Question

The pipeline of Fig. 9-2 has the foUowing propagation times: 40 ns for the operands to read from memory into registers R1 and R2, 45 ns for the signal to propagate through the multiplier, S ns for the transfer into R3, and 15 ns to add the two numbers into RS. a. What is the mWmum dock cycle time that can be used? b. A non pipeline system can perform the same operation by removing R3 and R4. How long will it take to multiply and add the operands without using the pipeline? c. Calculate the speedup of the pipeline for 10 tasks and again for 100 tasks. d. What Is the maximum speedup that can be ac:hleved?

Answer 1

Explanation:

The pipeline of Fig. 9-2 has the following propagation times: 40 ns for the operands to 10 read from memory into registers R1 and R2, 45 ns for the signal to propagate through the multiplier, S ns for the transfer into R3, and 15 ns to add the two numbers into RS.

a. What is the minimum dock cycle time that can be used?

b. A non pipeline system can perform the same operation by removing R3 and R4. How long will it take to multiply and add the operands without using the pipeline?

c. Calculate the speedup of the pipeline for 10 tasks and again for 100 tasks.

d. What Is the maximum

Answer 2

Answer:

Explanation:

Here is the answer