Basic MIPS implementation

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Unit 3
Processor and Control Unit
By
Mrs.M.Kavitha, AP/CSE
A.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 registers
Why? 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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Continue
• 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 slt
3. Execute memory-reference instructions: lw, sw
4. Execute branch/jump instructions: beq, j
1. Fetching instructions and incrementing the PC.

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Continue
2. Execute arithmetic-logical instructions: add, sub, and, or, and slt
The 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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Continue
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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Continue
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

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Example: Implementing Jumps

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Why a Single-Cycle Implementation Is Not Used Today
Example: Performance of Single-Cycle Machines
Calculate 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% loads
10% stores
45% ALU instructions
15% branches, and
5% jumps

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Continue
CPU clock cycle (option 1) = 600 ps.
CPU clock cycle (option 2) = 400 ×45% + 600×25% + 550 ×10% + 350 ×15% + 200×5%
= 447.5 ps.
Performance ratio = 34.1
5.447
600
=
45% ALU instructions
25% loads
10% stores
15% branches, and
5% 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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Continue
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  0
ALUSrcB  01
ALUOp  00 (for add)
PCSource  00
PCWrite  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  0
ALUSrcB  11
ALUOp  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;
ALUOp = 10
Branch:
if (A == B) PC <= ALUOut;
ALUOp = 01 (for subtraction)
PCSource = 01
PCWriteCond is asserted
Jump:
PC <= { PC[31:28], (IR[25:0],2’b00) };
PCWrite is asserted
PCSource = 10

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4. Memory access or R-type instruction completion step
Memory reference:
MDR <= Memory [ALUOut]; ⇒ MemRead, IorD=1
or
Memory [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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Continue
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 CPU
Using 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 = 0.25×5 + 0.10×4 + 0.52×4 + 0.11×3 + 0.02×3 = 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 countInstructio
n countInstructio
CPI
n countInstructio
n countInstructio
CPI
n countInstructio
CPIn countInstructio
n countInstructio
cyclesCPU clock
CPI
i
i
i
ii
ratioThe
∑
∑
×=
×
==

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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 terminology
I/O device request External Interrupt
Invoke the operating system from user
program
Internal Exception
Arithmetic overflow Internal Exception
Using an undefined instruction Internal Exception
Hardware 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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Continue
The multicycle datapath with the addition needed to implement exceptions

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Continue
The finite state machine with the additions to handle exception detection

Basic MIPS implementation

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Basic MIPS implementation