本文格式为Word版，下载可任意编辑计算机体系结构 计算机体系布局 第一章 1.11 Availability is the most important consideration for designing servers, followed closely by scalability and throughput. a. We have a single processor with a failures in time(FIT) of 100. What is the mean time to failure (MTTF) for this system? b. If it takes 1 day to get the system running again, what is the availability of the system? c. Imagine that the government, to cut costs, is going to build a supercomputer out of inexpensive computers rather than expensive, reliable computers. What is the MTTF for a system with 1000 processors? Assume that if one fails, they all fail. 答： a. 平均故障时间(MTTF)是一个稳当性度量方法，MTTF的倒数是故障率，一般以每10亿小时运行中的故障时间计算(FIT)。因此由该定义可知1/MTTF=FIT/109，所以MTTF=109/100=107。 b. 系统可用性=MTTF/(MTTF+MTTR)，其中MTTR为平均修复时间，在该题目中表示为系统重启时间。计算107/(107+24)约等于1. c. 由于一个处理器发生故障，其他处理器也不能使用，所以故障率为原来的1000倍，所以MTTF值为单个处理器MTTF的1/1000即107/1000=104。 1.14 In this exercise, assume that we are considering enhancing a machine by adding vector hardware to it. When a computation is run in vector mode on the vector hardware, it is 10 times faster than the normal mode of execution. We call the percentage of time that could be spent using vector mode the percentage of vectorization. a. Draw a graph that plots the speedup as a percentage of the computation performed in vector mode. Label the y-axis “Net speedup” and label the x-axis “Percent vectorization”. b. What percentage of vectorization is needed to achieve a speedup of 2? c. What percentage of the computation run time is spent in vector mode if a speedup of 2 is achieved? d. What percentage of vectorization is needed to achieved one-half the maximum speedup attainable from using vector mode? e. Suppose you have measured the percentage of vectorization of the program to be 70%. The hardware design group estimates it can speed up the vector hardware even more with significant additional investment. You wonder whether the compiler of vectorization would the compiler team need to achieve in order to equal an addition 2*speedup in the vector unit(beyond the initial 10*)? 答： a. 根据加速比定义可知，巩固加速比=10，假设令巩固比例为x，总 加速比为y，那么有y=1/(1-x+x/10)。x的取值范围为0，1；y的取值范围为0，10。如下图示： b. y=1/(1-x+x/10)；当y=2时，x=5/9=55.6%； c. (5/9)/10/(1/2)=1/9=11.1% d. 最大加速比理论上为10；最大加速比的一半就是5；y=1/(1-x+x/10)；当y=5时，x=8/9=88.9% e. 当前x=70%；y=1/(1-x+x/10)；可知y=2.7; 假设y=22.7=5.4；y=1/(1-x+x/10)；可知x=0.91； 其次章 2.8 The following questions investigate the impact of small and simple caches using CACTI and assume a 65nm(0.065m) technology. a. Compare the access times of 64KB caches with 64byte blocks and a single bank. What are the relative access times of two-way and four-way set associative caches in comparison to a direct mapped organization? b. Compare the access times of four-way set associative caches with 64 byte blocks and a single bank. What are the relative access times of 32KB and 64KB caches in comparison to a 16KB cache? c. For a 64KB cache, find the cache associativity between 1 and 8 with the lowest average memory access time given that misses per instruction for a certain workload suite is 0.00664 for direct mapped, 0.00366 for two-way set associative, 0.00987 for four-way set associative and 0.000266 for eight-way set associative cache. Overall, there are 0.3 data references per instruction. Assume cache misses take 10 ns in all models. To calculate the hit time in cycles, assume the cycle time output using CACTI, which corresponds to the maximum frequency a chche can operate without any bubbles in the pipeline. 答： a. 直接映射：0.86ns；两路组相联：1.12ns；四路组相联：1.37ns。 两路组相联访存时间是直接映射的1.12/0.86=1.30倍；四路组相联访存时间是直接映射的1.37/0.86=1.59倍。 b. 16KB cache 的访存时间为1.27ns，32KB cache为1.35ns，64KB cache为1.37ns。 32KB cache的访存时间是16KB cache访存时间的 1.35/1.27=1.06倍；64KB cache的访存时间是16KB cache访存时间的 1.37/1.27=1.08倍； c. 平均访存时间=命中率命中时间+缺失率缺失代价； DM缺失率=0.00664/0.3=2.2%； 2-way缺失率=0.00366/0.3=1.2%； 4-way缺失率=0.00987/0.3=0.33%； 8-way缺失率=0.000266/0.3=0.09%； DM访存所用时钟周期=0.86ns/0.5ns向上取整=2； 2-way访存所用时钟周期=1.12ns/0.5ns向上取整=3； 4-way访存所用时钟周期=1.37ns/0.83ns向上取整=2； 8-way访存所用时钟周期=2.03ns/0.79ns向上取整=3； DM缺失代价=10ns/0.5ns=20时钟周期； 2-way缺失代价=10ns/0.5ns=20时钟周期； 4-way缺失代价=10ns/0.83ns=13时钟周期； 8-way缺失代价=10ns/0.79ns=13时钟周期； DM平均访存时间=(1-0.22)2+0.022200.5=2.3960.5=1.2ns； 2-way平均访存时间=(1-0.012)3+0.012200.5=3.20.5=1.6ns； 4-way平均访存时间=(1-0.0033)2+0.0033130.83=2.0360.83=1.69ns 5