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EECS 150 - Components and Design Techniques for Digital Systems Lec 25 â Division & ECC

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### EECS 150 - Components and Design Techniques for Digital Systems Lec 25 – Division & ECC

David Culler

Electrical Engineering and Computer Sciences

University of California, Berkeley

http://www.eecs.berkeley.edu/~culler

http://www-inst.eecs.berkeley.edu/~cs150

Division

1001 Quotient

Divisor 1000 1001010 Dividend–1000 10 101 1010 –1000 10 Remainder (or Modulo result)

- At each step, determine one bit of the quotient
- If current “remainder” larger than division, subtract and set Q bit to 1
- Otherwise, bring down a bit of the dividend and set Q bit to 0
- Dividend = Quotient x Divisor + Remainder

sizeof(dividend) = sizeof(quotient) + sizeof(divisor)

- 3 versions of divide, successive refinement

EECS 150, Fa04, Lec 25-div & errors

32-bit DIVIDE HARDWARE Version 1

- 64-bit Divisor register, 64-bit adder/subtractor, 64-bit Remainder register, 32-bit Quotient register

Shift Right

Divisor

64 bits

Shift Left

Quotient

add/sub

32 bits

Write

Remainder

Control

64 bits

EECS 150, Fa04, Lec 25-div & errors

1. Subtract the Divisor register from the

Remainder register, and place the result

in the Remainder register.

2b. Restore the original value by adding the

Divisor register to the Remainder register, &

place the sum in the Remainder register. Also

shift the Quotient register to the left, setting

the new least significant bit to 0.

2a. Shift the

Quotient register

to the left setting

the new rightmost

bit to 1.

3. Shift the Divisor register right 1 bit.

Done

Start: Place Dividend in Remainder

Divide Alg. Version 1n+1 steps for n-bit Quotient & Rem.

Remainder Quotient Divisor000001110000 00100000

710 210

Remainder < 0

Remainder 0

Test Remainder

No: < n+1 repetitions

n+1

repetition?

Yes: n+1 repetitions (n = 4 here)

EECS 150, Fa04, Lec 25-div & errors

Example: initial condition

Shift Right

0010 0000

Divisor

Shift Left

0000

add/sub

Quotient

Write

0000 0111

Control

Remainder/Dividend

EECS 150, Fa04, Lec 25-div & errors

Example: iter 1, step 1

- 1: rem <= rem – div; [7 – 32 --> -25]
- Check rem <0?

Shift Right

0010 0000

Divisor

Shift Left

0000

add/sub

Quotient

Write

11100111

0000 0111

Control

Remainder/Dividend

EECS 150, Fa04, Lec 25-div & errors

Example: iter 1, step 2b

- rem <= rem + div;
- Shift Q left, bringing in 0

Shift Right

0010 0000

Divisor

Shift Left

0000

add/sub

Quotient

Write

00000111

Control

Remainder/Dividend

EECS 150, Fa04, Lec 25-div & errors

Example: iter 1, step 3

- Shift div right

Shift Right

00010000

Divisor

Shift Left

0000

add/sub

Quotient

Write

00000111

Control

Remainder/Dividend

EECS 150, Fa04, Lec 25-div & errors

Example: Iter 2, Steps 1,2b,3

- 1: rem <= rem – div; [7 – 16 --> -9]
- 2. check rem <0? rem <= rem + div; sll Q, Q0=0
- 3. srl div

Shift Right

00010000

00001000

Divisor

Shift Left

0000

0000

add/sub

Quotient

Write

00000111

Control

Remainder/Dividend

EECS 150, Fa04, Lec 25-div & errors

Example: Iter 3, Steps 1,2b,3

- 1: rem <= rem – div; [7 – 8 --> -1]
- 2. check rem <0? rem <= rem + div; sll Q, Q0=0
- 3. srl div

