00:01
So from the question, we started from a, which was for the default code, and we scheduled the instruction for the default code following the given pipeline relation by fast.
00:20
So, we started from the given code, we need to run cycle 1, cycle 2, cycle 3, cycle 4, cycle 5, and cycle 6.
00:38
So, the default code has 6 cycles per iteration, and for b, recording instruction to minimize source without unrolling.
00:58
So, recording instruction to minimize source, we need to consider the total dependency and the bypass tool.
01:06
So, here is the recorded schedule for cylinder 1, cylinder 2, cylinder 3, cylinder 4, and cylinder 5.
01:19
So, in this record schedule, we minimize source by ensuring that instruction depends on each other and scheduled as early as possible.
01:30
So, we still have 5 cycles per iteration, but the source has been minimized.
02:00
That's for b, and for c, unrolling the loop to eliminate source cycle...