Lecture 14: Processors CS 2011 Fall 2014, Dr. Rozier.

Lecture 14: Processors

CS 2011

Fall 2014, Dr. Rozier

BOMB LAB STATUS

MP2

Lab Phases: Recursive

• Phase 1 – Factorial

• Phase 2 - Fibonacci

Lab Phases: Arrays

• Phase 4 – Sum Array

• Phase 5 – Find Item

• Phase 6 – Bubble Sort

Lab Phases: Trees

Array representation: [1,2,3,4,5,6,7,0,0,0,0,0,0,0,0]

•Phase 7 – Tree Height

•Phase 8 – Tree Traversal[1,2,5,0,0,4,0,0,3,6,0,0,7,0,0]

1 2 3 4 5 6 7

PROCESSORS

What needs to be done to “Process” an Instruction?

• Check the PC• Fetch the instruction from memory• Decode the instruction and set control lines appropriately• Execute the instruction

– Use ALU– Access Memory– Branch

• Store Results• PC = PC + 4, or PC = branch target

CPU Overview

CPU Overview

Chapter 4 — The Processor — 10

Can’t just join wires together Use multiplexers

CPU + Control

Logic Design Basics

• Information encoded in binary– Low voltage = 0, High voltage = 1– One wire per bit– Multi-bit data encoded on multi-wire buses

• Combinational element– Operate on data– Output is a function of input

• State (sequential) elements– Store information

Combinational Elements

• AND-gate– Y = A & BAB

Y

I0I1

YMux

S

Multiplexer Y = S ? I1 : I0

A

B

Y+

A

B

YALU

F

Adder Y = A + B

Arithmetic/Logic Unit Y = F(A, B)

Storing Data?

S-R Latch

• S – set• R – reset

• Feedback keeps the bit “trapped”.

S-R Latch

Characteristic Table Excitation TableS R Q_next Action Q Q_next S R

0 0 Q hold 0 0 0 X

0 1 0 reset 0 1 1 0

1 0 1 set 1 0 0 1

1 1 X N/A 1 1 X 0

D Flip-Flop

• We can note in the S-R Latch that S is the complement of R in state changes

D Flip-Flop

• Feed D and ~D to a gated S-R Latch to create a one input synchronous SR-Latch

We’ll call it a D Flip-Flop, just to be difficult.

D Flip-Flop

• D – input signal• E – enable signal,

sometimes called clock or control

E/C D Q ~Q Notes

0 X Q_prev ~Q_prev

1 0 0 1

1 1 1 0

D Flip-Flop

• D – input signal• E – enable signal,

sometimes called clock or control

E/C D Q ~Q Notes

0 X Q_prev ~Q_prev No change

1 0 0 1 Reset

1 1 1 0 Set

Adding the Clock

More Realistic

Register File

Sequential Elements• Register: stores data in a circuit

– Uses a clock signal to determine when to update the stored value

– Edge-triggered: update when Clk changes from 0 to 1

D

Clk

QClk

D

Q

Sequential Elements• Register with write control

– Only updates on clock edge when write control input is 1

– Used when stored value is required later

D

Clk

Q

Write

Write

D

Q

Clk

Clocking Methodology• Combinational logic transforms data during

clock cycles– Between clock edges– Input from state elements, output to state

element– Longest delay determines clock period

Building a Datapath

• Datapath– Elements that process data and addresses

in the CPU• Registers, ALUs, mux’s, memories, …

• We will build a MIPS datapath incrementally– Refining the overview design

Pipeline

Fetch

Decode

IssueInteger

Multiply

Floating Point

Load Store

Write Back

Instruction Fetch

32-bit register

Increment by 4 for next instruction

ALU• Read two register operands• Perform arithmetic/logical operation• Write register result

Load/Store Instructions• Read register operands• Calculate address• Load: Read memory and update register• Store: Write register value to memory

Branch Instructions?

Datapath With Control

ALU Instruction

Load Instruction

Branch-on-Equal Instruction

Performance Issues

• Longest delay determines clock period– Critical path: load instruction– Instruction memory register file ALU data

memory register file

• Not feasible to vary period for different instructions

• Violates design principle– Making the common case fast

• We will improve performance by pipelining

Pipelining Analogy• Pipelined laundry: overlapping execution

– Parallelism improves performance

§4.5 An O

verview of P

ipelining Four loads: Speedup

= 8/3.5 = 2.3 Non-stop:

Speedup= 2n/0.5n + 1.5 ≈ 4= number of stages

MIPS Pipeline

• Five stages, one step per stage1. IF: Instruction fetch from memory2. ID: Instruction decode & register read3. EX: Execute operation or calculate address4. MEM: Access memory operand5. WB: Write result back to register

Pipeline Performance• Assume time for stages is

– 100ps for register read or write– 200ps for other stages

• Compare pipelined datapath with single-cycle datapath

Instr Instr fetch Register read

ALU op Memory access

Register write

Total time

lw 200ps 100 ps 200ps 200ps 100 ps 800ps

sw 200ps 100 ps 200ps 200ps 700ps

R-format 200ps 100 ps 200ps 100 ps 600ps

beq 200ps 100 ps 200ps 500ps

Pipeline PerformanceSingle-cycle (Tc= 800ps)

Pipelined (Tc= 200ps)

Pipeline Speedup

• If all stages are balanced– i.e., all take the same time

– Time between instructionspipelined

= Time between instructionsnonpipelined

Number of stages

• If not balanced, speedup is less• Speedup due to increased throughput

– Latency (time for each instruction) does not decrease

WRAP UP

For next time

Homework Exercises: 3.4.2, 3.4.4

3.10.1 – 3.10.5

Due Tuesday 11/4

Read Chapter 4.1-4.4