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Data representation and Arithmetic Algorithms

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Mumthas Shaikh
Mumthas Shaikh

Booth's algorithm is an efficient method for multiplying signed binary numbers. It treats positive and negative operands uniformly by recoding the multiplier and using addition and subtraction. The algorithm performs one addition or subtraction per bit of the multiplier by examining pairs of bits and selecting the shifted multiplicand or its negative. This minimizes operations compared to standard binary multiplication methods.

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Data representation and Arithmetic Algorithms
 Booth’s algorithm (signed Number Multiplication):  Treats both positive and negative operands uniformly. Algorithm : Let A—Accumalator Q – Multiplier register M- Multiplicand n iterations to be performed where n- no. of bits in the multiplier. Initialization : Initialize A to 0 Append a 0-bit in Q-1 position.
Do n times 1. Examine bits Q0Q-1 2. If Q0Q-1 =01 then a) Perform A=A+M b) RS(A.Q) If Q0Q-1 =10 then a) Perform A=A-M b) RS(A.Q) If Q0Q-1 =00/11 then a) RS(A.Q) After n iterations discard the LSB of Q register. Product-A.Q
Data representation and Arithmetic Algorithms
1. 5 x 2 (0101) (010) M A Q 0101 0000 0100 1. RS 0000 0010 2. A=A-M 1011 0010 RS 1101 1001 3. A=A+M 0010 1001 RS 0001 0100 -- Discard Product 0001 010 = (10)
1. 5 x 5 (0101) (0101) M A Q 0101 0000 01010 1. A=A-M 1011 01010 2. A=A+M 0010 10101 RS 1101 1001 3. A=A-M 1100 01010 RS 1110 00101 4. A=A+M 0011 00101 RS 0001 10010 -- Discard Product 00011001
3. -5 x 4 (1011) (0100) M A Q 1011 0000 01000 1. RS 0000 00100 2. RS 0000 00010 3. A=A-M 0101 00010 RS 0010 10001 4. A=A+M 1101 10001 RS 1110 11000 -- Discard Product 1110 1100
4. 8 x 10 (01000) (01010) M A Q 01010 00000 010000 1. RS 00000 001000 2. RS 00000 000100 3. RS 00000 000010 4. A=A-M 10110 000010 RS 11011 000001 5. A=A+M 00101 000001 RS 00010 100000-- Discard Product : 0001010000
 Booth’s algorithm is efficient when the multiplier Q contains a series of 1’s or 0’s since it minimizes on addition and subtraction operations.
 -1 times the shifted multiplicand is selected when moving from 0 to 1  +1 times the shifted multiplicand is selected when moving from 1 to 0 as the multiplier is scanned from right to left 1. +7 x +8 (00111) (01000) Multiplicand Multiplier +8---- 0 1 0 0 0 0 Implied zero +1 -1 0 0 0 (Recoded multiplier) 2’s complement of +7 ----- 11001
0 0 1 1 1 +1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x 0 0 0 0 0 0 0 0 x x 1 1 1 1 0 0 1 x x x 0 0 0 1 1 1 x x x x 1 0 0 0 0 1 1 1 0 0 0 (+56) Discard Product
2. +5 x +10 (00101) (01010) Multiplicand Multiplier +10---- 0 1 0 1 0 0 Implied zero +1 -1 +1 -1 0 (Recoded multiplier) 2’s complement of +5 ----- 11011
0 0 1 0 1 +1 -1 +1 -1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 1 1 x 0 0 0 0 0 1 0 1 x x 1 1 1 1 0 1 1 x x x 0 0 0 1 0 1 x x x x 11 0 0 0 0 1 1 0 0 1 0 (+50) Discard Product
3. +13 x -6 (01101) (11010) Multiplicand Multiplier -6--- 1 1 0 1 0 0 Implied zero 0 -1 +1 -1 0 (Recoded multiplier) 2’s complement of +13 ----- 10011
0 1 1 0 1 0 -1 +1 -1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 x 0 0 0 0 1 1 0 1 x x 1 1 1 0 0 1 1 x x x 0 0 0 0 0 0 x x x x 1 1 1 1 0 1 1 0 0 1 0 (-78) Discard Product
4. -5 x -5 (1011) (1011) Multiplicand Multiplier -5--- 1 0 1 1 0 Implied zero -1 +1 0 -1 (Recoded multiplier) 2’s complement of -5 ----- 0101
1 0 1 1 -1 +1 0 -1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 x 1 1 1 0 1 1 x x 0 0 1 0 1 x x x 1 0 0 0 1 1 0 0 1 (25) Discard Product
1011 Multiplicand (11) x 1101 Multiplier (13) 1011 Partial product(PP) 0000 PP 1011 PP 1011 PP 10001111 Product (143)  Note: if multiplier bit is 1 copy multiplicand otherwise zero  Note: need double length result
Unsigned Number multiplication Algorithm : M= 1011 Q= 1101 Let a variable x be initialized to 0000[4 bit binary number] 1. Since the bit q0 of multiplier is 1 x=x+ 1011= 0000 + 1011 = 1011 Right shift x ---- 0 1 0 1 | 1 2. Since the bit q1 of multiplier is 0 x= x+ 0000 = 0 1 0 1 | 1 Right shift x ---- 0 0 1 0 | 1 1 3. Since the bit q2 of multiplier is 1 x=x+ 1011= 0 0 1 0 | 1 1 1 0 1 1 1 1 0 1 | 1 1
Right shift x ---- 0 1 1 0 | 1 1 1 4. Since the bit q3 of multiplier is 1 x=x+ 1011= 0 1 1 0 | 1 1 1 1 0 1 1 10 0 0 1 | 1 1 1 Right shift x ---- 1 0 0 0 | 1 1 1 1 Product - 1 0 0 0 | 1 1 1 1
