From the link below:

https://www.chegg.com/homework-help/questions-and-answers/431-exercise-compare-performance-1-issue-2-issue-processors-taking-account-program-transfo-q25269311

I pasted the answer here in italics for convenience.

Code with stall introduced:

MOV X5, XZR

B ENT

TOP:

LSL X10, X5, #3

ADD X11, X1, X10

LDUR X12, [X11, #0]

LDUR X11, [X11, #8]

SUB X14, X12, X13

ADD X15, X2, X10

STUR X14, [X15, #0]

ADDI X5, X5, #2

ENT: CMP X5, X6

B.NE TOP

On a one issue machine with the stalls in the above requires 12 cycles/loop.

For a 2-issue machine it requires 11 cycles/loop.

=> Speed up = 12/11 = 1.09 which is a good improvement

As the first instruction is fetched in cycle5 in loop 1 for the instruction LSL and in first intruction in iteration2 cycle2 is fetched in cycle16 from this we know that the code takes 11 cycles/iteration on the 2-issue machine.

Why is there a second stall in the answer? Doesn't forwarding alleviate the need for a stall between the CMP and Branch instruction?

Thank you.