4- The propagation times for a pipeline system is given by segment1=40 ns, segment2=10 ns, segment3=30 ns and segment4= 5 ns.

a- What is the pipeline clock cycle?

b- How long it takes to perform n operations without pipelining using the clock cycle in (a)

c- Calculate the speedup of the pipeline for repeating the operation for 1000 tasks.

d- Repeat a-c if the largest segment is split into two segments.

a). Pipeline clock cycle is 40ns.

b) Time to execute n operations is 160n ns.

c). Speedup is 3.988036

d). In the case where the largest segment (segment 1) is split into 2 segments, there will be segment₁₁= 20ns and segement₁₂ =20ns.

Thu, the calculation are as follows.

a).Pipeline cycle is 30ns.

b).Time without pipeline is 150n ns.

c).speedup = 4.98008

Step-by-step explanation

a). Pipeline clock cycle is given by;

max(40,10,30,5)

maximum is 40

Thus, the answer is 40ns.

b) Time to execute n operations is given by.

Total segments= 4.

Thus time taken for n operations is calculated as;

n*4*40 = 160n ns

=160n ns.

c). Speedup is given by;

(time without pipeline)/ (time with pipeline).

= (4*40*1000) / 40(4+(1000-1))

=(4*40*1000) / 40(1003)

=3.988036

d). In the case where the largest segment (segment 1) is split into 2 segments, there will be segment₁₁= 20ns and segement₁₂ =20ns.

Thu, the calculation are as follows.

a).Pipeline cycle = max(20,20,10,30,5)

hence maximum is 30.

= 30ns.

b).Time without pipeline is given by.

total number of segments=5.

Thus,

5*30*n = 150n ns.

c). Speedup is given by;

Time without pipeline / time with pipeline.

=5*30*1000 / 30(5+999)

=4.98008.

speedup = 4.98008