Basic MIPS implementation

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Unit 3 Processor and Control Unit

By

Mrs.M.Kavitha, AP/CSEA.V.C College of Engineering

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A Basic MIPS Implementation• We're ready to look at an implementation of the MIPS• Simplified to contain only:

– memory-reference instructions: lw, sw – arithmetic-logical instructions: add, sub, and, or, slt– control flow instructions: beq, j

• Generic Implementation:– use the program counter (PC) to supply instruction address– get the instruction from memory– read registers– use the instruction to decide exactly what to do

• All instructions use the ALU after reading the registersWhy? memory-reference? arithmetic? control flow?

5.1 Introduction

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An Overview of the Implementation

• For most instructions: fetch instruction, fetch operands, execute, store.

• Missing Multiplexers, and some Control lines for read and write.

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An Overview of the Implementation

• The program counter gives the instruction address to the instruction memory.

• After the instruction is fetched ,the register operands required by an instruction are specified by fields of that instruction.

• Once the register operands have been fetched, they can be used to compute a memory address( for a load and store), to compute an arithmetic result( for an integer arithmetic-logical instruction) or a compare(for a branch).

• If the instruction is an arithmetic-logical instruction, the result from the ALU must be written to a register.

• If the operation is a load or store, the ALU result is used as an address to either store a value from memory into the registers. The result from the ALU or memory is written back into the register file.

• Branches require the use of the ALU output to determine the next instruction address which comes either from the ALU( where the PC and branch offset are summed) or from an adder that increments the current PC by 4.

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• The basic implementation of the MIPS subset including the necessary multiplexers and control lines.

• Multiplexer (data selector) selects from among several inputs based on the setting of its control lines. The control lines are set based on information taken from the instruction being executed.

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5.3 Building a Datapath

• Datapath– Elements that process data and addresses within the CPU

• Register file, ALUs, Adders, Instruction and Data Memories, …We need functional units (datapath elements) for:

1. Fetching instructions and incrementing the PC.2. Execute arithmetic-logical instructions: add, sub, and, or, and slt3. Execute memory-reference instructions: lw, sw 4. Execute branch/jump instructions: beq, j

1. Fetching instructions and incrementing the PC.

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2. Execute arithmetic-logical instructions: add, sub, and, or, and sltThe arithmetic logic instructions read operands from two registers,

perform an ALU operation on the contents of the registers and write the result to the register. So this instructions as R-type instructions.

add $t1, $t2, $t3 # t1 = t2 + t3

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3. Execute memory-reference instructions: lw, sw lw $t1, offset_value($t2)sw $t1, offset_value($t2)

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4. Execute branch/jump instructions: beq, j

beq $t1, $t2, offset

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Creating a Single Datapath

• Sharing datapath elements between two different instruction classes , we have connected multiple connections to the input of an element and used a multiplexer and control signals to select among the multiple inputs.

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Now we con combine all the pieces to make a simple datapath for the MIPS architecture:

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5.4 A Simple Control Implementation Scheme

The ALU Control

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Designing the Main Control Unit

beq $s1, $s2, L

rs rt4

31:26 25:21 20-16

address

15:0

+subtract

rs rt35 or 43

31:26 25:21 20-16

address

15:0

lw $s1, 100($s2)

sw $s1, 100($s2)

base

rs rt rd shamt0 funct

31:26 25:21 20-16 15:11 10:6 5:0

++

add $s1, $s2, $s3

s1

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Finalizing the Control

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Example: Implementing Jumps

000010 address

31:26 25:0

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Why a Single-Cycle Implementation Is Not Used Today

Example: Performance of Single-Cycle MachinesCalculate cycle time assuming negligible delays except:

– memory (200ps), – ALU and adders (100ps),– register file access (50ps)

Which of the following implementation would be faster:1. When every instruction operates in 1 clock cycle of fixes length.2. When every instruction executes in 1 clock cycle using a variable-length

clock.

To compare the performance, assume the following instruction mix:25% loads10% stores45% ALU instructions15% branches, and5% jumps

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CPU clock cycle (option 1) = 600 ps.

CPU clock cycle (option 2) = 400 45% + 60025% + 550 10% + 350 15% + 2005% = 447.5 ps.

Performance ratio = 34.15.447

600

45% ALU instructions 25% loads10% stores15% branches, and5% jumps

memory (200ps), ALU and adders (100ps),register file access (50ps)

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5.5 A Multicycle Implementation

• A single memory unit is used for both instructions and data.• There is a single ALU, rather than an ALU and two adders.• One or more registers are added after every major functional unit.

