Computer System Architecture Questions and Answers

Answer: B

Correct answer: parallel clusters

Parallel clusters allow more than one host to access the same data on shared storage at the same time, which needs a special distributed file system to arbitrate the accesses. A Beowulf cluster is a commodity-hardware compute cluster and is not defined by shared data access.

Answer: D

Correct answer: both a and b

In symmetric multiprocessing every processor is a peer: all of them run the operating system and all of them reach main memory over the same system bus. So both the bus and the memory are shared. The processors themselves are the things doing the sharing, not a resource that is shared.

Answer: A

Correct answer: hot standby mode

In asymmetric clustering one machine runs the application while the other does nothing but monitor it, ready to take over the instant it fails. A machine kept powered and ready in this way is in hot standby; plain standby would not imply that it is already running and watching.

Answer: A

Correct answer: Performance is increased when execution time is decreased

Performance is defined as the reciprocal of execution time, so the two move in opposite directions: halving the time doubles the performance. Any option in which both rise or both fall contradicts the reciprocal relation.

Answer: C

Correct answer: 4GHz

Clock rate is the reciprocal of the clock cycle time:

\[ \text{Clock rate}=\frac{1}{0.25\times10^{-9}\ \text{s}}=4\times10^{9}\ \text{Hz}=4\ \text{GHz} \]

A 0.5 ns cycle would give 2 GHz and a 0.125 ns cycle 8 GHz, which is what the other options correspond to.

Answer: C

Correct answer: Execution time

Execution time is the only measure that cannot be gamed: it counts the whole job on the actual machine. A higher clock rate may be spent on more cycles per instruction, and an instruction set says nothing about how fast the instructions run, so neither is a reliable measure of performance on its own.

Answer: A

Correct answer: MIPS = Instruction count/(Execution time)10^6

MIPS means millions of instructions per second, so the instruction count is divided by the execution time in seconds and then by 10⁶ to convert to millions:

\[ \text{MIPS}=\frac{\text{Instruction count}}{\text{Execution time}\times 10^{6}} \]

Dividing by 10³ would give thousands of instructions per second, and omitting the divisor gives plain instructions per second.

Answer: B

Correct answer: 1.5 times

Performance is the reciprocal of execution time, so the ratio of performances is the inverse ratio of times:

\[ \frac{\text{Performance}_A}{\text{Performance}_B}=\frac{\text{Time}_B}{\text{Time}_A}=\frac{15}{10}=1.5 \]

Computer A is therefore 1.5 times as fast as computer B.

Answer: B

Correct answer: Performance_X > Performance_Y

The statement is simply restated in symbols: if X performs better than Y then its performance figure is the larger one. Since performance is 1/execution time, this is the same as saying X's execution time is the shorter of the two.

Answer: C

Correct answer: Through put

Throughput measures how much work is completed in a given period - jobs per hour, transactions per second. Response time and execution time measure how long one piece of work takes, which is the opposite way of looking at the same machine: a system can improve throughput by running many tasks at once without making any single one faster.