Unsigned number Multiplication 2. 10 x 5 1010 101 (M) (Q) X= 0000 1. Since q0 is 1 X= X + 1010 = 1010 RS ---- 0101 | 0 2. Since q0 is 0 X=X + 0000 = 0101 | 0 RS ---- 0010 | 10 3. Since q0 is 1 X = X + 1010 0010 | 10 1010 1100|10 RS ----- 0110 | 010 Product – 0110010 ( 50 )
Data representation and Arithmetic Algorithms
Eg of normal division : 100010010/1101 10101 1101 |100010010 -1101 10000 -1101 1110 - 1101 1
1 0 1 0 1010 | 1 1 0 1 0 0 1 - 1 1 0 1 0 1 1 0 - 1 0 1 0 1 1 0 1 + 1 0 1 0 1 1 0 0 - 1 0 1 0 0 1 0 1 - 1 0 1 0 1 0 1 1 + 1 0 1 0 0 1 0 1 ----- Remainder
 Positive divisor --- Reg. M  Positive dividend - Reg Q  Reg A--0  After division is complete Quotient ----- Q Remainder ----- A A 0 bit is added at the left end of both A and M to serve as a sign bit for subtraction. Algorithm : Do n times ( n-- no. of bits in Q) 1. Shift A and Q left one binary position. 2. Subtract M from A, placing the answer back in A (A=A-M) 3. If the sign of A is 1 set qo to 0 and add M back to A (restore A ) , otherwise set q0 to 1.
1. 9 / 5 1001 101 M A Q 0101 0000 1001 1. Sh 0001 001 Sub (A=A-M) 1100 0010 +0101 0001 2. Sh 0010 010 Sub (A=A-M) 1101 0100 +0101 0010
M A Q 0101 3. Sh 0100 100 Sub (A=A-M) 1111 1000 +0101 0100 4. Sh 1001 000 Sub (A=A-M) 0100 0001 Remainder Quotient
2. 12 / 10 1100 1010 M A Q 01010 00000 1100 1. Sh 00001 100 Sub (A=A-M) 10111 1000 Restore 00001 2. Sh 00011 000 Sub (A=A-M) 11001 0000 Restore 00011
M A Q 01010 3. Sh 00110 000 Sub (A=A-M) 11100 0000 Restore 00110 4. Sh 01100 000 Sub (A=A-M) 00010 0001 Remainder Quotient
3. 4 / 5 100 101 M A Q 0101 0000 100 1. Sh 0001 00 Sub (A=A-M) 1100 000 Restore 0001 2. Sh 0010 00 Sub (A=A-M) 1101 000 Restore 0010
M A Q 0101 3. Sh 0100 00 Sub (A=A-M) 1111 000 Restore 0100 Remainder Quotient
 If A is (+) we shift left and subtract M from A i.e 2A-M  If A is (-) 1. Restore A---- A+M 2. Left Shift 2(A+M) 3. Subtract M from A 2(A+M)-M= 2A+M Algorithm : 1. Do n times (n---no. of bits in Q) If the sign of A is 0, shift A and Q left one binary position and subtract M from A, otherwise shift A and Q left and add M to A. If the sign of A is 0 , set q0 to 1 else set q0 to 0. 2. After n iterations , if the sign of A is 1, add M to A
 Eg : 8/3 M A Q 011 000 1000 1. Sh001 000 Sub 110 0000 2. Sh 100 000 Add 111 0000 3. Sh 110 000 Add 001 0001 4. Sh 010 001 Sub 111 0010 111 Quotient +011 010---Remainder
 Eg : 15/3 M A Q 011 000 1111 1. Sh001 111 Sub 110 1110 2. Sh 101 110 Add 000 1101 3. Sh 001 101 Sub 110 1010 4. Sh 101 010 Add 000 0101 Quotient Remainder
2 ways of representing decimal nos : 1. Fixed point representation The usual way of representating numbers is to write the number with the decimal point fixed in it’s correct position between the 2 appropriate digits eg : 13.75 or 345.78 This is called fixed point representation. Drawback: This representation becomes cumbersome when dealing with several very large or very small numbers. eg: 0.00000001375 or 1375000000000
2. Floating point representation Three numbers associated with a floating point number are : a. Mantissa M b. Exponent E c. Base B These three numbers together represent the real number M x BE Eg : 1.0 x 1018 1.0- Mantissa 18- Exponent 10- Base Base B – constant, Therefore a floating point number is stored as (M,E)
Suppose that M and E are both 3-bit sign and magnitude integers and B=2  M and E each can assume the values +/-0 , +/-1, +/-2 and +/-3.  All binary words of the form (M,E) = (x00,xxx) represent zero where x denotes either 0/1.  The smallest non zero positive number is : (001,111)= 1 x 2-3 = 0.125 The smallest non zero negative number is : (101,111)= -1 x 2-3 = -0.125 The largest representable positive number is : (011,011)= 3 x 23 =24 The largest representable negative number is : (111,011)= -3 x 23 = -24 Note : The left most bit which is the sign of the mantissa is also the sign of the floating point number
1.0 x 1018 0.1 x 1019 1000000 x 1012 0.000001 x 1024 ----- All represent the same number  Floating point representation is redundant i.e. the same no can be represented in more than one way.  It is generally desirable to have a unique/normal form for each representable number in a floating point system.  The mantissa is said to be normalized if the digit to the right of the binary point is non zero. eg: 0.1bbbbbb x 2 +/-E Hence 0.1 x 1019 is the unique normal form of 1.0 x 1018
