Each stage is 2ns. So, after 5 time units each of 2ns, the first instruction finishes (i.e., after 10ns), in every 2ns after that a new instruction gets finished. This is assuming no branch instructions. Now, once the pipeline is full, we can assume that the initial fill time doesn't matter our calculations and average execution time for each instruction is 2ns assuming no branch instructions.
(a) Now, we are given that 20% of instructions are branch (like JMP) and when a branch instruction is executed, no further instruction enters the pipeline. So, we can assume every 5th instruction is a branch instruction. So, with this assumption, total time to finish 5 instruction will be 5 * 2 + 8 = 18 ns (as when a branch instruction enters the pipeline and before it finishes, 4 pipeline stages will be empty totaling 4 * 2 = 8 ns, as it is mentioned in question that the next instruction fetch starts only when branch instruction completes). And this is the same for every set of 5 instructions, and hence the average instruction execution time = 18/5 = 3.6 ns
(b) This is just a complex statement. But what we need is to identify the % of branch instructions which cause a branch to be taken as others will have no effect on the pipeline flow.
20% of instructions are branch instructions. 80% of branch instructions are conditional.
That means .2*.8 = 16% of instructions are conditional branch instructions and it is given that 50% of those result in a branch being taken.
So, 8% of instructions are conditional branches being taken and we also have 20% of 20% = 4% of unconditional branch instructions which are always taken.
So, percentage of instructions where a branch is taken is 8+4 = 12% instead of 20% in (a) part.
So, in 100 instructions there will be 12 branch instructions. We can do a different calculation here as compared to (a) as 12 is not a divisor of 100. Each branch instruction causes a pipeline delay of 4*2 = 8 ns. So, 12 instructions will cause a delay of 12 * 8 = 96 ns. For 100 instructions, we need 100 * 2 = 200 ns without any delay and with delay we require 200 + 96 = 296 ns for 100 instructions.
So, average instruction execution time = 296/100 = 2.96 ns
(We can also use this method for part (a) which will give 100 * 2 + 20*8 = 360 ns for 100 instructions)