Shift Right

00000100

Divisor

Shift Left

0000

add/sub

Quotient

Write

00000111

Control

Remainder/Dividend

EECS 150, Fa04, Lec 25-div & errors

Example: Iter 4, Steps 1,2a,3

- 1: rem <= rem – div; [7 – 4 --> 3]
- 2. check rem <0? sll Q, Q0=1
- 3. shr div

Shift Right

00000100

00000010

Divisor

Shift Left

0001

0000

add/sub

Quotient

Write

00000111

00000011

Control

Remainder/Dividend

EECS 150, Fa04, Lec 25-div & errors

Example: Iter 5, Steps 1,2a,3

- 1: rem <= rem – div; [3 – 2 --> 1]
- 2. check rem <0? sll Q, Q0=1
- 3. shr div

Shift Right

00000010

00000001

Divisor

Shift Left

0011

0001

add/sub

Quotient

Write

0000 0011

00000001

Control

Remainder/Dividend

EECS 150, Fa04, Lec 25-div & errors

Version 1: Division 7/2

Iteration step quotient divisor remainder

0 Initial values 0000 0010 0000 0000 0111

1 1: rem=rem-div 0000 0010 0000 1110 0111

2b: rem<0 +div, sll Q, Q0=0 0000 0010 0000 0000 0111

3: shift div right 0000 0001 0000 0000 0111

2 1: rem=rem-div 0000 0001 0000 1111 0111

2b: rem<0 +div, sll Q, Q0=0 0000 0001 0000 0000 0111

3: shift div right 0000 0000 1000 0000 0111

3 1: rem=rem-div 0000 0000 1000 1111 1111

2b: rem<0 +div, sll Q, Q0=0 0000 0000 1000 0000 0111

3: shift div right 0000 0000 0100 0000 0111

4 1: rem=rem-div 0000 0000 0100 0000 0011

2a: rem0 sll Q, Q0=1 0001 0000 0100 0000 0011

3: shift div right 0001 0000 0010 0000 0011

5 1: rem=rem-div 0001 0000 0010 0000 0001

2a: rem0 sll Q, Q0=1 0011 0000 0010 0000 0001

3: shift div right 0011 0000 0001 0000 0001

EECS 150, Fa04, Lec 25-div & errors

Observations on Divide Version 1

- 1/2 bits in divisor always 0 1/2 of 2n-bit adder is wasted 1/2 of 2n-bit divisor is wasted
- Instead of shifting divisor to right, shift remainder to left?
- 1st step cannot produce a 1 in quotient bit (otherwise quotient 2n) switch order to shift first and then subtract, can save 1 iteration

EECS 150, Fa04, Lec 25-div & errors

DIVIDE HARDWARE Version 2

- 32-bit Divisor register, 32-bit ALU, 64-bit Remainder register, 32-bit Quotient register

Divisor

32 bits

Shift Left

Quotient

32 bits

add/sub

Shift Left

Remainder

Control

Write

64 bits

EECS 150, Fa04, Lec 25-div & errors

1. Shift the Remainder register left 1 bit.

2. Subtract the Divisor register from the left half of the Remainder register, & place the

result in the left half of the Remainder register.

3b. Restore the original value by adding the Divisor

register to the left half of the Remainder register,

&place the sum in the left half of the Remainder

register. Also shift the Quotient register to the left,

setting the new least significant bit to 0.

3a. Shift the

Quotient register

to the left setting

the new rightmost

bit to 1.

Done

Start: Place Dividend in Remainder

Divide Alg. Version 2Remainder Quotient Divisor

000001110000 0010

710 210

Remainder 0

Remainder < 0

Test Remainder

nth

repetition?

No: < n repetitions

Yes: n repetitions (n = 4 here)

EECS 150, Fa04, Lec 25-div & errors

Example V2: 7/2 initial conditions

Divisor

0010

Shift Left

0000

Quotient

add/sub

Shift Left

00000111

Control

Write

Remainder

EECS 150, Fa04, Lec 25-div & errors

Example V2: 7/2 iter 1

- 1. Shift left rem
- 2. sub: remLH <- remLH – div [0 – 2]
- Test remLH < 0
- 3b restore remLH <- remLH + div
- Shift 0 into Q

