Computer Architecture-I 1. Die Yield is given by the formula, Assignment 1 Solution Die Yield = Wafer Yield x (1 + (Defects per unit area x Die Area)/a) -a Let us assume a wafer yield of 100% and a 4 for current technology. a. Die yield for AMD Opteron, Die yield = (1 + (0.75 x 1.99)/4) -4 = 0.281 b. Die yield for 8-core SUN Niagara, Die yield = (1 + (0.75 x 3.80)/4) -4 = 0.116 c. The defect rate for both, the AMD Opteron and SUN Niagara is the same. But, the size of the die for Niagara is almost twice as that of AMD Opteron. Thus, the number of dies per wafer reduces significantly for the Niagara. Since the defect rate is same, the yield of Niagara suffers in comparison to Opteron. 4. Question 1.4 a. In order to compute the wattage for the server s power supply, we need to first calculate the power consumed by the entire system. i. Sun Niagara 8-core chip : Power Consumed at max load = 79W ii. 2 x 1GB 184-pin Kingston DRAM : Power consumed at max load = 2 x 3.7W = 7.4W iii. 2 x 7200rpm Hard Drive : Since, we are interested in max. load condition, we assume 0% idle time for the hard drive. Power = 7.9W Total power for 2 Drives = 15.8W Thus, the total power consumed by the system = (79+7.4+15.8) = 102.2W. Power Supply Efficiency = Power O/P / Power I/P Thus, PowerI /P = 102.2/0.7 = 146W This is the required power supply wattage for the system.
b. The hard drive is idle for 40% of the time. Power = (0.4 x 4) + (0.6 x 7.9) = 6.34W c. Since rpm is the only factor affecting idle time of a disk, the disk rpm is directly proportional to the read/seek and idle time of the disk. The disk with 7200 rpm has a read/seek of 60%. Then, for the same set of transactions, the 5400 rpm disk will take 4/3 more time than the 7200 rpm disk i.e. 80% read/seek. Thus, the 5400 rpm disk will idle for 20%. 6. Question 1.6 a. Performance/Power Ratio for each benchmark has been tabulated below, Benchmark Sun Fire T2000 IBM x346 SPECjbb 212.677 91.289 SPECWeb 42.427 9.926 b. If power is the main concern, the Sun Fire T2000 is a better choice since it has lower power consumption for both the benchmarks. c. It is true that For database benchmarks, the cheaper the system, the lower cost per database operation the system is. Even so, some server farms may go for expensive servers. These servers are equipped not only for better performance, but also lower power consumption. Power consumption is an ever-growing concern with large server farms which may consist of over 10000 processors and disks. Cheaper systems might yield a lower cost per operation which is desirable. But these systems may not be power efficient. The cost incurred due to excess power consumption, cooling costs is quite significant. Thus, it is necessary to weigh both these factors when making the choice. 9. Question 1.9 a. FIT = 100 Since FIT is given in billions of hours, MTTF = 10 9 /FIT = 10 9 /100 = 10 7 hours b. MTTR = 1 day = 24hours Availability = MTTF/ (MTTF + MTTR) = 0.9999
12. Question 1.12 a. Tabulated results for performance normalized to the Pentium D820 Chip Memory Performance Dhrystone Performance Athlon64 X2 4800+ 1.141 1.361235217 Pentium EE840 1.076 1.241327201 Pentium D820 1 1 Athlon64 X2 3800+ 0.980333333 1.12542707 Pentium 4 0.910333333 0.500722733 Athlon64 3000+ 0.984333333 0.501182654 Pentium 4 570 1.167 0.73653088 Processor X 2.333333333 0.328515112 b. Arithmetic Mean of Performance for each processor tabulated for both original and normalized performance values. Chip Arithmetic Mean for Original Results Arithmetic Mean for Normalized Results Athlon64 X2 4800+ 12070.5 1.251117608 Pentium EE840 11060.5 1.158663601 Pentium D820 9110 1 Athlon64 X2 3800+ 10035 1.052880201 Pentium 4 5176 0.705528033 Athlon64 3000+ 5290.5 