 Basically, there are two common signed number representations: 1. Sign and magnitude 000 +0 001 +1 010 +2 011 +3 100 -0 101 -1 110 -2 111 -3 Drawback : 2 representations for number 0 (+0 and -0)
2. Two's complement representation 000 +0 001 +1 010 +2 011 +3 100 -4 / +4 101 -3 110 -2 111 -1 Hence for a 3-bit number (+3 to -4) 0r (-3 to +4) 2n-1 -1 to -2n-1 -2n-1 +1 to 2n-1
An 8 bit register holds numbers in 2’s complement form with leftmost bit as the sign bit. 1. What is the largest (+) number that can be stored? Express the answer in decimal and binary format. 2. What is the largest (-) number that can be stored? Express the answer in decimal and binary format. 2n-1 -1 to -2n-1 127 to -128 01111111 10000000 Or -2n-1 +1 to 2n-1 -127 to +128 10000001 to 10000000
 An example of a floating point number in binary format (32 bit) Assumptions : 1. Implied base 2. 2. 7-bit signed exponent is expressed as a 2’s complement integer. 3. Left most bit represents sign of the number (0 means + ,1 means -) The 24 bit mantissa is considered to be a fraction with the binary point at it’s left end . To retain as many significant bits as possible, the fraction al mantissa is kept in a normalized form i.e it’s leftmost bit is always 1. 24-bit fractional mantissa7 –bit signed exponent S bit 32 bit
Eg: 0 0 0 0 1 0 0 1 001000000000000000000000 Unnormalized value + 0.001000000000000000000000 x 29 0 0 0 0 0 1 1 1 100000000000000000000000 + 0.100000000000000000000000 x 27 Normalized value Binary Normalization example
 Hence a 7-bit 2’s complement exponent has a range of -- 2n-1 -1 to -2n-1 2^6-1 to -2^6 +63 to -64  Some computers employ a different type of representation for exponents i.e. biased exponent representation  Here the n-bit exponent is expressed in excess 2(n-1) format.  Hence the 7-bit exponent is expressed in excess 64 format.  In this representation the sign bit is removed from being a separate entity.  A (+) no. called bias (64) is added to each exponent as the floating point number is formed, so that internally all exponents are positive.
 E’ ---new exponent  E’ = E + 64  Since -64 ≤ E ≤ +63  0 ≤ E’ ≤ +127 -64 is represented as ------- 0000000 -63 is represented as ------- 0000001 -62 is represented as ------- 0000010 ‘ ‘ ‘ ‘ 0 is represented as ------- 1000000 1 is represented as ------- 1000001 ‘ ‘ +63 is represented as ------- 1111111
 The IEEE Standard for Floating-Point Arithmetic (IEEE 754) is a technical standard established by the Institute of Electrical and Electronics Engineers (IEEE) and the most widely used standard for floating-point computation, followed by many CPU.  IEEE 754 defines 2 standards 1. Single precision no. (32 bits) 2. Double precision no. (64 bits)
 The standard for a 32 bit floating point number is ---- 1. Sign bit S 2. 8 bit exponents field E’ (in excess 127 representation) i.e the bias added here is 2n-1 -1 3. 23-bit mantissa field M 4. Base 2 SS E’E’ MM Sign of the number 0 ----- + 1 ----- - 8 bit signed exponent in excess 127 representation 23 bit fractional mantissa
 Since E varies from  -2n-1 +1 to 2n-1  i.e -127 ≤ E ≤ +128  E’ = E + 127  0 ≤ E’ ≤ +255 E E’ -127 00000000 represents the floating point no. zero -126 00000001 0 01111111 1 10000000 +128 11111111 represents ∞
Normalization in IEEE 754  In IEEE 754 format the mantissa is actually of the form 1.M where 1 is to the left of the binary point .  This I is implicit/hidden and is not stored along with the number but assumed to be there.  Use of hidden 1 means that the precision of a normalized number is effectively increased by 1 bit.  Eg :  Value represented --- M= 1.001010 E’= 40 E=E’-127 = 40 – 127 = -87 1.001010… X 2-87 Hence in IEEE 754 format the actual value of N N= (-1)S 2 E’-127 X (1.M) 00 0 0 1 0 1 0 0 0 00 0 1 0 1 0 0 0 0 001010………0001010………0
 -1023 ≤ E ≤ +1024  E’ = E + 1023  0 ≤ E’ ≤ +2047  Here the actual value of N is N= (-1)S 2 E’-1023 X(1.M) SS E’E’ MM Sign of the number 0 ----- + 1 ----- - 11 bit signed exponent in excess 1023 representation 52bit fractional mantissa
Convert the following nos. in IEEE- 754 single precision format. 1. 10.58 (1010.100101)2 1.010100101 x 23 E’=127+3= 130=10000010 S=0 M=010100101 00 1000001010000010 0 1 0 1 0 0 1 0 10 1 0 1 0 0 1 0 1
2. -8.08 = -(1000.000101) -(1.000000101 x 23 )2 E’ =127 + 3 =130 S=1 M= 000000101 3. 1.5 (1.1)2 1.1 x 20 E’= 0 +127 00 0111111101111111 10…..010…..0 11 1000001010000010 0 0 0 0 0 0 1 0 1……..0 0 0 0 0 0 1 0 1……..
