A single-cycle MIPS processor
 As previously discussed, an instruction set architecture is an interface
that defines the hardware operations which are available to software.
 Any instruction set can be implemented in many different ways. Over the
next few weeks we’ll compare two important implementations.
— In a basic single-cycle implementation all operations take the same
amount of time—a single cycle.
— In a pipelined implementation, a processor can overlap the execution
of several instructions, potentially leading to big performance gains.
July 31, 2017
1
Single-cycle implementation
 In lecture, we will describe the implementation a simple MIPS-based
instruction set supporting just the following operations.
Arithmetic:
Data Transfer:
Control:
add
sub
lw
sw
and
or
slt
beq
— We use MIPS because it is significantly easier to implement than x86.
 Today we’ll build a single-cycle implementation of this instruction set.
— All instructions will execute in the same amount of time; this will
determine the clock cycle time for our performance equations.
— We’ll explain the datapath first, and then make the control unit.
July 31, 2017
A single-cycle MIPS processor
2
Computers are state machines
 A computer is just a big fancy state machine.
— Registers, memory, hard disks and other storage form the state.
— The processor keeps reading and updating the state, according to the
instructions in some program.
 Computers are state machines (or finite automata).
CPU
State
July 31, 2017
A single-cycle MIPS processor
3
John von Neumann
 In the old days, “programming” involved actually changing a machine’s
physical configuration by flipping switches or connecting wires.
— A computer could run just one program at a time.
— Memory only stored data that was being operated on.
 Then around 1944, John von Neumann and others got the idea to encode
instructions in a format that could be stored in memory just like data.
— The processor interprets and executes instructions from memory.
— One machine could perform many different tasks, just by loading
different programs into memory.
— The “stored program” design is often called a Von Neumann machine.
July 31, 2017
A single-cycle MIPS processor
4
Memories
 It’s easier to use a Harvard architecture at first, with
programs and data stored in separate memories.
 To fetch instructions and read & write words, we
need these memories to be 32-bits wide (buses are
represented by dark lines here), so these are 230 x 32
memories. (We will ignore byte addressability for the
moment.)
 Blue lines represent control signals. MemRead and
MemWrite should be set to 1 if the data memory is to
be read or written respectively, and 0 otherwise.
— When a control signal does something when it is
set to 1, we call it active high (vs. active low)
because 1 is usually a higher voltage than 0.
 For today, we will assume you cannot write to the
instruction memory.
— Pretend it’s already loaded with a program,
which doesn’t change while it’s running.
July 31, 2017
A single-cycle MIPS processor
Read Instruction
address
[31-0]
Instruction
memory
MemWrite
Read
address
Read
data
Write
address
Write
data
Data
memory
MemRead
5
Instruction fetching
 The CPU is always in an infinite loop, fetching instructions from memory
and executing them.
 The program counter or PC register holds the address of the current
instruction.
 MIPS instructions are each four bytes long, so the PC should be
incremented by four to read the next instruction in sequence.
2004
Add
2004
4
P
C
2000
Read Instruction
address
[31-0]
Instruction
memory
July 31, 2017
A single-cycle MIPS processor
6
Encoding R-type instructions
 A few weeks ago, we saw encodings of MIPS instructions as 32-bit values.
 Register-to-register arithmetic instructions use the R-type format.
— op is the instruction opcode, and func specifies a particular arithmetic
operation (see the back of the textbook).
— rs, rt and rd are source and destination registers.
op
rs
rt
rd
shamt
func
6 bits
5 bits
5 bits
5 bits
5 bits
6 bits
 An example instruction and its encoding:
add $s4, $t1, $t2
July 31, 2017
000000 01001
01010
A single-cycle MIPS processor
10100
00000 1000000
7
Registers and ALUs
 R-type instructions must access registers and an ALU.
 Our register file stores thirty-two 32-bit values.
— Each register specifier is 5 bits long.
— You can read from two registers at a time.
— RegWrite is 1 if a register should be written.
RegWrite
Read
register 1
Read
data 1
Read
register 2
Read
data 2
Write
register
Write
data
Registers
 Here’s a simple ALU with five operations, selected by
a 3-bit control signal ALUOp.