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Replacing the three ALUs of the single-cycle by a single ALU means that the single ALU must accommodate all the inputs that used to go to the three different ALUs.

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Breaking the Instruction Execution into Clock Cycles

1. Instruction fetch step

IR <= Memory[PC];PC <= PC + 4;

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IR <= Memory[PC];

To do this, we need: MemRead Assert IRWrite Assert IorD 0-------------------------------PC <= PC + 4;ALUSrcA 0ALUSrcB 01ALUOp 00 (for add)

PCSource 00PCWrite set

The increment of the PC and instruction memory access can occur in parallel, how?

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2. Instruction decode and register fetch step– Actions that are either applicable to all instructions– Or are not harmful

A <= Reg[IR[25:21]];B <= Reg[IR[20:16]];ALUOut <= PC + (sign-extend(IR[15-0] << 2 );

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A <= Reg[IR[25:21]];B <= Reg[IR[20:16]];Since A and B are

overwritten on every cycle Done

ALUOut <= PC + (sign-extend(IR[15-0]<<2);

This requires:ALUSrcA 0ALUSrcB 11ALUOp 00 (for add)

branch target address will be stored in ALUOut.

The register file access and computation of branch target occur in parallel.

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3. Execution, memory address computation, or branch completion

Memory reference:ALUOut <= A + sign-extend(IR[15:0]);

Arithmetic-logical instruction:ALUOut <= A op B;

Branch:if (A == B) PC <= ALUOut;

Jump:PC <= { PC[31:28], (IR[25:0], 2’b00) };

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Memory reference:ALUOut <= A + sign-extend(IR[15:0]); ALUSrcA = 1 && ALUSrcB = 10 ALUOp = 00

Arithmetic-logical instruction:ALUOut <= A op B;ALUSrcA = 1 && ALUSrcB = 00 ALUOp = 10

Branch:if (A == B) PC <= ALUOut;ALUSrcA = 1 && ALUSrcB = 00 ALUOp = 01 (for subtraction)PCSource = 01PCWriteCond is asserted

Jump:PC <= { PC[31:28], (IR[25:0],2’b00) };PCWrite is assertedPCSource = 10

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4. Memory access or R-type instruction completion step

Memory reference:MDR <= Memory [ALUOut]; MemRead, IorD=1orMemory [ALUOut] <= B; MemWrite, IorD=1

Arithmetic-logical instruction (R-type):

Reg[IR[15:11]] <= ALUOut; RegDst=1,RegWrite, MemtoReg=0

5. Memory read completion step

Load:Reg[IR[20:16]] <= MDR; RegDst=0, RegWrite, MemtoReg=1

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Summary of the steps taken to execute any instruction class

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Defining the Control

Two different techniques to specify the control:– Finite state machine– Microprogramming

Example: CPI in a Multicycle CPUUsing the SPECINT2000 instruction mix, which is: 25% load, 10% store, 11% branches, 2% jumps, and 52% ALU.What is the CPI, assuming that each state in the multicycle CPU requires 1 clock cycle?

Answer:The number of clock cycles for each instruction class is the following:

Load: 5 25% Stores: 4 10% ALU instruction: 4 52% Branches: 3 11% Jumps: 3 2%

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Example Continue

The CPI is given by the following:

is simply the instruction frequency for the instruction class i. We can therefore substitute to obtain:

CPI = 0.255 + 0.104 + 0.524 + 0.113 + 0.023 = 4.12

This CPI is better than the worst-case CPI of 5.0 when all instructions take the same number of clock cycles.

n countInstruction countInstructio

CPIn countInstruction countInstructioCPI

n count InstructioCPI n countInstructio

n countInstructiocyclesCPU clock CPI

i

ii

ii

ratio The

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Defining the Control (Continue)

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The complete finite state machine control

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• Finite state machine controllers are typically implemented using a block of combinational logic and a register to hold the current state.

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5.6 Exceptions

• Exceptions• Interrupts

Type of event From where? MIPS terminologyI/O device request External InterruptInvoke the operating system from user program Internal Exception

Arithmetic overflow Internal ExceptionUsing an undefined instruction Internal ExceptionHardware malfunction Either Exception or interrupt

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How Exception Are Handled

To communicate the reason for an exception:1. a status register ( called the Cause register)2. vectored interrupts

Exception type Exception vector address (in hex)Undefined instruction C000 0000hex

Arithmetic overflow C000 0020hex

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How Control Checks for Exception

Assume two possible exceptions: Undefined instruction Arithmetic overflow

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The multicycle datapath with the addition needed to implement exceptions

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The finite state machine with the additions to handle exception detection