Aspire Faculty ID #18606 · Topic: UGC NET Computer Science Dec 2022 Shift II (Paper II) · Just now
UGC NET Computer Science Dec 2022 Shift II (Paper II)

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?

Solution

Average CPI for unpipelined machine is:

$=4(0.40)+4(0.20)+5(0.40)$

$=1.6+0.8+2$

$=4.4$

Unpipelined average instruction time:

$=4.4 \times 10$

$=44$ nsec

For pipelined machine, clock time becomes:

$10+1=11$ nsec

Speedup:

$=\frac{44}{11}$

$=4$

So, speedup is $4$ times.

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