计算机组成与设计硬件软件接口课后习题答案资料下载.pdf

1、 Part II:#Solutions Guide 52 Instructors Manual for Computer Organization and Design 1.1 q 1.2 u 1.3 f 1.4 a 1.5 c 1.6 d 1.7 i 1.8 k 1.9 j 1.10 o 1.11 w 1.12 p 1.13 n 1.14 r 1.15 y 1.16 s 1.17 l 1.18 g 1.19 x 1.20 z 1.21 t 1.22 b 1.23 h 1.24 m 1.25 e 1.26 v 1.27 j 1.28 b 1.29 f 1.30 j 1.31 i 1.32 e

2、1 Solutions Part II:#Solutions Guide 53 1.33 d 1.34 g 1.35 c 1.36 g 1.37 d 1.38 c 1.39 j 1.40 b 1.41 f 1.42 h 1.43 a 1.44 a 1.45 Time for Time for 1.46 As discussed in section 1.4,die costs rise very fast with increasing die area.Con-sider a wafer with a large number of defects.It is quite likely th

3、at if the die area is verysmall,some dies will escape with no defects.On the other hand,if the die area is verylarge,it might be likely that every die has one or more defects.In general,then,die areagreatly affects yield（as the equations on page 48 indicate）,and so we would expect thatdies from wafe

4、r B would cost much more than dies from wafer A.1.47 The die area of the Pentium processor in Figure 1.16 is 91 mm 2 and it containsabout 3.3 million transistors,or roughly 36,000 per square millimeter.If we assume theperiod has an area of roughly.1 mm 2,it would contain 3500 transistors（this is cer

5、tainlya very rough estimate）.Similar calculations with regard to Figure 1.26 and the Intel4004 result in 191 transistors per square millimeter or roughly 19 transistors.1.48 We can write Dies per wafer=f（Die area）1）and Yield=f（Die area）2）and thusCost per die=f（Die area）3）.More formally,we can write:

6、#1.49 No solution provided.1.50 From the caption in Figure 1.16 we have 198 dies at 100%yield.If the defectdensity is 1 per square centimeter,then the yield is approximated by 1/（1+1 .91/2）2）=.47.Thus 198 .47=93 dies with a cost of$1000/93=$10.75 per die.1.51 Defects per area.12-revolution12-=rev154

7、00-minutesrev-60 ondssecminute-5.56 ms=12-revolution12-=rev17200-minutesrev-60 ondssecminute-4.17 ms=Cost per dieCost per waferDies per waferyield -=Dies per waferWafer areaDie area-=Yield11Defect per areaDie area 2 +（）2-=54 Instructors Manual for Computer Organization and Design 1.521.531.54 No sol

8、ution provided.1.55 No solution provided.1.56 No solution provided.1980Die area0.16Yield0.48Defect density17.041992Die area0.97Yield0.48Defect density1.981992+1980Improvement8.62Yield11Defects per areaDie area 2 +（）2-=Part II:#Solutions Guide 55 2.1 For program 1,M2 is 2.0（10/5）times as fast as M1.F

9、or program 2,M1 is 1.33（4/3）times as fast as M2.2.2 Since we know the number of instructions executed and the time it took to executethe instructions,we can easily calculate the number of instructions per second whilerunning program 1 as（200 10 6）/10=20 10 6 for M1 and（160 10 6）/5=32 10 6 forM2.2.3

10、We know that Cycles per instruction=Cycles per second/Instructions per sec-ond.For M1 we thus have a CPI of 200 10 6 cycles per second/20 10 6 instructionsper second=10 cycles per instruction.For M2 we have 300/32=9.4 cycles per instruc-tion.2.4 We are given the number of cycles per second and the n

11、umber of seconds,so wecan calculate the number of required cycles for each machine.If we divide this by theCPI well get the number of instructions.For M1,we have 3 seconds 200 10 6 cy-cles/second=600 10 6 cycles per program/10 cycles per instruction=60 10 6 in-structions per program.For M2,we have 4

12、 seconds 300 10 6 cycles/second=1200 10 6 cycles per program/9.4 cycles per instruction=127.7 10 6 instructions per pro-gram.2.5 M2 is twice as fast as M1,but it does not cost twice as much.M2 is clearly the ma-chine to purchase.2.6 If we multiply the cost by the execution time,we are multiplying tw

13、o quantities,for each of which smaller numbers are preferred.For this reason,cost times executiontime is a good metric,and we would choose the machine with a smaller value.In theexample,we get$10,000 10 seconds=100,000 for M1 vs.$15,000 5 seconds=75,000for M2,and thus M2 is the better choice.If we u

14、sed cost divided by execution time andassume we choose the machine with the larger value,then a machine with a ridiculous-ly high cost would be chosen.This makes no sense.If we choose the machine with thesmaller value,then a machine with a ridiculously high execution time would be cho-sen.This too m

15、akes no sense.2.7 We would define cost-effectiveness as performance divided by cost.This is essen-tially（1/Execution time）（1/Cost）,and in both cases larger numbers are more cost-effective when we multiply.2.8 We can use the method in Exercise 2.7,but the execution time is the sum of the twoexecution

16、 times.So M1 is slightly more cost-effective,specifically 1.04 times more.2 SolutionsExecutions per second per dollar for M111310,000 -1130,000-=Executions per second per dollar for M21915,000 -1135,000-=56 Instructors Manual for Computer Organization and Design 2.9 We do this problem by finding the amount of time that program 2 can be run in anhour and using that for executions per second,the throughput measure.With performance measured by throughput for program 2,machine M2 is=1.2times faster than