Question:

A non-pipelined system takes 7.5 ns to process an instruction with a clock cycle of 1.5 ns. The same instruction can be processed in a 5-segment pipeline.

The instruction set architecture of the non-pipelined system is replaced with a CISC design. Empirical tests show that for a typical program, 81% of the instructions take only 3 ns to complete and 19% of the instructions take 7.5 ns. Determine the average time taken to process 13 instructions with this system to one decimal place.

Answer:

The average time per instruction is given by

fi + ti

where

fi = fraction of instruction of class i

ti = execution time for fraction fi

Putting the values, we get the average time per instruction

= (0.81 * 3 ns) + (0.19 * 7.5 ns)

= 3.855 ns

Thus, time to execute 13 instructions

= (Number of instructions) * (Average execution time per instruction)

= 13 * (3.855 ns)

= 50.115 ns

I do not understand why they did (0.81 * 3 ns) + (0.19 * 7.5 ns), how does doing this give you the average time per instruction?