ALU
July 31, 2017
ALUOp
Function
000
001
010
110
111
and
or
add
subtract
slt
A single-cycle MIPS processor
ALUOp
8
Executing an R-type instruction
1. Read an instruction from the instruction memory.
2. The source registers, specified by instruction fields rs and rt, should be
read from the register file.
3. The ALU performs the desired operation.
4. Its result is stored in the destination register, which is specified by field
rd of the instruction word.
RegWrite
Read Instruction
address
[31-0]
I [25 - 21]
I [20 - 16]
Instruction
memory
I [15 - 11]
Read
register 1
Read
register 2
Write
register
Write
data
op
31
July 31, 2017
rs
26
25
Read
data 1
20
Zero
Result
Read
data 2
ALUOp
Registers
rt
21
ALU
rd
16
15
shamt
11 10
A single-cycle MIPS processor
6 5
func
0
9
Encoding I-type instructions
 The lw, sw and beq instructions all use the I-type encoding.
— rt is the destination for lw, but a source for beq and sw.
— address is a 16-bit signed constant.
op
rs
rt
address
6 bits
5 bits
5 bits
16 bits
 Two example instructions:
lw
$t0, –4($sp)
100011 11101
01000
1111 1111 1111 1100
sw
$a0, 16($sp)
101011 11101
00100
0000 0000 0001 0000
July 31, 2017
A single-cycle MIPS processor
10
Accessing data memory
 For an instruction like lw $t0, –4($sp), the base register $sp is added to
the sign-extended constant to get a data memory address.
 This means the ALU must accept either a register operand for arithmetic
instructions, or a sign-extended immediate operand for lw and sw.
 We’ll add a multiplexer, controlled by ALUSrc, to select either a register
operand (0) or a constant operand (1).
r1
0
M
u
x
r2
1
RegDst
July 31, 2017
r1
0
M
u
x
r1
r2
r2
1
RegDst
A single-cycle MIPS processor
11
Accessing data memory
 For an instruction like lw $t0, –4($sp), the base register $sp is added to
the sign-extended constant to get a data memory address.
 This means the ALU must accept either a register operand for arithmetic
instructions, or a sign-extended immediate operand for lw and sw.
 We’ll add a multiplexer, controlled by ALUSrc, to select either a register
operand (0) or a constant operand (1).
RegWrite
Read Instruction
address
[31-0]
MemWrite
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
0
I [15 - 11]
M
u
x
1
Read
register 2
Write
register
Write
data
Read
data 1
Zero
Read
data 2
Registers
Result
M
u
x
ALUOp
ALUSrc
Sign
extend
(FFFC)hex
(A37C)hex
July 31, 2017
0
1
RegDst
I [15 - 0]
ALU
Read
address
Read
data
Write
address
Write
data
Data
memory
MemToReg
1
M
u
x
0
MemRead
(FFFFFFFC)hex
(0000A37C)hex
A single-cycle MIPS processor
12
MemToReg
 The register file’s “Write data” input has a similar problem. It must be
able to store either the ALU output of R-type instructions, or the data
memory output for lw.
 We add a mux, controlled by MemToReg, to select between saving the
ALU result (0) or the data memory output (1) to the registers.
RegWrite
Read Instruction
address
[31-0]
MemWrite
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
0
I [15 - 11]
M
u
x
1
Read
register 2
Write
register
Write
data
Read
data 1
Zero
Read
data 2
Registers
July 31, 2017
0
Result
M
u
x
1
ALUOp
ALUSrc
RegDst
I [15 - 0]
ALU
Read
address
Read
data
Write
address
Write
data
Data
memory
MemToReg
1
M
u
x
0
MemRead
Sign
extend
A single-cycle MIPS processor
13
RegDst
 A final annoyance is the destination register of lw is in rt instead of rd.
op
rs
rt
address
lw $rt, address($rs)
 We’ll add one more mux, controlled by RegDst, to select the destination
register from either instruction field rt (0) or field rd (1).
RegWrite
Read Instruction
address
[31-0]
MemWrite
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
0
I [15 - 11]
M
u
x
1
Read
register 2
Write
register
Write
data
Read
data 1
Zero
Read
data 2
Registers
July 31, 2017
0
Result
M
u
x
1
ALUOp
ALUSrc
RegDst
I [15 - 0]
ALU
Read
address
Read
data
Write
address
Write
data
Data
memory
MemToReg
1
M
u
x
0
MemRead
Sign
extend
A single-cycle MIPS processor
14
Branches
 For branch instructions, the constant is not an address but an instruction
offset from the current program counter to the desired address.
