solution to midterm question 3

3)  pipeline

For each code sequence below, state whether it must stall, can avoid stalls using only forwarding, or can execute without stalling or forwarding.

3.1) sequence 1

lw x10, 0(x10)
add x11, x10, x10

answer 
 
stall (after lw) because data dependency on x10, lw will get data at stage 4 (mem) but add needs this data at stage 3, forwarding will not help because forwarding can deliver data at stage 3 (alu) which (mem) did not yet get the data.

timeline in the pipeline:

           
lw x10   fet  dec  alu  mem  wb
add x11       fet  dec  alu  mem  wb

3.2) sequence 2

add x11, x10, x10
addi x12, x10, 5
addi x14, x11, 5

answer

between add x11 and addi x12 there is no data dependency.
between addi x12 and addi x14 there is no data dependency.
between add x11 and addi x14 there is data dependency on x11 but these instructions are already two clocks apart.
conclusion: there is no stall

in case of add x11 and addi x14, forwarding will not help because it can only forward to the next instruction (not two instructions)

timeline in the pipeline:

add x11    fet dec alu  mem  wb
addi x12       fet dec  alu  mem wb
addi x14           fet  dec  alu mem wb


3.3) sequence 3

addi x11, x10, 1
addi x12, x10, 2
addi x13, x10, 3
addi x14, x10, 4
addi x15, x10, 5

answer

all five instructions have no data dependency to each other.
conclusion: no stall