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Computer System Architecture Questions and Answers

41.

To increase performance of computer its through put is increased by

Answer: A

Correct answer: Replacing processor with faster version

Throughput is work completed per unit time, so it rises when the machine gets through each task faster or can run more tasks at once. A faster processor does the first directly. Changing the operating system, the cache or the I/O may help in particular cases, but the processor is the general answer the definition points to.

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42.

Computer A having clock cycle time of 250 ps and cycle per instruction of 2.0 for some programs, and computer B having clock cycle time of 500 ps and a cycle per instruction of 1.2 for same program. Which one is faster for this program

Answer: A

Correct answer: Computer A

Compare the time per instruction, which is the clock cycle time multiplied by the cycles per instruction:

\[ A: 250\ \text{ps}\times 2.0=500\ \text{ps} \qquad B: 500\ \text{ps}\times 1.2=600\ \text{ps} \]

A takes 500 ps per instruction against B's 600, so A is the faster machine by a factor of 600/500 = 1.2. Note that B has the better CPI and A the better clock; neither figure decides the question by itself.

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43.

When PC having Clock rate of 2 and CPU clock cycle for a program is 4 then Execution time of this computer for a program will be

Answer: D

Correct answer: 2

Execution time is the number of CPU clock cycles divided by the clock rate:

\[ \text{Execution time}=\frac{\text{CPU clock cycles}}{\text{Clock rate}}=\frac{4}{2}=2 \]

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44.

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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45.

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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46.

Where the result of an arithmetic and logical operation are stored ?

Answer: A

Correct answer: In Accumulator

The accumulator is the register that holds one operand of an arithmetic or logic operation and receives the result. Cache and ROM are memory rather than working registers, and the instruction register holds the instruction being executed, not its result.

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47.

What is an exception condition in a computer system caused by an event external to the CPU known as ?

Answer: C

Correct answer: Interrupt

An interrupt is a signal raised by something outside the CPU - a device, a timer, an error condition - that causes the processor to break off what it is doing and attend to it. A halt stops the processor rather than diverting it, and a process is a program in execution.

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48.

Whenever CPU detects an interrupt, what it do with current state ?

Answer: A

Correct answer: Save it

Before servicing an interrupt the CPU saves its current state - the program counter and the registers - so that when the service routine finishes it can resume the interrupted program exactly where it left off. Discarding the state would make return impossible, and finishing the current program first would defeat the purpose of interrupting.

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49.

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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50.

Which one among the following is an important data transfer technique ?

Answer: C

Correct answer: DMA

Direct Memory Access lets an I/O device transfer a block of data straight to or from memory without the CPU handling each word, which is why it is the standard high-speed transfer technique. CAD and CAM are application areas, not transfer methods.

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