Transcript CS 230 Chapter 3 Arithmetic for Computers

CS 230: Computer Organization
and Assembly Language
Aviral Shrivastava
Department of Computer Science and Engineering
School of Computing and Informatics
Arizona State University
Slides courtesy: Prof. Yann Hang Lee, ASU, Prof. Mary
Jane Irwin, PSU, Ande Carle, UCB
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Announcements
• Quiz 2
• Project 2
• Quiz 3
– Thursday, Oct 06, 2009
– Complete Chapter 3
• Project 3
– Implement an assembler
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Floating Point (Real) Numbers
• We need a way to represent
– numbers with fractions, e.g., 3.1416
– very small numbers, e.g., .000000001
– very large numbers, e.g., 3.15576  109
• Representation:
– sign, exponent, significand: (–1)sign  significand  2exponent
– more bits for significand gives more accuracy
– more bits for exponent increases range
• IEEE 754 floating point standard:
– single precision: 8 bit exponent, 23 bit significand
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IEEE 754 floating-point standard
• IEEE Floating Point: (-1)S  1.M  2 E-127
– Mantissa (sign and magnitude -- S, M)
– Exponent (excess 127 binary number -- E)
S
1 bit
E
8 bits
M
23 bits
• Example:
–
–
–
–
decimal: -.75 = - ( ½ + ¼ )
binary: -.11 = -1.1 x 2-1
floating point: exponent = 126 = 01111110
IEEE single precision:
1 01111110 10000000000000000000000
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Remember
• For FP Number <SEM>
– Value = (-1)S*1.M*2E-127
• 1st bit is the sign bit
• Exponent = E-127
– Why?
• Mantissa is after the decimal
– 1.1100111  Mantissa is 1100111
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IEEE FP Numbers
• Example: 0 1000 0011 1101 0000 0000 0000 0000 000
–
S = 0 (positive mantissa), E = 131 - 127 = 4, and M = .1101
– the number = 1.1101 x 24 = 11101 = 29
• Example: 1 0111 1011 1101 0000 0000 0000 0000 000
–
S = 1 (negative mantissa), E = 123 - 127 = -4, and M = .1101
– the number = -1.1101 x 2-4 = -0.00011101 = -(2-4+ 2-5+ 2-6+ 2-8)
= -0.11328125
• Example: -10.609375
10.609375 = (1010.100111)2 = 1.010100111 x 23
– S = 1, E = 3+127 = 130 = 1000 0010,
and M = 010100111 ...
– FP: 1 1000 0010 0101 0011 1000 0000 0000 000
0
1
0
0
1
1
1
.609375 x 2
.21875 x 2
.4375 x 2
.875 x 2
.75 x 2
.5 x 2
.0
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FP Addition
Compute 9.999*101 + 1.610*10-1, assuming
only 3 places in decimal
1. Make the exponents same
• 9.999*101 + 0.01610*101
2. Add the significands
• 10.015*101
3. Normalize
• 1.0015*102
4. Round-off
• 1.002*102
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FP Addition
• Example
1.1101 x 23 + 1.001 x 2-1
= 1.1101 x 23 + 0.0001001x23 =1.1110001 x 23
• Steps: X(m,e) + Y(m,e)
– Shift the smaller operand --- align binary point
•compute Ye-Xe, right shift Xm
 Xm’ if
Ye > Xe
– Add magnitudes (Xm’ + Ym)
– Normalize and round-off if necessary
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FP Subtraction
• 0.510 – 0.437510
• Step 1: Convert into FP Representation
– 0.510 = 1.000 x 2-1
– -0.437510 = -1.110x2-2
• Step 2: Shift significand of smaller number
– -1.110x2-2 = -0.111x2-1
• Step 3: Subtract the significands
– 1.000 x 2-1 - 0.111x2-1 = 0.001x2-1
• Step 4: Normalize the sum, check for over/underflow
– = 0.001x2-1 = 1.000x2-4
• Step 5: Round off
– 1.000x2-4
• Step 6: Result
– 1.000x2-4 = 0.0625
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FP Multiply
Compute 1.110*1010 *9.200*10-5 assuming digit
significands
1. Add the exponents
• Exponent = 10-5 = 5
2. Multiply the significands
• 1.110*9.200 = 10.212000
3. Normalize
• 1.0212000*106
4. Round-off
• 1.021*106
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FP Multiply
• Assume 4-bit mantissa
• compute 0 0111 1110 0101 x 0 0111 1101 1110
– add exponent and adjust the bias:
0111 1110 + 0111 1101 - 0111 1111
= 0111 1100
– multiply 1.0101 and 1.1110  10.0111 0110
– normalize and detect underflow or overflow:
exponent: 0111 1101, mantissa: 1.0011 1011 0
– round the product and the result is
0 0111 1101 0100
• compute
(1.0101 x 2-1 ) x (1.1110 x 2-2) = .65625 x .46875 =0.3076171875
= 10.01110110 x 2-3
= 1.001110110 x 2-2 = .3076171875
= 0 0111 1101 0100 = .3125
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IEEE 754 Floating Point Numbers
• Special Representation
– When exponent is 1111 1111 or 0000 0000