4. 21 21 = (10101)2 = 1.0101 x 24 E’ = 127 + 4 =131 M= 0101.... S=0 00 1000001110000011 0 1 0 1 0………00 1 0 1 0………0
5. 0.021= 0.00000011001 = 1.1001 x 2 -7 E’= 127-7= 120 M= 1001 S= 0
 Give the value represented by the following IEEE single precision no. E’= 10000001 = 129 E= E’ -127 = 129-127 = 2 (1.01)2 = (1.25)10 X= -1.25 x 22 = -5 11 1000000110000001 0 1 0………00 1 0………0
Convert the following no. in IEEE- 754 double precision format. (0.0625)10 = 0.0625 x 2 =0.125 ---- 0 0.125 x 2 = 0.25 ---- 0 0.25 x 2 = 0.5 ---- 0 0.5 x 2 =1 ---- 1 (0.0625)10 = (0.0001)2 = 1.0 x 2 -4 E’ = E + 1023 = (1019)10 M = 00000…… S =0
 X= 0.3 x 10 2  Y=0.2 x 10 3  X+Y = ( 0.03 x 103 ) + (0.2 x 103 )= 0.23 x 103  X-Y= ( 0.03 x 103 ) - (0.2 x 103 )= -0.17 x 103
 Floating point arithmetic with binary numbers: 1) 1.10101 x 24 + 1.00101 x 26 (0.0110101 x 26 +1.00101 x 26 ) = 1.1001001 x 26 2) 1.10011 x 24 + 1.00101 x 24 = 10.11000 x 24 (significand/mantissa overflow) Normalize the result 1.011000 x 25 3) 1.00110x 2254 + 1.10001 x 2254 10.10111 x 2254 1.010111 x 2255 exponent =255 signifies exponent overflow
4 ) 1.11011 x 26 – 1.11001 x 26 = 0.00011 x 26 Normalize the result – 1.1 x 22 When decrementing the exponent check for exponent underflow 5) 1.00111 x 210 + 1.10111 x 235 0.000000000000000000000000100111 x 235 + 1.10111 x 235 =1.101110000000000000000000100111x 235 (significand underflow) Result= 1.10111 x 235
 Floating-point subtraction example: E=10001010; S=1.11100000000000000000000 - E=10001010; S=1.11000000000000000000000 E=10001010; S=0.00100000000000000000000 Difference E=10000111; S=1.00000000000000000000000 Normalization
Floating point arithmetic (+/-)  Check for zeros  Align significands (adjusting exponents)  Add or subtract significands  Normalize result
Flowchart for floating point addition and subtraction
 Check for zero  Add/subtract exponents  Multiply/divide significands (watch sign)  Normalize  Round  All intermediate results should be in double length storage
 Eg: Multiply  21.44  by 7.24  Convert into IEEE 754 format  21.44 = 10101.0111 x 2^0 = 1.01010111 x 2^4 M= 01010111 E’= 131  7.24=111.0011 x 2^0 =1.110011 x 2^2 M= 110011 E’= 129
 For multiplication- Add the exponents and multiply the corresponding mantissas Step 1 : Add exponents  129+131= 260  When the exponents are added the resulting exponents is doubly biased.  Hence correct the resulting sum by subtracting one bias from the resulting exponent.  260-127=133.
Step 2 : Multiply the mantissa.  Considering only the first 4 digits of the mantissas  1.010 x 1.110 = 10.001100 Step 3 : Normalize the mantissa  1.0001100 x 2 ^ 134
 Points to Note : 1. Exponent overflow Xe = 120 Ye = 122 X’e= 120 + 127 = 247 Y’e= 122+127=249 X’e + Y’e =496 Subtract a bias : 496-127 = 369 Range of E’ from 0 to 255 Hence there is an exponent overflow
2. Exponent underflow Xe = -60 Ye = -74 X’e= -60 + 127 = 67 Y’e= -74 +127=53 X’e + Y’e =120 Subtract a bias 120-127 = -7 Range of E’ from 0 to 255 Hence there is an exponent underflow
Floating Point Multiplication
 For division :  X=Xm x 2^Xe  Y=Ym x 2^Ye Z=(Xm/Ym) x 2^(Xe-Ye) Note when the divisor exponent (Xe+127) is subtracted from dividend exponent (Ye+127), it removes the bias which must be added back in i.e ((Xe+127) -(Ye+127)) = Xe-Ye Hence correction-- (Xe-Ye) +127 Case 1 : Assume Xe= 120 , Ye= -2 Xe’= 147 Ye’= +125 Xe’-Ye’=272 Correction : 272 + 127 =399 ------- Exponent overflow
Floating Point Division

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Data representation and Arithmetic Algorithms

  • 2.  Booth’s algorithm (signed Number Multiplication):  Treats both positive and negative operands uniformly. Algorithm : Let A—Accumalator Q – Multiplier register M- Multiplicand n iterations to be performed where n- no. of bits in the multiplier. Initialization : Initialize A to 0 Append a 0-bit in Q-1 position.