Divisor

0010

Shift Left

0000

0000

Quotient

add/sub

Shift Left

00001110

11101110

00001110

00000111

Control

Write

Remainder

EECS 150, Fa04, Lec 25-div & errors

Announcements

- Dec 9, last day of class will do HKN survey
- Final mid term: Dec 15 12:30 – 3:30 in 1 LECONTE
- EECS in the news
- Recovery Oriented Computing
- Joint with Stanford!
- Availability = MTBF / (MTBF + Time to Repair)
- Historically focus on increasing MTBF
- ROC focuses on reducing Time to Restart:
- Isolation and Redundancy
- System-wide support for Undo
- Integrated Diagnostic Support.
- Online Verification of Recovery Mechanisms.
- Design for high modularity, measurability, and restartability.
- Dependability/Availability Benchmarking.

Dave Patterson

EECS 150, Fa04, Lec 25-div & errors

Observations on Divide Version 2

- Eliminate Quotient register by combining with Remainder as shifted left.
- Start by shifting the Remainder left as before.
- Thereafter loop contains only two steps because the shifting of the Remainder register shifts both the remainder in the left half and the quotient in the right half
- The consequence of combining the two registers together and the new order of the operations in the loop is that the remainder will shifted left one time too many.
- Thus the final correction step must shift back only the remainder in the left half of the register

EECS 150, Fa04, Lec 25-div & errors

DIVIDE HARDWARE Version 3

- 32-bit Divisor register, 32-bit adder/subtractor, 64-bit Remainder register, (0-bit Quotient reg)

Divisor

32 bits

32-bit ALU

“HI”

“LO”

Shift Left

(Quotient)

Remainder

Control

Write

64 bits

EECS 150, Fa04, Lec 25-div & errors

Start: Place Dividend in Remainder

1. Shift the Remainder register left 1 bit.

2. Subtract the Divisor register from the left half of the Remainder register, & place the

result in the left half of the Remainder register.

3b. Restore the original value by adding the Divisor

register to the left half of the Remainder register,

&place the sum in the left half of the Remainder

register. Also shift the Remainder register to the

left, setting the new least significant bit to 0.

3a. Shift the

Remainder register

to the left setting

the new rightmost

bit to 1.

Done. Shift left half of Remainder right 1 bit.

Divide Algorithm Version 3Remainder Divisor0000 0111 0010

710 210

Remainder < 0

Remainder 0

Test Remainder

*upper-half

No: < n repetitions

nth

repetition?

Yes: n repetitions (n = 4 here)

EECS 150, Fa04, Lec 25-div & errors

Observations on Divide Version 3

- Same Hardware as shift and add multiplier: just 63-bit register to shift left or shift right
- Signed divides: Simplest is to remember signs, make positive, and complement quotient and remainder if necessary
- Note: Dividend and Remainder must have same sign
- Note: Quotient negated if Divisor sign & Dividend sign disagreee.g., –7 ÷ 2 = –3, remainder = –1

EECS 150, Fa04, Lec 25-div & errors

Error Correction Codes (ECC)

- Memory systems generate errors (accidentally flipped-bits)
- DRAMs store very little charge per bit
- “Soft” errors occur occasionally when cells are struck by alpha particles or other environmental upsets.
- Less frequently, “hard” errors can occur when chips permanently fail.
- Problem gets worse as memories get denser and larger
- Where is “perfect” memory required?
- servers, spacecraft/military computers, ebay, …
- Memories are protected against failures with ECCs
- Extra bits are added to each data-word
- used to detect and/or correct faults in the memory system
- in general, each possible data word value is mapped to a unique “code word”. A fault changes a valid code word to an invalid one - which can be detected.

EECS 150, Fa04, Lec 25-div & errors

non-code

Sparse population of code words (2M << 2N)

- with identifiable signature

Correcting Code Concept- Detection: bit pattern fails codeword check
- Correction: map to nearest valid code word

Space of possible bit patterns (2N)

EECS 150, Fa04, Lec 25-div & errors

Each data value, before it is written to memory is “tagged” with an extra bit to force the stored word to have even parity:

Each word, as it is read from memory is “checked” by finding its parity (including the parity bit).

b7b6b5b4b3b2b1b0p

b7b6b5b4b3b2b1b0p

+

+

c

Simple Error Detection CodingParity Bit

- A non-zero parity indicates an error occurred:
- two errors (on different bits) is not detected (nor any even number of errors)
- odd numbers of errors are detected.
- What is the probability of multiple simultaneous errors?

