Consider the operation of a machine with the data path in th

Consider the operation of a machine with the data path in the figure below. Suppose that loading the ALU input registers takes 10 ns, running the ALU takes 10 ns, and storing the result back in the register scratchpad takes 10 ns. (a) What is the maximum number of MIPS this machine is capable of in the absence of pipelining? (b) What is the speedup benefit possible from pipelining?

Solution

Solution:

Given:Loading the ALU input registers takes 10ns,running ALU takes 10 ns and storing takes 10ns

So,Loading of A ans B takes 10+10=20ns..................................(I)

Calculating A+B takes 10ns.....................................................(II)

Storing the result takes 10 ns....................................................(III)

Normal average execution time = (20+10+10)= 50 ns.

a) Maximum number of MIPS

So, MIPS=10^(-6)/(40×10^(-9))=25

Maximum number of MIPS=25.ans

b)Speedup benefits

Considering the overhead with pipeline be 2ns,then

Clock time with pipeline = 10 + 2 = 12 ns.

Assuming everything is neglected or included in overhead, with pipeline we complete 1 instruction every clock cycle. i.e.,

MIPS=10^(6)/(12×10^(9))=83.33

Speedup benefits=83-25=58 ans