  • 3. Do n times 1. Examine bits Q0Q-1 2. If Q0Q-1 =01 then a) Perform A=A+M b) RS(A.Q) If Q0Q-1 =10 then a) Perform A=A-M b) RS(A.Q) If Q0Q-1 =00/11 then a) RS(A.Q) After n iterations discard the LSB of Q register. Product-A.Q
  • 5. 1. 5 x 2 (0101) (010) M A Q 0101 0000 0100 1. RS 0000 0010 2. A=A-M 1011 0010 RS 1101 1001 3. A=A+M 0010 1001 RS 0001 0100 -- Discard Product 0001 010 = (10)
  • 6. 1. 5 x 5 (0101) (0101) M A Q 0101 0000 01010 1. A=A-M 1011 01010 2. A=A+M 0010 10101 RS 1101 1001 3. A=A-M 1100 01010 RS 1110 00101 4. A=A+M 0011 00101 RS 0001 10010 -- Discard Product 00011001
  • 7. 3. -5 x 4 (1011) (0100) M A Q 1011 0000 01000 1. RS 0000 00100 2. RS 0000 00010 3. A=A-M 0101 00010 RS 0010 10001 4. A=A+M 1101 10001 RS 1110 11000 -- Discard Product 1110 1100
  • 8. 4. 8 x 10 (01000) (01010) M A Q 01010 00000 010000 1. RS 00000 001000 2. RS 00000 000100 3. RS 00000 000010 4. A=A-M 10110 000010 RS 11011 000001 5. A=A+M 00101 000001 RS 00010 100000-- Discard Product : 0001010000
  • 9.  Booth’s algorithm is efficient when the multiplier Q contains a series of 1’s or 0’s since it minimizes on addition and subtraction operations.
  • 10.  -1 times the shifted multiplicand is selected when moving from 0 to 1  +1 times the shifted multiplicand is selected when moving from 1 to 0 as the multiplier is scanned from right to left 1. +7 x +8 (00111) (01000) Multiplicand Multiplier +8---- 0 1 0 0 0 0 Implied zero +1 -1 0 0 0 (Recoded multiplier) 2’s complement of +7 ----- 11001
  • 11. 0 0 1 1 1 +1 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 x 0 0 0 0 0 0 0 0 x x 1 1 1 1 0 0 1 x x x 0 0 0 1 1 1 x x x x 1 0 0 0 0 1 1 1 0 0 0 (+56) Discard Product
  • 12. 2. +5 x +10 (00101) (01010) Multiplicand Multiplier +10---- 0 1 0 1 0 0 Implied zero +1 -1 +1 -1 0 (Recoded multiplier) 2’s complement of +5 ----- 11011
  • 13. 0 0 1 0 1 +1 -1 +1 -1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 0 1 1 x 0 0 0 0 0 1 0 1 x x 1 1 1 1 0 1 1 x x x 0 0 0 1 0 1 x x x x 11 0 0 0 0 1 1 0 0 1 0 (+50) Discard Product
  • 14. 3. +13 x -6 (01101) (11010) Multiplicand Multiplier -6--- 1 1 0 1 0 0 Implied zero 0 -1 +1 -1 0 (Recoded multiplier) 2’s complement of +13 ----- 10011
  • 15. 0 1 1 0 1 0 -1 +1 -1 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 1 1 x 0 0 0 0 1 1 0 1 x x 1 1 1 0 0 1 1 x x x 0 0 0 0 0 0 x x x x 1 1 1 1 0 1 1 0 0 1 0 (-78) Discard Product
  • 16. 4. -5 x -5 (1011) (1011) Multiplicand Multiplier -5--- 1 0 1 1 0 Implied zero -1 +1 0 -1 (Recoded multiplier) 2’s complement of -5 ----- 0101
  • 17. 1 0 1 1 -1 +1 0 -1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 x 1 1 1 0 1 1 x x 0 0 1 0 1 x x x 1 0 0 0 1 1 0 0 1 (25) Discard Product
  • 18. 1011 Multiplicand (11) x 1101 Multiplier (13) 1011 Partial product(PP) 0000 PP 1011 PP 1011 PP 10001111 Product (143)  Note: if multiplier bit is 1 copy multiplicand otherwise zero  Note: need double length result
  • 19. Unsigned Number multiplication Algorithm : M= 1011 Q= 1101 Let a variable x be initialized to 0000[4 bit binary number] 1. Since the bit q0 of multiplier is 1 x=x+ 1011= 0000 + 1011 = 1011 Right shift x ---- 0 1 0 1 | 1 2. Since the bit q1 of multiplier is 0 x= x+ 0000 = 0 1 0 1 | 1 Right shift x ---- 0 0 1 0 | 1 1 3. Since the bit q2 of multiplier is 1 x=x+ 1011= 0 0 1 0 | 1 1 1 0 1 1 1 1 0 1 | 1 1
  • 20. Right shift x ---- 0 1 1 0 | 1 1 1 4. Since the bit q3 of multiplier is 1 x=x+ 1011= 0 1 1 0 | 1 1 1 1 0 1 1 10 0 0 1 | 1 1 1 Right shift x ---- 1 0 0 0 | 1 1 1 1 Product - 1 0 0 0 | 1 1 1 1
  • 21. Unsigned number Multiplication 2. 10 x 5 1010 101 (M) (Q) X= 0000 1. Since q0 is 1 X= X + 1010 = 1010 RS ---- 0101 | 0 2. Since q0 is 0 X=X + 0000 = 0101 | 0 RS ---- 0010 | 10 3. Since q0 is 1 X = X + 1010 0010 | 10 1010 1100|10 RS ----- 0110 | 010 Product – 0110010 ( 50 )
  • 23. Eg of normal division : 100010010/1101 10101 1101 |100010010 -1101 10000 -1101 1110 - 1101 1
  • 24. 1 0 1 0 1010 | 1 1 0 1 0 0 1 - 1 1 0 1 0 1 1 0 - 1 0 1 0 1 1 0 1 + 1 0 1 0 1 1 0 0 - 1 0 1 0 0 1 0 1 - 1 0 1 0 1 0 1 1 + 1 0 1 0 0 1 0 1 ----- Remainder
  • 25.  Positive divisor --- Reg. M  Positive dividend - Reg Q  Reg A--0  After division is complete Quotient ----- Q Remainder ----- A A 0 bit is added at the left end of both A and M to serve as a sign bit for subtraction. Algorithm : Do n times ( n-- no. of bits in Q) 1. Shift A and Q left one binary position. 2. Subtract M from A, placing the answer back in A (A=A-M) 3. If the sign of A is 1 set qo to 0 and add M back to A (restore A ) , otherwise set q0 to 1.