EECS 150, Fa04, Lec 25-div & errors

Use more parity bits to pinpoint bit(s) in error, so they can be corrected.

Example: Single error correction (SEC) on 4-bit data

use 3 parity bits, with 4-data bits results in 7-bit code word

3 parity bits sufficient to identify any one of 7 code word bits

overlap the assignment of parity bits so that a single error in the 7-bit work can be corrected

Procedure: group parity bits so they correspond to subsets of the 7 bits:

p1 protects bits 1,3,5,7

p2 protects bits 2,3,6,7

p3 protects bits 4,5,6,7

1 2 3 4 5 6 7

p1 p2 d1 p3 d2 d3 d4

Bit position number

001 = 110

011 = 310

101 = 510

111 = 710

010 = 210

011 = 310

110 = 610

111 = 710

100 = 410

101 = 510

110 = 610

111 = 710

p1

p2

p3

Hamming Error Correcting CodeNote: number bits from left to right.

EECS 150, Fa04, Lec 25-div & errors

Example: c = c3c2c1= 101

error in 4,5,6, or 7 (by c3=1)

error in 1,3,5, or 7 (by c1=1)

no error in 2, 3, 6, or 7 (by c2=0)

Therefore error must be in bit 5.

Note the check bits point to 5

By our clever positioning and assignment of parity bits, the check bits always address the position of the error!

c=000 indicates no error

eight possibilities

1 2 3 4 5 6 7

p1 p2 d1 p3 d2 d3 d4

Note: parity bits occupy power-of-two bit positions in code-word.

On writing to memory:

parity bits are assigned to force even parity over their respective groups.

On reading from memory:

check bits (c3,c2,c1) are generated by finding the parity of the group and its parity bit. If an error occurred in a group, the corresponding check bit will be 1, if no error the check bit will be 0.

check bits (c3,c2,c1) form the position of the bit in error.

Hamming Code ExampleEECS 150, Fa04, Lec 25-div & errors

Interactive Quiz

1 2 3 4 5 6 7 positions

001 010 011 100 101 110 111

P1 P2 d1 P3 d2 d3 d4 role

- You receive:
- 1111110
- 0000010
- 1010010
- What is the correct value?

Position of error = C3C2C1

Where Ci is parity of groupi

EECS 150, Fa04, Lec 25-div & errors

Overhead involved in single error correction code:

let p be the total number of parity bits and d the number of data bits in a p + d bit word.

If p error correction bits are to point to the error bit (p + d cases) plus indicate that no error exists (1 case), we need:

2p >= p + d + 1,

thus p >= log(p + d + 1)

for large d, p approaches log(d)

8 data => 4 parity

16 data => 5 parity

32 data => 6 parity

64 data => 7 parity

Adding on extra parity bit covering the entire word can provide double error detection

1 2 3 4 5 6 7 8

p1 p2 d1 p3 d2 d3 d4 p4

On reading the C bits are computed (as usual) plus the parity over the entire word, P:

C=0 P=0, no error

C!=0 P=1, correctable single error

C!=0 P=0, a double error occurred

C=0 P=1, an error occurred in p4 bit

Hamming Error Correcting CodeTypical modern codes in DRAM memory systems:

64-bit data blocks (8 bytes) with 72-bit code words (9 bytes).

