Consider an unpipelined machine with 10nsec clock cycles which uses four…
2022
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.
- A.
2 times
- B.
4 times
- C.
6 times
- D.
8 times
Attempted by 98 students.
Show answer & explanation
Correct answer: B
Key idea: compute the average unpipelined instruction time, then compare it to the pipelined instruction time (accounting for the 1 ns clock overhead).
Average cycles per instruction (CPI) for the unpipelined machine = 0.4*4 + 0.2*4 + 0.4*5 = 4.4 cycles.
Unpipelined average instruction time = 4.4 cycles * 10 ns per cycle = 44 ns.
Pipelined machine (ideal) has CPI = 1. The pipelined clock period = original clock (10 ns) + overhead (1 ns) = 11 ns, so pipelined instruction time = 11 ns.
Speedup = (Unpipelined instruction time) / (Pipelined instruction time) = 44 ns / 11 ns = 4.
Final answer: 4 times
A video solution is available for this question — log in and enroll to watch it.