  • 26. 1. 9 / 5 1001 101 M A Q 0101 0000 1001 1. Sh 0001 001 Sub (A=A-M) 1100 0010 +0101 0001 2. Sh 0010 010 Sub (A=A-M) 1101 0100 +0101 0010
  • 27. M A Q 0101 3. Sh 0100 100 Sub (A=A-M) 1111 1000 +0101 0100 4. Sh 1001 000 Sub (A=A-M) 0100 0001 Remainder Quotient
  • 28. 2. 12 / 10 1100 1010 M A Q 01010 00000 1100 1. Sh 00001 100 Sub (A=A-M) 10111 1000 Restore 00001 2. Sh 00011 000 Sub (A=A-M) 11001 0000 Restore 00011
  • 29. M A Q 01010 3. Sh 00110 000 Sub (A=A-M) 11100 0000 Restore 00110 4. Sh 01100 000 Sub (A=A-M) 00010 0001 Remainder Quotient
  • 30. 3. 4 / 5 100 101 M A Q 0101 0000 100 1. Sh 0001 00 Sub (A=A-M) 1100 000 Restore 0001 2. Sh 0010 00 Sub (A=A-M) 1101 000 Restore 0010
  • 31. M A Q 0101 3. Sh 0100 00 Sub (A=A-M) 1111 000 Restore 0100 Remainder Quotient
  • 32.  If A is (+) we shift left and subtract M from A i.e 2A-M  If A is (-) 1. Restore A---- A+M 2. Left Shift 2(A+M) 3. Subtract M from A 2(A+M)-M= 2A+M Algorithm : 1. Do n times (n---no. of bits in Q) If the sign of A is 0, shift A and Q left one binary position and subtract M from A, otherwise shift A and Q left and add M to A. If the sign of A is 0 , set q0 to 1 else set q0 to 0. 2. After n iterations , if the sign of A is 1, add M to A
  • 33.  Eg : 8/3 M A Q 011 000 1000 1. Sh001 000 Sub 110 0000 2. Sh 100 000 Add 111 0000 3. Sh 110 000 Add 001 0001 4. Sh 010 001 Sub 111 0010 111 Quotient +011 010---Remainder
  • 34.  Eg : 15/3 M A Q 011 000 1111 1. Sh001 111 Sub 110 1110 2. Sh 101 110 Add 000 1101 3. Sh 001 101 Sub 110 1010 4. Sh 101 010 Add 000 0101 Quotient Remainder
  • 35. 2 ways of representing decimal nos : 1. Fixed point representation The usual way of representating numbers is to write the number with the decimal point fixed in it’s correct position between the 2 appropriate digits eg : 13.75 or 345.78 This is called fixed point representation. Drawback: This representation becomes cumbersome when dealing with several very large or very small numbers. eg: 0.00000001375 or 1375000000000
  • 36. 2. Floating point representation Three numbers associated with a floating point number are : a. Mantissa M b. Exponent E c. Base B These three numbers together represent the real number M x BE Eg : 1.0 x 1018 1.0- Mantissa 18- Exponent 10- Base Base B – constant, Therefore a floating point number is stored as (M,E)
  • 37. Suppose that M and E are both 3-bit sign and magnitude integers and B=2  M and E each can assume the values +/-0 , +/-1, +/-2 and +/-3.  All binary words of the form (M,E) = (x00,xxx) represent zero where x denotes either 0/1.  The smallest non zero positive number is : (001,111)= 1 x 2-3 = 0.125 The smallest non zero negative number is : (101,111)= -1 x 2-3 = -0.125 The largest representable positive number is : (011,011)= 3 x 23 =24 The largest representable negative number is : (111,011)= -3 x 23 = -24 Note : The left most bit which is the sign of the mantissa is also the sign of the floating point number
  • 38. 1.0 x 1018 0.1 x 1019 1000000 x 1012 0.000001 x 1024 ----- All represent the same number  Floating point representation is redundant i.e. the same no can be represented in more than one way.  It is generally desirable to have a unique/normal form for each representable number in a floating point system.  The mantissa is said to be normalized if the digit to the right of the binary point is non zero. eg: 0.1bbbbbb x 2 +/-E Hence 0.1 x 1019 is the unique normal form of 1.0 x 1018
  • 39.  Basically, there are two common signed number representations: 1. Sign and magnitude 000 +0 001 +1 010 +2 011 +3 100 -0 101 -1 110 -2 111 -3 Drawback : 2 representations for number 0 (+0 and -0)
  • 40. 2. Two's complement representation 000 +0 001 +1 010 +2 011 +3 100 -4 / +4 101 -3 110 -2 111 -1 Hence for a 3-bit number (+3 to -4) 0r (-3 to +4) 2n-1 -1 to -2n-1 -2n-1 +1 to 2n-1