L:
beq
add
add
j
add
$at, $0, L
$v1, $v0, $0
$v1, $v1, $v1
Somewhere
$v1, $v0, $v0
 The target address L is three instructions past the beq, so the encoding of
the branch instruction has 0000 0000 0000 0011 for the address field.
000100
00001
00000
0000 0000 0000 0011
op
rs
rt
address
 Instructions are 4 bytes long, so the actual memory offset is 4*3=12 bytes.
July 31, 2017
A single-cycle MIPS processor
15
The steps in executing a beq
1.
2.
3.
4.
Fetch the instruction, like beq $at, $0, offset, from memory.
Read the source registers, $at and $0, from the register file.
Compare the values by subtracting them in the ALU.
If the subtraction result is 0, the source operands were equal and the PC
should be loaded with the target address, PC + 4 + (offset x 4).
5. Otherwise the branch should not be taken, and the PC should just be
incremented to PC + 4 to fetch the next instruction sequentially.
July 31, 2017
A single-cycle MIPS processor
16
Branching hardware
We need a second adder, since the ALU
is already doing subtraction for the beq.
0
M
u
x
Add
PC
4
Multiply constant
by 4 to get offset.
Add
1
 PCSrc=1 branches
to PC+4+(offset4).
 PCSrc=0 continues
to PC+4.
Shift
left 2
PCSrc
RegWrite
Read Instruction
address
[31-0]
MemWrite
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
0
I [15 - 11]
M
u
x
1
Read
register 2
Write
register
Write
data
Read
data 1
Zero
Read
data 2
Registers
July 31, 2017
0
Result
M
u
x
1
ALUOp
ALUSrc
RegDst
I [15 - 0]
ALU
Read
address
Read
data
Write
address
Write
data
Data
memory
MemToReg
1
M
u
x
0
MemRead
Sign
extend
A single-cycle MIPS processor
17
The final datapath
0
M
u
x
Add
PC
4
Add
1
Shift
left 2
PCSrc
RegWrite
Read Instruction
address
[31-0]
MemWrite
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
0
I [15 - 11]
M
u
x
1
Read
register 2
Write
register
Write
data
Read
data 1
Zero
Read
data 2
Registers
July 31, 2017
0
Result
M
u
x
1
ALUOp
ALUSrc
RegDst
I [15 - 0]
ALU
Read
address
Read
data
Write
address
Write
data
Data
memory
MemToReg
1
M
u
x
0
MemRead
Sign
extend
A single-cycle MIPS processor
18
Control
 The control unit is responsible for setting all the control signals so that
each instruction is executed properly.
— The control unit’s input is the 32-bit instruction word.
— The outputs are values for the blue control signals in the datapath.
 Most of the signals can be generated from the instruction opcode alone,
and not the entire 32-bit word.
 To illustrate the relevant control signals, we will show the route that is
taken through the datapath by R-type, lw, sw and beq instructions.
July 31, 2017
A single-cycle MIPS processor
19
R-type instruction path
 The R-type instructions include add, sub, and, or, and slt.
 The ALUOp is determined by the instruction’s “func” field.
0
M
u
x
Add
PC
4
Add
1
Shift
left 2
PCSrc
RegWrite
Read Instruction
address
[31-0]
MemWrite
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
0
I [15 - 11]
M
u
x
1
Read
register 2
Write
register
Write
data
Read
data 1
Zero
Read
data 2
Registers
July 31, 2017
0
Result
M
u
x
1
ALUOp
ALUSrc
RegDst
I [15 - 0]
ALU
Read
address
Read
data
Write
address
Write
data
Data
memory
MemToReg
1
M
u
x
0
MemRead
Sign
extend
A single-cycle MIPS processor
20
lw instruction path
 An example load instruction is lw $t0, –4($sp).
 The ALUOp must be 010 (add), to compute the effective address.
0
M
u
x
Add
PC
4
Add
1
Shift
left 2
PCSrc
RegWrite
Read Instruction
address
[31-0]
MemWrite
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
0
I [15 - 11]
M
u
x
1
Read
register 2
Write
register
Write
data
Read
data 1
Zero
Read
data 2
Registers
July 31, 2017
0
Result
M
u
x
1
ALUOp
ALUSrc
RegDst
I [15 - 0]
ALU
Read
address
Read
data
Write
address
Write
data
Data
memory
MemToReg
1
M
u
x
0
MemRead
Sign
extend
A single-cycle MIPS processor
21
sw instruction path
 An example store instruction is sw $a0, 16($sp).
