computer_organization_solutions - Solution* for Chapter 1...

computer_organization_solutions
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Unformatted text preview: Solution* for Chapter 1 Exercise* Solutions for Chapter 1 Exercises 1.1 5, CPU 1.2 1, abstraction 1.3 3, bit 1.4 8, computer family 1.5 19, memory 1.6 10, datapath 1.7 9, control 1.8 11, desktop (personal computer) 1.9 15, embedded system 1.10 22, server 1.11 18, LAN 1.12 27, WAN 1.13 23, supercomputer 1.14 14, DRAM 1.15 13, defect 1.16 6, chip 1.17 24, transistor 1.18 12, DVD 1.19 28, yield 1.20 2, assembler 1.21 20, operating system 1.22 7, compiler 1.23 25, VLSI 1.24 16, instruction 1.25 4, cache 1.26 17, instruction set architecture Solutions for Chapter 1 Exercises 1.27 21, semiconductor 1.28 26, wafer 1.29 i 1.30 b 1.31 e 1.32 i 1.33 h 1.34 d 1.35 f 1.36 b 1.37 c 1.38 f 1.39 d 1.40 a 1.41 c 1.42 i 1.43 e 1.44 g 1.45 a 1.46 Magnetic disk: Time for 1/2 revolution =1/2 rev x 1/7200 minutes/rev X 60 seconds/ minutes 3 4.17 ms Time for 1/2 revolution = 1/2 rev x 1/10,000 minutes/rev X 60 seconds/ minutes = 3 ms Bytes on center circle = 1.35 MB/seconds X 1/1600 minutes/rev x 60 seconds/minutes = 50.6 KB Bytes on outside circle = 1.35 MB/seconds X 1/570 minutes/rev X 60 seconds/minutes = 142.1 KB 1.48 Total requests bandwidth = 30 requests/sec X 512 Kbit/request = 15,360 Kbit/sec < 100 Mbit/sec. Therefore, a 100 Mbit Ethernet link will be sufficient. Solution* for Chapter X Exarclsm 1.49 Possible solutions: Ethernet, IEEE 802.3, twisted pair cable, 10/100 Mbit Wireless Ethernet, IEEE 802.1 lb, no medium, 11 Mbit Dialup, phone lines, 56 Kbps ADSL, phone lines, 1.5 Mbps Cable modem, cable, 2 Mbps 1.50 a. Propagation delay = mis sec Transmission time = LIR sec End-to-end delay =m/s+L/R b. End-to-end delay =mls+ LJR+t c. End-to-end delay = mis + 2I/R + f/2 1.51 Cost per die = Cost per wafer/(Dies per wafer x Yield) = 6000/( 1500 x 50%) = 8 Cost per chip = (Cost per die + Cost_packaging + Cost_testing)/Test yield = (8 + 10)/90% = 20 Price = Cost per chip x (1 + 40%) - 28 If we need to sell n chips, then 500,000 + 20« = 28», n = 62,500. 1.52 CISCtime = P x 8 r = 8 P r n s RISC time = 2Px 2T= 4 PTns RISC time = CISC time/2, so the RISC architecture has better performance. 1.53 Using a Hub: Bandwidth that the other four computers consume = 2 Mbps x 4 = 8 Mbps Bandwidth left for you = 10 - 8 = 2 Mbps Time needed = (10 MB x 8 bits/byte) / 2 Mbps = 40 seconds Using a Switch: Bandwidth that the other four computers consume = 2 Mbps x 4 = 8 Mbps Bandwidth left for you = 10 Mbps. The communication between the other computers will not disturb you! Time needed = (10 MB x 8 bits/byte)/10 Mbps = 8 seconds Solutions for Chapter 1 EXWCIMS 1.54 To calculate d = axfc-axc, the CPU will perform 2 multiplications and 1 subtraction. Time needed = 1 0 x 2 + 1 x 1 = 2 1 nanoseconds....
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