EECS 150, Fa04, Lec 25-div & errors

Summary

- Arithmetic
- Addition, subtraction, multiplication, & division follow the rules you learned as a child, but in base 2
- Iterative Multiplication and Division are essentially same circuit
- Clever book keeping so n-bit add/sub using n-bit register and 2n-bit shift register
- tricks to avoid sub/check/add restore???
- See “Booth’s Alg” tricks below for multiply (if time)
- Error coding
- Map data word into larger code word
- Hamming distance is number of bit flips between symbols
- Parity detects errors of hamming distance 1
- Overlapping parity detects 2 and corrects 1 (SECDED)

EECS 150, Fa04, Lec 25-div & errors

Motivation for Booth’s Algorithm

- Example 2 x 6 = 0010 x 0110: 0010 x 0110 + 0000 shift (0 in multiplier) + 0010 add (1 in multiplier) + 0010 add (1 in multiplier) + 0000 shift (0 in multiplier) 00001100
- If ALU can subtract as well as add, get same result as follows: 6 = – 2 + 8 0110 = – 00010 + 01000 = 11110 + 01000
- For example

0010 x 0110 0000 shift (0 in multiplier)– 0010 sub(first 1 in multiplier)0000 shift (mid string of 1s) + 0010 add (prior step had last 1) 00001100

EECS 150, Fa04, Lec 25-div & errors

Booth Multiplier: an Introduction

- Recode each 1 in multiplier as “+2-1”
- Converts sequences of 1 to 10…0(-1)
- Might reduce the number of 1’s

0 0 1 1 11 1 1 0 0

+1 -1

+1 -1

+1 -1

+1 -1

+1 -1

+1 -1

0 1 0 0 0 0 0 -1 0 0

EECS 150, Fa04, Lec 25-div & errors

Recoding (Encoding) Example

- If you use the last row in multiplication, you should get exactly the same result as using the first row (after all, they represent the same number!)

0 1 1 0 111 0 0 0 1 0

(+1 -1) (+1 -1) (+1 -1)

(+1 -1)(+1 -1)

(+1 -1)

+1 0 -1 +1 0 0 -1 0 0 +1 -1 0

EECS 150, Fa04, Lec 25-div & errors

Booth Multiplication Example

0 0 1 1 0 6x

0 1 1 1 0 14

+1 0 0 -1 0

0 0 0 0 0

1 1 0 1 0 (-6)

0 0 0 0 0

0 0 0 0 0

0 0 1 1 0

0 0 1 0 1 0 1 0 0 84

1 1 1

EECS 150, Fa04, Lec 25-div & errors

Booth’s Alg: Implementation Approach

end of run

beginning of run

Current Bit Bit to the Right Explanation Example Op

1 0 Begins run of 1s 0001111000 sub

1 1 Middle of run of 1s 0001111000 none

0 1 End of run of 1s 0001111000 add

0 0 Middle of run of 0s 0001111000 none

Originally for Speed (when shift is faster than add, it is advantageous to replace adds and subs with shifts)

Basic idea: replace a string of 1s in multiplier with an initial subtract for rightmost 1 in a run of 1’s, then later add back a 1 for the bit to the left of the last 1 in the run

middle of run

0 1 1 1 1 0

–1

+ 10000

01111

EECS 150, Fa04, Lec 25-div & errors

Booth’s Example (2 x 7)

Operation Multiplicand Product next?

0. initial value 0010 0000 01110 10 -> sub

1a. P = P - m 1110 + 1110 1110 01110 shift P (sign ext)

1b. 0010 1111 00111 11 -> nop, shift

2. 0010 1111 10011 11 -> nop, shift

3. 0010 1111 11001 01 -> add

4a. 0010 + 0010 0001 11001 shift

4b. 0010 0000 1110 0 done

EECS 150, Fa04, Lec 25-div & errors

Booth’s Example (2 x -3)

Operation Multiplicand Product next?

0. initial value 0010 0000 11010 10 -> sub

1a. P = P - m 1110 +1110 1110 11010shift P (sign ext)

1b. 0010 1111 01101 01 -> add + 0010

2a. 0001 01101shift P

2b. 0010 0000 10110 10 -> sub + 1110

3a. 0010 1110 10110 shift

3b. 0010 1111 0101 111 -> nop

4a 1111 0101 1 shift

4b. 0010 1111 10101 done

EECS 150, Fa04, Lec 25-div & errors

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