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Computer System Architecture Questions and Answers

31.

If CPU and I/O interface share a common bus than transfer of data between two units is known as ?

Answer: C

Correct answer: Synchronous

When the CPU and the interface share a common bus they are driven by the same clock, so the transfer needs no separate timing agreement between them - that is what makes it synchronous. Asynchronous transfer is used precisely when the two units have independent clocks and must signal each other with control lines.

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32.

Another type of multiple-CPU system is the

Answer: C

Correct answer: Clustered System

A clustered system is the other way of putting several CPUs to work: two or more complete computers, each with its own memory, joined over a network and sharing storage so that they act as one. A minicomputer and a supercomputer are classes of machine, not ways of combining CPUs.

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33.

Valid and unimpeachable measurement of performance of any computer is

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.

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34.

A set of standard CPU-intensive, floating points and integers benchmarks are referred to as

Answer: A

Correct answer: SPEC benchmark

SPEC - the Standard Performance Evaluation Corporation - publishes the suites of CPU-intensive integer and floating-point programs used to compare machines. The other three names are not benchmark suites.

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35.

In asymmetric clustering other machines perform operations while one machine is in

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.

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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.

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37.

What is the unit which decodes and translates each instruction and generates the necessary enable signals for ALU and other units called

Answer: B

Correct answer: Control Unit

The control unit fetches each instruction, decodes it, and issues the enable and timing signals that make the ALU, registers and memory carry it out. The ALU performs the operations rather than directing them, and the CPU is the whole of which the control unit is one part.

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38.

Which devices among the following are usually designed on the complex electromechanical principle ?

Answer: B

Correct answer: Peripheral devices

Peripherals - printers, keyboards, disk drives, tape units - are electromechanical and electromagnetic devices of some complexity, in contrast with the purely electronic circuits of the CPU and memory. Printing devices and input devices are both classes of peripheral, so naming either of them alone is too narrow.

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39.

Symmetric multiprocessing architecture of computer system uses shared

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.

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40.

Computer B has an execution time of 110 ns for a program and computer A has an execution time of 1001 ns for the same program. Then

Answer: B

Correct answer: B is 9.1 times as fast as A for programs

The machine with the shorter execution time is the faster one, and the ratio of the times gives the factor:

\[ \frac{1001\ \text{ns}}{110\ \text{ns}}=9.1 \]

B takes 110 ns against A's 1001 ns, so it is B that is 9.1 times as fast, not A.

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