Problem 1

Here, we assume that our branch prediction is perfect and we have no stalls caused by branching. Assume that the numbering of clock cycles starts from 1 and the first instruction in the given code fragment is fetched in the first clock cycle. Then we have the following table:

Instruction	IF	ID	EX	MEM	WB	Comments
L.S F2,100(R1)	5	6	7	8	9
L.S F3,500(R1)	6	7	8	9	10
SUB.S F5,F3,F2	7	8-9	10-14	15	16	Stall for RAW hazard with F3 in ID stage.
ADD.S F5,F5,F4	8-9	10-14	15-19	20	21	Stall for RAW hazard with F5 in ID stage.
S.S 1000(R1),F5	10-14	15	16-20	21	22	Stall in EX stage for both RAW hazard with F5 and structural hazard with MEM.
DADDI R1,R1,#4	15	16-20	21	22	23
DADDI R5,R1,#-400	16-20	21	22	23	24
BNEZ R5,Loop	21	22	23	24	25

For one iteration of the loop, we have the total number of clock cycles is 21 and we have eight instructions. Then

$$CPI_{One Iteration} = \frac{21}{8} = 2.625$$

For the entire loop, we have the last four clock cycles of each iteration are the first four clock cycles of the next iteration. So, the number of clock cycles of the entire loop will be $21 + (21 - 4) \times 99$ because we have 100 iterations. We have 800 instructions for the entire loop. Then

$$CPI_{Entire Loop} = \frac{21 + (21 - 4)\times 99}{800} = 2.13$$

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