See the answer sheet.

1 Hypothetical OOO processor

For this assignment, we will consider an out-of-order processor which has two execution units for arithmetic instructions, and one execution unit for memory instructions.

1.0.1 This processor’s pipeline

Instructions in this processor are executed as follows:

first, they go through a set of in-order pipeline stages, two a time:
- fetch, where instructions are read from the pipeline stage. For this exercise, assume branch prediction is perfect so two new instructions are fetched every cycle
- decode, where the machine is interpreted
- rename, where register renaming takes place and instructions are placed into an instruction queue, available to be issued as early as the next cycle
then, each cycle instructions are issued from the instruction queue:
- each cycle, the instruction queue is scanned for instructions whose operands are ready. (For this exercise, we’ll only account for their register operands being ready and assume that we do not need to worry about data memory-related dependencies.)
  - operands for an instruction are considered ready during the cycle in which they finish execution
- the first memory instruction available and first two non-memory instructions available are selected to be issued. (If fewer instructions are available, as many are issued as possible.)
- during the cycle an instruction is issued, its register value inputs are read or forwarded
when an instruction is issued, in the following cycles it is executed on an execution unit
- for memory instructions, this execution unit is a pipelined data cache. It receives one instruction per cycle and produces the result after three cycles. To simplify this exercise, we will assume no cache misses.
- for non-memory instructions, this execution unit is one of two arithmetic logic units. It receives one instruction per cycle and produces the result near the end of this cycle.
in the cycle after an instruction is executed it is written back to the register file in one cycle
finally, up to two instructions are committed each cycle using information placed in a reorder buffer during the issue stage. An instruction is only committed if all instructions fetched before have also committed.

Given these stages, an non-memory instruction will take 7 cycles to go through the pipeline if it does not need to wait for it operands or wait for earlier instructions to commit (fetch, decode, rename, issue/register read, execute, writeback, commit). A memory instruction will take 9 cycles (two extra cycles of `execution’ because of the speed of the data cache).

When operands are not available, an instruction will spend extra cycles waiting in the instruction queue between its rename and issue stage. When an instruction is written back but some prior instructions have not done so, then the instruction will spend extra cycles waiting in the reorder buffer before it is committed.

1.0.2 This processor’s registers

The processor has 15 logical registers %r01 through %r15, but these are implemented using 64 physical registers and register renaming. We call the physical registers %x01 through %x64.

We will show instructions in a three-argument form like (regardless of whether register renaming has taken place):

add %r01, %r02 -> %r03

this indicates to add %r01 and %r02 and put the result into %r03.

Answer the questions on the answer sheet linked above.

1.1 Part 1: register renaming

Assume initially that %r01 is assigned to %x01, %r02 to %x02, and so on. Registers %x16 through %x63 are available for register renaming.

Rewrite the following instructions using the %x01 taking into account how register renaming might occur that would allow all of these instructions to be placed in the instruction queue at once:

add %r01, %r01-> %r01  
imul %r02, %r03-> %r04 
add %r01, %r01-> %r01  
add %r02, %r04-> %r04  
add %r03, %r01-> %r01  
imul %r01, %r04-> %r01 
movq 0x0(%r01)-> %r01
imul %r01, %r03-> %r03 
movq %r03, 0x0(%r01)

Record your answer at the answer sheet linked above.

1.2 Part 2: instruction dispatch

Suppose the instruction queue for this processor contains the following instructions after renaming

A. add %x05, %x06 -> %x16
B. sub %x16, %x07 -> %x17
C. movq 0x100(%x05) -> %x18
D. movq 0x108(%x05) -> %x19
E. imul %x01, %x02 -> %x20
F. add %x05, %x08 -> %x21
G. sub %x18, %x19 -> %x22
H. add %x01, %x16 -> %x23
I. movq 0x200(%x22) -> %x24
J. imul %x08, %x16 -> %x25
K. addq %x25, %x20 -> %x26
L. movq %x26, 0x200(%x22)

Initially the value of registers %x01 through %x15 is available.

Complete the timeline to show how all the instructions can be issued and executed:

                cycle 1    2    3   4    5    6   7    8     9    10   11    12 
execution unit:       
arithmetic 1          A
arithmetic 2          E
memory stage 1        C
memory stage 2             C
memory stage 3                  C

Record your answer at the answer sheet linked above.

1.3 Part 3: pipeline diagram

Complete a pipeline diagram for the processor running the following instructions (shown in the form from before register renaming):

1. add %r02, %r01 -> %r01
2. add %r02, %r01 -> %r01
3. add %r02, %r03 -> %r03
4. add %r04, %r05 -> %r05
5. add %r05, %r01 -> %r01
6. add %r01, %r01 -> %r01
7. add %r03, %r05 -> %r05
8. add %r02, %r04 -> %r04

Identify the stages as:

F for when the instruction is feteched
D for when an instruction is decoded
R for when an instruction is renamed and dispatched into an instruction queue
I for when an instruction is issued from an instruction queue and its registers are read
E for when an instruction is being executed (after it is issued)
W for when an instruction’s results are written back to the physical register file
C for when an instruction is committed

Leave the specified stage blank when an instruction is not being processed in any of the above ways (such as when the instruction is in the instruction queue waiting to be issued or any cycles between when an instruction is written back and when it is committed).

The first three rows are:

1 F D R I E W C
2 F D R   I E W C
3   F D R I E W C

complete the remaining rows of the table at the answer sheet linked above.

You may assume that register values computed by previous instructions are available and that physical registers are free such that stages will not be delayed for missing previous register values or lack of available physical registers.