throughput of pipeline

Question

using a sequential implementation, it takes a total of 320 ns for each instruction, 300ns for combinational logic to complete and 20 ns to store result. This means that a throughput will be 3.12 millions instructions/second. assuming you switch to a 3 stage pipeline by splitting the combinational logic into 3 equal parts and all reg. take 20ns to store results.Assuming pipeline never starts, what will be the improvement in throughput?

2 instructions/sec            3 instructions/sec                   4 instructions/sec

Answer 1

Time in pipeline=100+20+20=140

Throughput=1/140ns=7.14

So improvement=7.14-3.12=4.02

Comments

but the register time will be counted only once i think it should be 100 +20 only. correct me if i m wrong .

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Same doubt...how is it 100+20+20 ?