Solution Manual for Structured Computer Organization 6E 6th Edition.

Solution Manual for Structured Computer Organization 6E 6th Edition.

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PROBLEM SOLUTIONS FOR CHAPTER 2 5 SOLUTIONS TO CHAPTER 2 PROBLEMS 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 for- ev er. 3. You cannot say anything for sure. If computer 1 has a five-stage pipeline, it can issue up to 500 million instructions/second. If computer 2 is not pipelined, it cannot do any better than 200 million instructions/sec. Thus with- out more information, you cannot say which is faster. 4. On-chip memory does not affect the first three principles. Having only LOAD s and STORE s touch memory is no longer required. There is no particular rea- son 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. The monastery resembles Fig. 2-7, with one master and many slaves. 6. The access time for registers is a few nanoseconds. For optical disk it is a few hundred milliseconds. The ratio here is about 10 8 . 7. Sixty-four 6-bit numbers exist, so 4 trits are needed. In general, the number of trits, k , needed to hold n bits is the smallest value of k such that 3 k 2 n . 8. A pixel requires 6 + 6 + 6 = 18 bits, so a single visual frame is 1. 8 × 10 7 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 for a few min- utes when no station is broadcasting and see if you can memorize the color bit pattern in the noise. 9. With 44,000 samples per second of 16 bits each, we have a data rate of 704 kbps. 10. 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 a 25-KB ROM. This estimate is an upper bound, because many of the nucleotides are used for purposes other than coding genes.
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