 The ALUOp must be 010 (add), again to compute the effective address.
0
M
u
x
Add
PC
4
Add
1
Shift
left 2
PCSrc
RegWrite
Read Instruction
address
[31-0]
MemWrite
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
0
I [15 - 11]
M
u
x
1
Read
register 2
Write
register
Write
data
Read
data 1
Zero
Read
data 2
Registers
July 31, 2017
0
Result
M
u
x
1
ALUOp
ALUSrc
RegDst
I [15 - 0]
ALU
Read
address
Read
data
Write
address
Write
data
Data
memory
MemToReg
1
M
u
x
0
MemRead
Sign
extend
A single-cycle MIPS processor
22
beq instruction path
 One sample branch instruction is beq $at, $0, offset.
 The ALUOp is 110 (subtract), to test for equality.
The branch may
or may not be
taken, depending
on the ALU’s Zero
output
0
M
u
x
Add
PC
4
Add
1
Shift
left 2
PCSrc
RegWrite
Read Instruction
address
[31-0]
MemWrite
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
0
I [15 - 11]
M
u
x
1
Read
register 2
Write
register
Write
data
Read
data 1
Zero
Read
data 2
Registers
July 31, 2017
0
Result
M
u
x
1
ALUOp
ALUSrc
RegDst
I [15 - 0]
ALU
Read
address
Read
data
Write
address
Write
data
Data
memory
MemToReg
1
M
u
x
0
MemRead
Sign
extend
A single-cycle MIPS processor
23
Control signal table
Operation RegDst RegWrite ALUSrc ALUOp
MemWrite
MemRead
MemToReg
add
1
1
0
010
0
0
0
sub
1
1
0
110
0
0
0
and
1
1
0
000
0
0
0
or
1
1
0
001
0
0
0
slt
1
1
0
111
0
0
0
lw
0
1
1
010
0
1
1
sw
X
0
1
010
1
0
X
beq
X
0
0
110
0
0
X
 sw and beq are the only instructions that do not write any registers.
 lw and sw are the only instructions that use the constant field. They also
depend on the ALU to compute the effective memory address.
 ALUOp for R-type instructions depends on the instructions’ func field.
 The PCSrc control signal (not listed) should be set if the instruction is beq
and the ALU’s Zero output is true.
July 31, 2017
A single-cycle MIPS processor
24
Generating control signals
 The control unit needs 13 bits of inputs.
— Six bits make up the instruction’s opcode.
— Six bits come from the instruction’s func field.
— It also needs the Zero output of the ALU.
 The control unit generates 10 bits of output, corresponding to the signals
mentioned on the previous page.
 You can build the actual circuit by using big K-maps, big Boolean algebra,
or big circuit design programs.
 The textbook presents a slightly different control unit.
RegDst
RegWrite
Read Instruction
address
[31-0]
I [31 - 26]
ALUSrc
ALUOp
I [5 - 0]
MemWrite
Control
Instruction
memory
MemRead
MemToReg
PCSrc
Zero
July 31, 2017
A single-cycle MIPS processor
25
Summary
 A datapath contains all the functional units and connections necessary to
implement an instruction set architecture.
— For our single-cycle implementation, we use two separate memories,
an ALU, some extra adders, and lots of multiplexers.
— MIPS is a 32-bit machine, so most of the buses are 32-bits wide.
 The control unit tells the datapath what to do, based on the instruction
that’s currently being executed.
— Our processor has ten control signals that regulate the datapath.
— The control signals can be generated by a combinational circuit with
the instruction’s 32-bit binary encoding as input.
 On next slides, we’ll see the performance limitations of this single-cycle
machine and discuss how to improve upon it.
July 31, 2017
A single-cycle MIPS processor
26
The single-cycle design from last time
0
M
u
x
Add
PC
4
Add
1
Shift
left 2
A control unit (not
shown) generates all
the control signals
from the instruction’s
“op” and “func” fields.
