Download Homework 2 Solutions - Computer Architecture and Organization | ECE 337 and more Assignments Computer Architecture and Organization in PDF only on Docsity! ECE 337 Fall 2008 HW 2 Solution Problems on Textbook: Exercises 1,2, 3, 4, 5, 8, 9, 10, 11, 12, 25, 32, 33 in Chapter 2 1. The data path cycle is 20 nsec. The maximum number of data path cycles/sec is thus 50 million. The best the machine could do is thus 50 MIPS. 2. The program counter must be incremented to point to the next instruction. If this step were omitted, the computer would execute the initial instruction forever. 3. You cannot say anything for sure. If computer 1 has a five-stage pipeline, it can issue up to 5×100=500 million instructions/second. If computer 2 is not pipelined, it cannot do any better than 200 million instructions/sec. Thus without more information, you cannot say which one is faster. 4. On-chip memory does not affect the first three principles. Having only LOADs and STOREs touch memory is no longer required. There is no particular reason not to have a memory-to-memory architecture if memory references are as fast as register references. Likewise, the need for many registers becomes less in this environment. 5. A pipeline processor is better. Array processors are useful only if the problem contains inherent parallelism (perform the same operations on multiple data elements). 8. Sixty-four 6-bit numbers exist, so 4 trits are needed ( 63log 2 3.785 4⎡ ⎤ = =⎡ ⎤⎢ ⎥⎢ ⎥ ). In general, the number of trits, k, needed to hold n bits is the smallest value of k such that 3 2k n≥ 9. A pixel requires 6 + 6 + 6 = 18 bits, so a single visual frame is 1.8 ×107 bits. With 10 frames a second, the gross data rate is 180 Mbps. Unfortunately, the brain’s processing rate is many orders of magnitude less than this. As an experiment, try watching the random noise on a color television when no station is broadcasting and see if you can memorize the color bit pattern in the noise for a few minutes. 10. With 44,000 samples per second of 16 bits each, we have a data rate of 704 kbps. 11. There are 2 bits per nucleotide, so the information capacity of the human genome is about 6 gigabits. Dividing this number by 30,000, we get about 200,000 bits per gene. Just think of a gene as an 25-KB ROM. This estimate
