = (0.8 × 10^6 × 1.25 + 0.6 × 10^6 × 1 + 0.6 × 10^6 × 2.5)/(3 × 10^9 ) = 1.03 × 10–3^ s CPI(P2) = 1.03 × 10–3^ × 3 × 10^9 /(2 ×10^6 ) = 1. MIPS(P1) = 3 × 10^9 /(1.54 × 10^6 ) = 1.94 × 10^3
1.14.
a. T(P1) = 5.86 × 10–4^ s (see problem 1.14.5) performance(P1) = 1/T(P1) = 1.7 × 10^3 T(P2) = 1.78 × 10–3^ s (see problem 1.14.5) performance(P2) = 1/T(P2) = 5.6 × 10^2 perf(P1) > perf(P2), MIPS(P1) > MIPS(P2), MFLOPS(P1) < MFLOPS(P2) b. T(P1) = 1.93 × 10–3^ s (see problem 1.14.5) performance(P1) = 1/T(P1) = 5.1 × 10^2 T(P2) = 1.03 × 10–3^ s (see problem 1.14.5) performance(P2) = 1/T(P2) = 9.7 × 10^2 perf(P1) < perf(P2), MIPS(P1) < MIPS(P2), MFLOPS(P1) > MFLOPS(P2)
Solution 1. 1.15.
a. T (^) fp = 70 × 0.8 = 56 s. Tnew= 56 + 85 + 55 + 40 = 236 s. Reduction: 5.6% b. Tfp = 40 × 0.8 = 32 s. Tnew= 32 + 90 + 60 + 20 = 202 s. Reduction: 3.8%
1.15.
a. T (^) new = 250 × 0.8 = 200 s, Tfp + Tl/s + Tbranch = 165 s, T (^) int = 35 s. Reduction time INT: 58.8% b. T (^) new = 210 × 0.8 = 168 s, Tfp + Tl/s + Tbranch = 120 s, T (^) int = 48 s. Reduction time INT: 46.6%
Chapter 1 Solutions S
1.15.
a. Tnew = 250 × 0.8 = 200 s, Tfp + Tint + Tl/s = 210 s. NO b. Tnew = 210 × 0.8 = 168 s, Tfp + Tint + Tl/s = 190 s. NO
1.15.
Clock cyles = CPIfp × No. FP instr. + CPIint × No. INT instr. + CPIl/s × No. L/S instr. + CPIbranch × No. branch instr.
T (^) cpu = clock cycles/clock rate = clock cycles/2 × 109
a. 2 processors: clock cycles = 4,096 × 10^6 ; Tcpu = 2.048 s b. 16 processors: clock cycles = 512 × 10^6 ; Tcpu = 0.256 s
To half the number of clock cycles by improving the CPI of FP instructions:
CPIimproved fp × No. FP instr. + CPIint × No. INT instr. + CPI (^) l/s × No. L/S instr. + CPI (^) branch × No. branch instr. = clock cycles/
CPIimproved fp = (clock cycles/2 − (CPIint × No. INT instr. + CPI (^) l/s × No. L/S instr. + CPI (^) branch × No. branch instr.))/No. FP instr.
a. 2 processors: CPIimproved fp = (2,048 – 3,816)/280 < 0 ==> not possible b. 16 processors: CPIimproved fp = (256 – 462)/50 < 0 ==> not possible
1.15.5 Using the clock cycle data from 1.15.4:
To half the number of clock cycles improving the CPI of L/S instructions:
CPIfp × No. FP instr. + CPIint × No. INT instr. + CPIimproved l/s × No. L/S instr. + CPI (^) branch × No. branch instr. = clock cycles/
CPIimproved l/s = (clock cycles/2 − (CPIfp × No. FP instr. + CPIint × No. INT instr. + CPI (^) branch × No. branch instr.))/No. L/S instr.
a. 2 processors: CPIimproved l/s = (2,048 – 1,536)/640 = 0. b. 16 processors: CPIimproved l/s = (256 – 198)/80 = 0.
1.15.
Clock cyles = CPIfp × No. FP instr. + CPIint × No. INT instr. + CPI (^) l/s × No. L/S instr. + CPI (^) branch × No. branch instr.
T (^) cpu = clock cycles/clock rate = clock cycles/2 × 109
Chapter 1 Solutions S
1.16.5 Geometric mean of computing time ratios = 0.62. Multiplying this by the computing time for a 64-processor system gives a computing time for a 128-processor system of 11.474 ms.
Geometric mean of routing time ratios = 1.19. Multiplying this by the routing time for a 64-processor system gives a routing time for a 128-processor system of 30.9 ms.
1.16.6 Computing time = 201/0.62 = 324 ms. Routing time = 0, since no communica- tion is required.
Author Query
AQ 1: Page S2: As meant t/o? AQ 2: Page S3: As meant t/o? AQ 3: Page S4: Close up t/o? AQ 4: Page S12: Inserted heading OK? AQ 5: Page S18: Blank cells as meant?
