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Floating Point
(Multiplication)
Disusun Oleh:
Tri Fena Rohawati (22.05.3770)
Definisi
• Floating point adalah sebuah format
bilangan yang dapat digunakan untuk
merepresentasikan sebuah nilai yang
sangat besar atau sangat kecil.
Representation of Floating Point Numbers in
Single Precission
IEEE 754 standard
Value = N = (-1)S X 2 E-127 X (1.M)
S
E
M
Representation (cont..)
Di mana : S = sign bit (0->positif, 1->negatif)
E = Exponent
M = Mantissa
Floating point Conversion Example
Bilangan desimal 0.7510 akan dikonversikan dalam bentuk IEEE 754
32 bit format….
(0.75)10
0 = 0 (converted into biner)
.75 =
Calculation
Integer
.75 * 2
1
.5 * 2
1
Fraction
.5
0
Jadi, 0.75 = 11
(0.75)10 = 0.11
0.11=1.1 * 2-1 (normalized a binary number)
Floating point Conversion Example (cont…)
1.1 * 2-1 (normalized a binary number)
hidden
The mantissa is positive so the sign S is given by:
S=0
The biased exponent E is given by E = e + 127
E = -1 + 127 = 12610 =
011111102
Fractional part of mantissa M:
M = .10000000000000000000000
(in 23 bits)
The IEEE 754 single precision representation is given by:
0 01111110 10000000000000000000000
S
1 bits
E
8 bits
M
23 bits
Simplified Floating Point
Multiplication Flowchart
Floating Point Multiplication Example
Multiply the following two numbers represented in the IEEE 754 single
precision format: X = -1810 represented as:
1 10000011
00100000000000000000000
and Y = 9.510 represented as:
0 100000100 00110000000000000000000
(1) Value of one or both operands = 0? No, continue with step 2
(2) Compute the sign: S = Xs XOR Ys = 1 XOR 0 = 1
(3) Multiply the mantissas: The product of the 24 bit mantissas is 48 bits with
two bits to the left of the binary point:
(01).0101011000000….000000
Truncate to 24 bits:
hidden ® (1).01010110000000000000000
(4) Compute exponent of result:
Xe + Ye - 12710 = 1000 0011 + 1000 0010 - 0111111 = 1000 0110
(5) Result mantissa needs normalization? No
(6) Overflow? No. Underflow? No
Result
1 10000110 01010101100000000000000
IEEE 754 Single precision Multiplication
Rounding terjadi dalam floating point multiplication saat mantissa
berubah dari 48bit jadi 24 bit.
Overflow
• Overflow terjadi ketika jumlah exponents lebih dari 127
• Saat ini exponent mulai dari 128 ( E=225 ) dan mantissa di set = 0
• Overflow terjadi saat hasil tidak memiliki tanda sesuai dengan tanda
operand ( hasil negatif akan didapatkan saat menambah2 angka
positif )
• Digunakan two’s complement untuk adding or subtracting numbers
• A carry occurs when the result of an addition or subtraction,
considering the operands and result as unsigned numbers, does not
fit in the result.
IEEE 754 Single precision Multiplication
Underflow
• Underflow terjadi saat penjumlahan dari exponent lebih dari (-)126,
angka negatif yang di definisikan dalam bias adalah (-)127
• Angka yang bukan nol adalah 2 – 149, untuk mempertahankan
ketepatan tiap 1 bit dalam mantissa.
• Saat ini, exponent di set (-)127 (E=0)
-If M = 0, the number is exactly zero.
-If M is not zero, then a denormalized number is
indicated which has an exponent of -127 and a
hidden bit of 0.
IEEE 754 Single precision Multiplication
Underflow
Arithmetic underflow (or "floating point underflow", "floating
underflow", "underflow") adalah kondisi dimana hasil dari floating
point operation lebih kecil dalam (mendekati nol, angka
positif/negatif)
Contoh :
For example, an eight-bit two's complement exponent can
represent multipliers of 2 - 128 to 2127. A result less than 2 128 would cause underflow.