  • 41. An 8 bit register holds numbers in 2’s complement form with leftmost bit as the sign bit. 1. What is the largest (+) number that can be stored? Express the answer in decimal and binary format. 2. What is the largest (-) number that can be stored? Express the answer in decimal and binary format. 2n-1 -1 to -2n-1 127 to -128 01111111 10000000 Or -2n-1 +1 to 2n-1 -127 to +128 10000001 to 10000000
  • 42.  An example of a floating point number in binary format (32 bit) Assumptions : 1. Implied base 2. 2. 7-bit signed exponent is expressed as a 2’s complement integer. 3. Left most bit represents sign of the number (0 means + ,1 means -) The 24 bit mantissa is considered to be a fraction with the binary point at it’s left end . To retain as many significant bits as possible, the fraction al mantissa is kept in a normalized form i.e it’s leftmost bit is always 1. 24-bit fractional mantissa7 –bit signed exponent S bit 32 bit
  • 43. Eg: 0 0 0 0 1 0 0 1 001000000000000000000000 Unnormalized value + 0.001000000000000000000000 x 29 0 0 0 0 0 1 1 1 100000000000000000000000 + 0.100000000000000000000000 x 27 Normalized value Binary Normalization example
  • 44.  Hence a 7-bit 2’s complement exponent has a range of -- 2n-1 -1 to -2n-1 2^6-1 to -2^6 +63 to -64  Some computers employ a different type of representation for exponents i.e. biased exponent representation  Here the n-bit exponent is expressed in excess 2(n-1) format.  Hence the 7-bit exponent is expressed in excess 64 format.  In this representation the sign bit is removed from being a separate entity.  A (+) no. called bias (64) is added to each exponent as the floating point number is formed, so that internally all exponents are positive.
  • 45.  E’ ---new exponent  E’ = E + 64  Since -64 ≤ E ≤ +63  0 ≤ E’ ≤ +127 -64 is represented as ------- 0000000 -63 is represented as ------- 0000001 -62 is represented as ------- 0000010 ‘ ‘ ‘ ‘ 0 is represented as ------- 1000000 1 is represented as ------- 1000001 ‘ ‘ +63 is represented as ------- 1111111
  • 46.  The IEEE Standard for Floating-Point Arithmetic (IEEE 754) is a technical standard established by the Institute of Electrical and Electronics Engineers (IEEE) and the most widely used standard for floating-point computation, followed by many CPU.  IEEE 754 defines 2 standards 1. Single precision no. (32 bits) 2. Double precision no. (64 bits)
  • 47.  The standard for a 32 bit floating point number is ---- 1. Sign bit S 2. 8 bit exponents field E’ (in excess 127 representation) i.e the bias added here is 2n-1 -1 3. 23-bit mantissa field M 4. Base 2 SS E’E’ MM Sign of the number 0 ----- + 1 ----- - 8 bit signed exponent in excess 127 representation 23 bit fractional mantissa
  • 48.  Since E varies from  -2n-1 +1 to 2n-1  i.e -127 ≤ E ≤ +128  E’ = E + 127  0 ≤ E’ ≤ +255 E E’ -127 00000000 represents the floating point no. zero -126 00000001 0 01111111 1 10000000 +128 11111111 represents ∞
  • 49. Normalization in IEEE 754  In IEEE 754 format the mantissa is actually of the form 1.M where 1 is to the left of the binary point .  This I is implicit/hidden and is not stored along with the number but assumed to be there.  Use of hidden 1 means that the precision of a normalized number is effectively increased by 1 bit.  Eg :  Value represented --- M= 1.001010 E’= 40 E=E’-127 = 40 – 127 = -87 1.001010… X 2-87 Hence in IEEE 754 format the actual value of N N= (-1)S 2 E’-127 X (1.M) 00 0 0 1 0 1 0 0 0 00 0 1 0 1 0 0 0 0 001010………0001010………0
  • 50.  -1023 ≤ E ≤ +1024  E’ = E + 1023  0 ≤ E’ ≤ +2047  Here the actual value of N is N= (-1)S 2 E’-1023 X(1.M) SS E’E’ MM Sign of the number 0 ----- + 1 ----- - 11 bit signed exponent in excess 1023 representation 52bit fractional mantissa
  • 51. Convert the following nos. in IEEE- 754 single precision format. 1. 10.58 (1010.100101)2 1.010100101 x 23 E’=127+3= 130=10000010 S=0 M=010100101 00 1000001010000010 0 1 0 1 0 0 1 0 10 1 0 1 0 0 1 0 1
  • 52. 2. -8.08 = -(1000.000101) -(1.000000101 x 23 )2 E’ =127 + 3 =130 S=1 M= 000000101 3. 1.5 (1.1)2 1.1 x 20 E’= 0 +127 00 0111111101111111 10…..010…..0 11 1000001010000010 0 0 0 0 0 0 1 0 1……..0 0 0 0 0 0 1 0 1……..