0.742757994 Pentium 4 570 7355.5 0.95176544 Processor X 6000 1.330924223 c. From the table above, one can draw a conflicting conclusion in reference to the performance of Processor X. If one examines the performance given in the first column, it is clear that the processors viz. Athlon64 X2 4800+, Pentium EE840, Pentium D820, Athlon64 X2 3800+ and the Pentium 4 570 are all faster than Processor X. This is contrary to the results in the second column where Processor X is faster than all of the said processors. d. Geometric mean for Dhrystone benchmark for the single and dual core processors is given below: Geometric mean (Single Core) = 0.4964 Geometric mean (Dual Core) = 1.1743
e. The scatter graph for Dhrystone performance Vs Memory Performance is given below: f. The scatter graph clearly indicates that the dual core processors outperform their single core counterparts in Dhrystone performance. The Dhrystone benchmark is an integer benchmark which primarily exercises the logical/arithmetic functionality in CPU. The dramatic improvement in Dhrystone performance can be justified simply by the fact that there are 2 cores available for computation instead of 1. It can also be seen that there is no major improvement in memory performance. This is because the latency in memory is not related to number of CPU cores available. Thus, even if the processor is a dual core, the latency in memory load/store operations is similar to the single core. The only exception to this is the memory performance of Processor X which is fictitiously high.
13. Question 1.13 a. It is given that 40% of operations are memory centric and 60% are CPU-centric. Following table gives the weighted execution times for the benchmarks. Chip Memory Benchmark Execution Time Dhrystone Benchmark Weighted Arithmetic Mean Athlon64 X2 4800+ 0.000292141 4.82672E -05 0.00015 Pentium EE840 0.000309789 5.29297E -05 0.00016 Pentium D820 0.000333333 6.5703E-05 0.00017 Athlon64 X2 3800+ 0.00034002 5.83805E -05 0.00017 Pentium 4 0.000366166 0.000131216 0.00023 Athlon64 3000+ 0.000338639 0.000131096 0.00021 Pentium 4 570 0.000285633 8.92061E -05 0.00017 Processor X 0.000142857 0.0002 0.00018 b. Since the application suite is CPU-intensive, we consider the Dhrystone performance of the two CPUs in comparison. Speed-up from Pentium 4 570 to Athlon64 X2 4800+ can be measured as the ratio of their Dhrystone performance. Hence Speed-up = 20718/11210 = 1.848 c. Let the required ratio of memory-processor computation be a. Then, for equal performance, we can consider the following equation. 3501a + 11210(1-a) = 3000a + 15220(1-a) Thus, 4511a = 4010 i.e. a = 0.89 Thus, the performance of Pentium 4 570 equals Pentium D 820 when there are 89% memory operations and 11% processor operations.
14. Question 1.14 According to Amdahl s Law, speed up is given by, Speed-up system = (Execution Time) old /(Execution Time) new = 1/((1 Fraction enhanced ) + (Fraction enhanced /Speed-up enhanced )) a. The first application is run in isolation and 40% of it is parallelizable. Thus, Fraction enhanced = 0.4. Also, since the new processor is a dual core, Speed-up enhanced = 2. Then, the overall speed-up is given by the formula above. Speed-up system = 1.25 b. The second application is run in isolation and 99% of it is parallelizable. Hence, we have: Fraction enhanced = 0.4 Speed-up enhanced = 2 Thus, Speed-up system = 1.98 c. Now both, the first and second application are running on the system. Since, the first application uses 80% of system resources, only 40% of 80% (= 32%) will be enhanced by a factor of 2. Thus, Speed-up system = 1.19 d. Similar to the solution above, 99% of 20% (= 19.8%) will be enhanced by a factor of 2. Thus, Speed-up system = 1.10