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Consider an unpipelined machine with 10nsec clock cycles which uses four cycles for ALU operations and branches where as five cycles for memory operation. Assume that the relative frequencies of these operations are: 40%, 20% and 40%, respectively. Due to clock skew and setup pipeline let us consider that the machine adds one nsec overhead to the clock. How much speedup is observed in the instruction execution rate when a pipelined machine is considered. 

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Answer (Detailed Solution Below)

Option 2 : 4 times
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The correct answer is 4 times

EXPLANATION:

Operation Frequency Clock   
ALU 40% = 0.40 4
BRANCH 20% = 0.20 4
MEMORY 40% = 0.40 5

Total Clock Cycle = 0.40 *4 + 0.20 * 4 + 0.40 * 5 

Total Clock Cycle = 1.6 + 0.8 + 2

Total Clock Cycle = 4.4 Clock Cycle

Here, 

Non-pipeline time = 4.4 clock cycle  * 10 nsec

Non-pipeline time = 44nsec

Pipeline time = Clock time + overhead time = 10nsec + 1nsec = 11nsec

Speedup= Time taken by non-pipeline machine / Time taken by pipeline ​machine

Speedup = 44/11 = 4
​So, the correct answer is 4 times

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