PCSrc
RegWrite
Read Instruction
address
[31-0]
MemWrite
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
0
I [15 - 11]
M
u
x
1
Read
register 2
Write
register
Write
data
Read
data 1
Zero
Read
data 2
Registers
July 31, 2017
0
Result
M
u
x
1
ALUSrc
RegDst
I [15 - 0]
ALU
ALUOp
Read
address
Read
data
1
Data
memory
0
Write
address
Write
data
MemToReg
M
u
x
MemRead
Sign
extend
Multicycle datapath
27
The example add from last time
 Consider the instruction add $s4, $t1, $t2.
000000
01001
01010
10100
00000
100000
op
rs
rt
rd
shamt
func
 Assume $t1 and $t2 initially contain 1 and 2 respectively.
 Executing this instruction involves several steps.
1. The instruction word is read from the instruction memory, and the
program counter is incremented by 4.
2. The sources $t1 and $t2 are read from the register file.
3. The values 1 and 2 are added by the ALU.
4. The result (3) is stored back into $s4 in the register file.
July 31, 2017
Multicycle datapath
28
How the add goes through the datapath
PC+4
0
M
u
x
Add
PC
4
Add
1
Shift
left 2
PCSrc
RegWrite
Read Instruction
address
[31-0]
I [25 - 21] 01001
I [20 - 16] 01010
Instruction
memory
0
I [15 - 11]
M
u
x
1
10100
MemWrite
Read
register 1
Read
register 2
Write
register
Write
data
Read
data 1
Read
data 2
Registers
I [15 - 0]
July 31, 2017
ALU
00...10
Zero
0
Result
M
u
x
1
ALUSrc
RegDst
MemToReg
00...01
Sign
extend
Multicycle datapath
ALUOp
Read
address
Read
data
1
Data
memory
0
Write
address
Write
data
M
u
x
MemRead
00...11
29
Performance of Single-cycle Design
CPU timeX,P = Instructions executedP * CPIX,P * Clock cycle timeX
= N
July 31, 2017
* 1 * ???
Performance
30
Edge-triggered state elements
 In an instruction like add $t1, $t1, $t2, how do we know
$t1 is not updated until after its original value is read?
 We’ll assume that our state elements are positive edge
triggered, and are updated only on the positive edge of a
clock signal.
— The register file and data memory have explicit write
control signals, RegWrite and MemWrite. These units
can be written to only if the control signal is asserted
and there is a positive clock edge.
— In a single-cycle machine the PC is updated on each
clock cycle, so we don’t bother to give it an explicit
write control signal.
CPU
RegWrite
Read
register 1
Read
data 1
Read
register 2
Read
data 2
Write
register
Registers
Write
data
MemWrite
Read
address
Read
data
Write
address
Write
data
Data
memory
MemRead
PC
State
July 31, 2017
Multicycle datapath
31
The datapath and the clock
1. On a positive clock edge, the PC is updated with a new address.
2. A new instruction can then be loaded from memory. The control unit sets
the datapath signals appropriately so that
— registers are read,
— ALU output is generated,
— data memory is read or written, and
— branch target addresses are computed.
3. Several things happen on the next positive clock edge.
— The register file is updated for arithmetic or lw instructions.
— Data memory is written for a sw instruction.
— The PC is updated to point to the next instruction.

In a single-cycle datapath everything in Step 2 must complete within one
clock cycle, before the next positive clock edge.
How long is that clock cycle?
July 31, 2017
Multicycle datapath
32
Compute the longest path in the add instruction
PC+4
0
M
u
x
Add
PC
4
Add
2 ns
2 ns
RegWrite
Read Instruction
address
[31-0]
Read
register 1
I [20 - 16]
2 ns
0
I [15 - 11]
M
u
x
1
RegDst
I [15 - 0]
July 31, 2017
PCSrc
0 ns
MemWrite
I [25 - 21]
Instruction
memory
1
Shift
left 2
0 ns
Read
data 1
Zero
Read
register 2
Read
data 2
Write
register
Write
data
ALU
Registers
0
Result
M
u
x
1
ALUOp
2 ns
1 ns
ALUSrc
Sign
extend
0 ns
Multicycle datapath
Read
address
Read
data
1
Data
memory
0
Write
address
Write
data
MemToReg
M
u
x
0 ns
MemRead
2 ns
33
The slowest instruction...
