Consider an unpipelined machine with 10 nsec 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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Correct answer: B. 4 times
Explanation
Unpipelined average = 10 ns × (0.4×4 + 0.2×4 + 0.4×5) = 44 ns. The pipelined machine completes one instruction per 11 ns clock (10 + 1 overhead). Speed-up = 44/11 = 4.
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