William Sandqvist [email protected] It could be pedagogical to study a small 4 bit nibble computer in order to understand how real numbers are stored and

William Sandqvist [email protected]

It could be pedagogical to study a small 4 bit nibble computer in order to understand how real numbers are stored and handled!

Normaly 32 bit registers are used, but the principles from the nibble computer will remain the same.


The first microprocessor

Intel 4004 1971 was a 4 bit (nibble) processor. The customer Busicom wanted a dedicated calculator chip, but the disobedient engineers created a general purpose processor.


4.3 Imagine a ”Nibble” computer

• Computer registers are ”rings”.

A four bit register can hold 24 = 16 numbers.

Either 8 positive (+0…+7) and 8 negative (-1…-8) ”signed numbers”, or 16 (0…F) ”unsigned numbers”.

If a result is bigger than 16, the register overflows. (In some processors indicated with a Carry-flag).


Real numbers in Nibble computer

)2().1()1(][ 100123

123 bbb bbbbb

• Imagine a ”Nibble” floating point format!

• Imagine a ”Nibble” fixed point format, Q3!

125.025.05.0

2222][

0123

30

21

12

030123

bbbb

bbbbbbbb


Q3 [b3b2b1b0]

Fixed point format

The Q-format is the same as the signed numbers but downscaled:

-1 < Q3 < 1-b0

The step between numbers is the weight of b0 0.125

Fixed point circle!Fixed point number line!


Floating point (Nibble)Floating point format

Sign b3

Significand 1.b0

1.02 = 1.010

1.12 = 1.510

Exponent b2b1-1 (exess1)

200-1 = 0.5201-1 = 1210-1 = 2211-1 = 4

+ 1.0 0.5 = +0.5+ 1.0 1 = +1+ 1.0 2 = +2+ 1.0 4 = +4

+ 1.5 0.5 = +0.75+ 1.5 1 = +1.5+ 1.5 2 = +3+ 1.5 4 = +6

- 1.0 0.5 = -0.5- 1.0 1 = -1- 1.0 2 = -2- 1.0 4 = -4

- 1.5 0.5 = -0.75- 1.5 1 = -1.5- 1.5 2 = -3- 1.5 4 = -6

Floating point number line!


Quantisation error

Maximum Quantisation Error MQE:

Fixed point: (2-3)/2 = 2-4 = 0.125/2 = 0.0625

Float: (6-4)/2 = 1


The number Zero

In fixed point formats the number zero is represented as [0000]

In floating point format, there is no representation for zero!

We could use, 0000 as +0 and 1000 as –0, as in the IEEE standard format!

The IEEE 32 and 64 bit float standard

Floating point circle!

http://en.wikipedia.org/wiki/IEEE_754


Converting Real numbers to Binary

Hawe you forgotten how?

http://www.everything2.net/title/Converting+real+numbers+to+binary


4.4 Six bit float)2().1()1(][ 1

012012345345 bbb bbbbbbbbb

Significand 1.b2b1b0

1.0002 = 1.00010

1.0012 = 1.12510

1.0102 = 1.2510

1.0112 = 1.37510 1.1002 = 1.510 1.1012 = 1.62510 1.1102 = 1.7510

1.1112 = 1.87510

Exponent b4b3-1 (exess1)00-1 = -1 200-1 = 0.501-1 = 0 201-1 = 110-1 = 1 210-1 = 211-1 = 2 211-1 = 4

• 0.2510 = 0.012 = 1.02 2-2 exponent -2 not representable

• 0.812510 = 0.11012 = 1.10122-1 representable

• -1,37510 = -1.0112 = -1.0112 20 representable

• 4.2510 = 100.012 = 1.00012 22 significand 1.0001 not representable

• 7.510 = 111.12 = 1.1112 22 representable


Float AdditionAlgorithm:1. Check for zeroes2. Align significands3. Add/Sub significands4. Normalize result

a = 0011112 0 01 111 + 01 1.111b = 0100102 0 10 010 + 10 1.010

Align significands. 10 > 01 shift smaller number a to right to get same exponent:Significand 0.1111 exponent 10

Add significands: Normalize result:

10.0011 exp 10 1.00011 exp 11Rounding: 1.00011 ~ 1.001

Result: 0 11 001

a = 1.875 b = 2.5 a+b = 4.5 (4.375)


Float MultiplicationSimpler than addition!