 If all instructions must complete within one clock cycle, then the cycle
time has to be large enough to accommodate the slowest instruction.
 For example, lw $t0, –4($sp) is the slowest instruction needing __ns.
— Assuming the circuit latencies below.
Read Instruction
address
[31-0]
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
2 ns
0
I [15 - 11]
M
u
x
1
0 ns
I [15 - 0]
Read
data 1
Zero
Read
register 2
Read
data 2
Write
register
Write
data
ALU
0
M
u
x
Registers
1
0 ns
1 ns
Result
2 ns
Read
address
Read
data
1
Data
memory
0
Write
address
Write
data
M
u
x
0 ns
2 ns
Sign
extend
0 ns
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Multicycle datapath
34
The slowest instruction...
 If all instructions must complete within one clock cycle, then the cycle
time has to be large enough to accommodate the slowest instruction.
 For example, lw $t0, –4($sp) needs 8ns, assuming the delays shown here.
reading the instruction memory
2n
s
reading the base register $sp
8ns
1n
Read Instruction
address
[31-0]
s
I [25computing
- 21]
memory
address $sp-4
Read
Read
register 1
I [20 - 16]
Instruction
memory
2 ns
data 1
ALU
s
Read
0
data 2
Writedata memory
M
reading
the
I [15 - 11]
Read
register 2
u
x
register
Write
data
Registers
s 1
0 ns
1 ns
storing
data
back
to
$t0
I [15 - 0]
Sign
Zero
0
M
u
x
1
July 31, 2017
Result
2 ns2n
0 ns
Multicycle datapath
Read
address
Read
data
1
Data
memory
0
Write
address
Write
data
M
u
x
0 ns
2 ns
0 ns
extend
s
2n
1n
35
...determines the clock cycle time
 If we make the cycle time 8ns then every instruction will take 8ns, even
if they don’t need that much time.
 For example, the instruction add $s4, $t1, $t2 really needs just 6ns.
reading the instruction memory
reading registers $t1 and $t2
computing $t1 + $t2
storing the result into $s0
Read Instruction
address
[31-0]
I [25 - 21]
Read
register 1
I [20 - 16]
Instruction
memory
2 ns
0
I [15 - 11]
M
u
x
1
0 ns
I [15 - 0]
Read
data 1
6 ns
ALU
Zero
Read
register 2
Read
data 2
Write
register
Write
data
2 ns
1 ns
2 ns
1 ns
0
M
u
x
Registers
1
0 ns
1 ns
Result
2 ns
Read
address
Read
data
1
Data
memory
0
Write
address
Write
data
M
u
x
0 ns
2 ns
Sign
extend
0 ns
July 31, 2017
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36
How bad is this?
 With these same component delays, a sw instruction would need 7ns, and
beq would need just 5ns.
 Let’s consider the gcc instruction mix from p. 189 of the textbook.
Instruction Frequency
Arithmetic
Loads
Stores
Branches
48%
22%
11%
19%
 With a single-cycle datapath, each instruction would require 8ns.
 But if we could execute instructions as fast as possible, the average time
per instruction for gcc would be:
(48% x 6ns) + (22% x 8ns) + (11% x 7ns) + (19% x 5ns) = 6.36ns
 The single-cycle datapath is about 1.26 times slower!
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37
It gets worse...
 We’ve made very optimistic assumptions about memory latency:
— Main memory accesses on modern machines is >50ns.
• For comparison, an ALU on an AMD Opteron takes ~0.3ns.
 Our worst case cycle (loads/stores) includes 2 memory accesses
— A modern single cycle implementation would be stuck at <10Mhz.
— Caches will improve common case access time, not worst case.
 Tying frequency to worst case path violates first law of performance!!
— “Make the common case fast” (we’ll revisit this often)
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38
Summary
 Performance is one of the most important criteria in judging systems.
— Here we’ll focus on Execution time.
 Our main performance equation explains how performance depends on
several factors related to both hardware and software.
CPU timeX,P = Instructions executedP * CPIX,P * Clock cycle timeX
 It can be hard to measure these factors in real life, but this is a useful
guide for comparing systems and designs.
 A single-cycle CPU has two main disadvantages.
— The cycle time is limited by the worst case latency.
— It isn’t efficiently using its hardware.
 Next time, we’ll see how this can be rectified with pipelining.
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Multicycle datapath
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