2- The following pipeline have initially seven segments: t 120 nsec, t 100 nsec

Question

2- The following pipeline have initially seven segments: t 120 nsec, t 100 nsec ts-140 nsec, t-180 nsec, ts-80 nsec, to 60 nsec and tr 40 Find the following parameters and insert them in the given table. a- The pipeline clock cycle, execution time for 200 tasks and the speed up ratio. b- Repeat (a) when ta is split into 3 identical segments. c- Repeat (a) when ta and to are combined into one segment d- Repeat (a) when ta is split into 3 identical segments while ta and ts are combined into one segment Pipeline Name Number of Segments Clock cyale Pipeline execution time Pipeline speed-up ratio n4k-1

Explanation / Answer

(a) In non-uniform pipeline, cycle time = max (segments delay + Buffer dealy)

since, there is no buffer presents here, so cycle time (tp)= max (segments delay)

                                                                             =max(120, 100,140,180,80,60,40) =180nsec

since, number of tasks= 200

execution time of the pipeline (in nsec) = (n+k-1)tp , where n = number of tasks and k =number of segments

                                      = (200 + 7 -1) *180 = 206 *180 = 37080 nsec   -----------------(i)

when pipeline contains non-uniform stage delay, then

speed up ratio (S) = (execution time in non-pipeline) / (execution time in pipeline)

                          =n * tn / (( n + k -1)* tp)         where tn = (t1 +t2 + t3 +....+ t7) = 720 nsec

*Note: splitting or combining stages doesn't change the non-pipeline execution time,

so, non-pipeline execution time = n * tn = 200 *720 nsec = 144000 nsec -----------------(ii)

so, S = 200 *720 / 37080 = 3.8834                      (see (i) and (ii) given above)

(b) Given, t4 is splitted into 3 identical parts, (say t41, t42, t43).

so, t41 = t42 = t43 = t4 /3 = 180/ 3=60 nsec

now, number of segments (k) = 7-1 +3 = 9

cycle time (tp) = max(segments delay) = max(120, 100,140,60,60,60,80,60,40 ) = 140 nsec

Execution time of pipeline (in nsec) = (n + k-1)*tp = (200 +9 -1)*140

                                                     = 208 *140 nsec = 29120 nsec -----------(iii)

speed up ratio (S) = n * tn / ((n + k -1)*tp ) =144000 / 29120 = 4.9450        (see (ii) and (iii))

(c) Given, t2 and t3 are combining into single segment, (say t23)

so, t23 = t2 +t3 = 100 +140 =240 nsec

now, number of segments (k) = 7-1 = 6

cycle time (tp) = max(segments delay) = max(120, 240,180,80,60,40 ) = 240 nsec

Execution time of pipeline (in nsec) = (n + k-1)*tp = (200 +6 -1)*240

                                                   = 205 *240 nsec = 49200 nsec   ------------(iv)

speed up ratio (S) = n * tn / ((n + k -1)*tp ) =144000 / 49200 = 2.9268        (see (ii) and (iv))

(d)Given, t4 is splitted into 3 identical segments (say, t41, t42, t43) and t2, t3 are combined into single segment (say t23).

then , t41 = t42 = t43 = t4 /3 =180/3 =60 nsec

         t23 = t2 + t3 =100 +140 = 240 nsec

now, number of segments (k) = (7-1 +3) - 1 = 8

cycle time (tp) = max(segments delay) = max(120, 240,60,60,60,80,60,40 ) = 240 nsec

Execution time of pipeline (in nsec) = (n + k-1)*tp = (200 +8 -1)*240

                                                   = 207 *240 nsec = 49680 nsec ----------------(v)

speed up ratio (S) = n * tn / ((n + k -1)*tp ) =144000 / 49680 = 2.8985        (see (ii) and (v))

So, now fill up the table,

Pipeline name number of segments (k) clock cycle (nsec) pipeline execution time(nsec) Pipeline speedup ratio a 7 180 37080 3.8834 b 9 140 29120 4.9450 c 6 240 49200 2.9268 d 8 240 49680 2.8985

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