  • 53. 4. 21 21 = (10101)2 = 1.0101 x 24 E’ = 127 + 4 =131 M= 0101.... S=0 00 1000001110000011 0 1 0 1 0………00 1 0 1 0………0
  • 54. 5. 0.021= 0.00000011001 = 1.1001 x 2 -7 E’= 127-7= 120 M= 1001 S= 0
  • 55.  Give the value represented by the following IEEE single precision no. E’= 10000001 = 129 E= E’ -127 = 129-127 = 2 (1.01)2 = (1.25)10 X= -1.25 x 22 = -5 11 1000000110000001 0 1 0………00 1 0………0
  • 56. Convert the following no. in IEEE- 754 double precision format. (0.0625)10 = 0.0625 x 2 =0.125 ---- 0 0.125 x 2 = 0.25 ---- 0 0.25 x 2 = 0.5 ---- 0 0.5 x 2 =1 ---- 1 (0.0625)10 = (0.0001)2 = 1.0 x 2 -4 E’ = E + 1023 = (1019)10 M = 00000…… S =0
  • 57.  X= 0.3 x 10 2  Y=0.2 x 10 3  X+Y = ( 0.03 x 103 ) + (0.2 x 103 )= 0.23 x 103  X-Y= ( 0.03 x 103 ) - (0.2 x 103 )= -0.17 x 103
  • 58.  Floating point arithmetic with binary numbers: 1) 1.10101 x 24 + 1.00101 x 26 (0.0110101 x 26 +1.00101 x 26 ) = 1.1001001 x 26 2) 1.10011 x 24 + 1.00101 x 24 = 10.11000 x 24 (significand/mantissa overflow) Normalize the result 1.011000 x 25 3) 1.00110x 2254 + 1.10001 x 2254 10.10111 x 2254 1.010111 x 2255 exponent =255 signifies exponent overflow
  • 59. 4 ) 1.11011 x 26 – 1.11001 x 26 = 0.00011 x 26 Normalize the result – 1.1 x 22 When decrementing the exponent check for exponent underflow 5) 1.00111 x 210 + 1.10111 x 235 0.000000000000000000000000100111 x 235 + 1.10111 x 235 =1.101110000000000000000000100111x 235 (significand underflow) Result= 1.10111 x 235
  • 60.  Floating-point subtraction example: E=10001010; S=1.11100000000000000000000 - E=10001010; S=1.11000000000000000000000 E=10001010; S=0.00100000000000000000000 Difference E=10000111; S=1.00000000000000000000000 Normalization
  • 61. Floating point arithmetic (+/-)  Check for zeros  Align significands (adjusting exponents)  Add or subtract significands  Normalize result
  • 62. Flowchart for floating point addition and subtraction
  • 63.  Check for zero  Add/subtract exponents  Multiply/divide significands (watch sign)  Normalize  Round  All intermediate results should be in double length storage
  • 64.  Eg: Multiply  21.44  by 7.24  Convert into IEEE 754 format  21.44 = 10101.0111 x 2^0 = 1.01010111 x 2^4 M= 01010111 E’= 131  7.24=111.0011 x 2^0 =1.110011 x 2^2 M= 110011 E’= 129
  • 65.  For multiplication- Add the exponents and multiply the corresponding mantissas Step 1 : Add exponents  129+131= 260  When the exponents are added the resulting exponents is doubly biased.  Hence correct the resulting sum by subtracting one bias from the resulting exponent.  260-127=133.
  • 66. Step 2 : Multiply the mantissa.  Considering only the first 4 digits of the mantissas  1.010 x 1.110 = 10.001100 Step 3 : Normalize the mantissa  1.0001100 x 2 ^ 134
  • 67.  Points to Note : 1. Exponent overflow Xe = 120 Ye = 122 X’e= 120 + 127 = 247 Y’e= 122+127=249 X’e + Y’e =496 Subtract a bias : 496-127 = 369 Range of E’ from 0 to 255 Hence there is an exponent overflow
  • 68. 2. Exponent underflow Xe = -60 Ye = -74 X’e= -60 + 127 = 67 Y’e= -74 +127=53 X’e + Y’e =120 Subtract a bias 120-127 = -7 Range of E’ from 0 to 255 Hence there is an exponent underflow
  • 70.  For division :  X=Xm x 2^Xe  Y=Ym x 2^Ye Z=(Xm/Ym) x 2^(Xe-Ye) Note when the divisor exponent (Xe+127) is subtracted from dividend exponent (Ye+127), it removes the bias which must be added back in i.e ((Xe+127) -(Ye+127)) = Xe-Ye Hence correction-- (Xe-Ye) +127 Case 1 : Assume Xe= 120 , Ye= -2 Xe’= 147 Ye’= +125 Xe’-Ye’=272 Correction : 272 + 127 =399 ------- Exponent overflow