Algorithm: 1. Check for zeroes2. Add exponents and subtract Bias3. Multiply significands4. Normalize

Exponents: 01 10 Bias = 1exp = 01+10 -1 = 10

Multiply significands

Normalize result:

10.010110 exp 10 1.0010110 exp 11

Rounding: 1.0010110 ~ 1.001

Result: 0 11 001

a = 1.875 b = 2.5 a*b = 4.5 (4.6875)


IEEE – 32 bit float

Dec IEEE-754

By using the exess–127 format for the exponent, we can compare floats as if they were sign and magnitude integers!

http://babbage.cs.qc.edu/IEEE-754/Decimal.html





5-4 Cache conflict misses

for (i = 0; i < 50; i++) for(j = 0; j < 4; j++) x[i][j] = a[i][j] * c[i];

Data cache 1 k Word (Word = 32 bit). Cache line 4 Words.

Conflict misses depending on memory placement of x and a?


Memory – Cache adressmapping

Memorys are often Byte-organized, but we could draw it as if it was organized in Memory-lines with the same size as the Cache-line.

Direct addressmapping:

Memory-line: i Cache-line: j = i % K

If the CPU needs a data-item in a line, the whole line is transfered fom memory to cache.


a and x always conflict miss …

Conflict miss if the distans between the startaddress in memory of a and of x is a multiple of 1 k Words, 4096 Bytes.

for(j = 0; j < 4; j++) x[i][j] = a[i][j] * c[i];


a and x conflict miss 1:4 …for(j = 0; j < 4; j++) x[i][j] = a[i][j] * c[i];

Conflict miss 1:4 if the distans between the startaddress in memory of a and of x is a multiple of 1 k Words 3 Words, 4096 Bytes 12 Bytes.


a and x no conflict miss …for(j = 0; j < 4; j++) x[i][j] = a[i][j] * c[i];

No conflict miss if the distans between the startaddresses in memory of a and of x is not a multiple of 1 k Words, and not a multiple of 1 k Words 3, 2, 1 Words!



8-1 Arbitration3 Masters and 4 Slaves are connected to a classical bus.

Read access pattern:

M1 M2 M3

Rd S1 Rd S2 Rd S4

Rd S1 Rd S3 Rd S3

Rd S2

Sequential execution, 7 cycles needed.

1. M1 Rd S12. M2 Rd S23. M3 Rd S44. M2 Rd S35. M3 Rd S36. M2 Rd S27. M1 Rd S1

Many other alternatives are possible!


Slave side arbitrationWeighted round robin slave-side arbitration.

S1 S2 S3 S4

M1 1 - - -

M2 - 1 2 -

M3 - - 1 1

Best case execution time (3 cycles)

1. M1 Rd S1 ; M2 Rd S2 ; M3 Rd S4

2. M1 Rd S1 ; M2 Rd S3 ;

3. ; M2 Rd S2 ; M3 Rd S3

Worst case execution time (4 cycles)

1. M1 Rd S1 ; M2 Rd S2 ; M3 Rd S4

2. M1 Rd S1 ; ; M3 Rd S3

3. ; M2 Rd S3 ;

4. ; M2 Rd S2 ;

Connections and weights:

M1 M2 M3

Rd S1 Rd S2 Rd S4

Rd S1 Rd S3 Rd S3

Rd S2

Documents

William Sandqvist [email protected] It could be pedagogical to study a small 4 bit nibble computer in order to understand how real numbers are stored and