Type
Exponent
Fraction
Zeroes
0
0
De-normalized
numbers
0
non zero
Normalized
numbers
1 to 2e − 2
any
Infinities
2e − 1
0
NaNs
2e − 1
non zero
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IEEE 754 FP Representation
• There are two Zeroes
– +0 (s is 0) and −0 (s is 1)
• There are two Infinities
– +∞ (s is 0) and −∞ (s is 1)
• Numbers closest to zero
– is a denormalized value
– all 0s in the Exp field and the binary value 1 in the Fraction field
– ±2−149 ≈ ±1.4012985×10−45
• Normalized numbers closest to zero
– binary value 1 in the Exp field and 0 in the fraction field)
– ±2−126 ≈ ±1.175494351×10−38
• Finite numbers furthest from zero
– 254 in the Exp field and all 1s in the fraction field
– ±(1-2-24)×2128[2] ≈ ±3.4028235×1038
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Inaccuracy in FPs
Exponent
Minimum
Maximum
Precision
0
1
1.999999880791
1.19209289551e-7
1
2
3.99999976158
2.38418579102e-7
2
4
7.99999952316
4.76837158203e-7
10
1024
2047.99987793
1.220703125e-4
11
2048
4095.99975586
2.44140625e-4
23
8388608
16777215
1
24
16777216
33554430
2
127
1.7014e38
3.4028e38
2.02824096037e31
• Double precision arithmetic
• Double word – 64-bits
• 11 bit exponent, 52 bit mantissa
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FP numbers
• Limited precision:
(- 1.5 x 1038 + 1.5 x 1038 ) + 1 = 0 + 1 = 1
-1.5 x 1038 + (1.5 x 1038 + 1) = -1.5 x 1038 + 1.5 x 1038 = 0
• A view of FP representation, (assume 4 bit precision)
de-normal numbers
0
2-126
2-125
2-124
• There is a big gap between two large consecutive numbers:
– gap = 2-23 x 2e-127
• Add de-normal numbers (exponent=0, mantissa >0) to fill
the gap between 0 and 2-126
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The Patriot Missile Failure
• On February 25, 1991, during the
Gulf War, an American Patriot
Missile battery in Dharan, Saudi
Arabia, failed to track and
intercept an incoming Iraqi Scud
missile. The Scud struck an
American Army barracks, killing 28
soldiers and injuring around 100
other people
http://www.ima.umn.edu/~arnold/disasters/patriot.html
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The Patriot Missile Failure - I
• Computed time on 1/10th of seconds
• In 24-bit precision arithmetic
• Truncation instead of round-off
– 1/10 = 0.0001100110011001100110011001100
• Patriot Stored
–
0.00011001100110011001100
• Error
–
0.0000000000000000000000011001100
– = 0.000000095 decimal
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The Patriot Missile Failure
• Patrioit battery was up for more than 100
hours
• Accumulated Error:
– 0.000000095×100×60×60×10=0.34 1/10ths of a
second
• Speed of scud is 1,676 meters per second,
and so travels more than half a kilometer
in this time
• Completely out of range
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Explosion of the Ariane 5
On June 4, 1996 an unmanned Ariane 5 rocket launched by the
European Space Agency exploded just forty seconds after its liftoff from Kourou, French Guiana. The rocket was on its first
voyage, after a decade of development costing $7 billion. The
destroyed rocket and its cargo were valued at $500 million. A
board of inquiry investigated the causes of the explosion and in
two weeks issued a report. It turned out that the cause of the
failure was a software error in the inertial reference system.
Specifically a 64 bit floating point number relating to the
horizontal velocity of the rocket with respect to the platform was
converted to a 16 bit signed integer. The number was larger than
32,767, the largest integer storeable in a 16 bit signed integer, and
thus the conversion failed.
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The Number Agenda
• Unsigned Numbers
– Representation, (d31 d30 … d2 d1 d0)2 = d31*231 + d30*230 + … d2*22 + d1* 21 + d0*20
– Addition, Subtraction
– Multiplication, Division
• Signed Numbers
(d31 d30 … d2 d1 d0)2 = (-1)*231 + d30*230 + … d2*22 + d1* 21 + d0*20
– Representation
– Addition, Subtraction
– Multiplication, Division
• Floating Point Numbers (For large and fractions)
– Representation (d31 d30 … d2 d1 d0)2 = (-1)*d31 * 1.d22… d0 * 2(d30…d23 - 127)
– Addition, Subtraction
– Multiplication
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Yoda says…
• Luke: I can’t believe it.
• Yoda: That is why you fail
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