CSECE 365 Computer Architecture

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CS/ECE 365 Computer Architecture

• Soundararajan Ezekiel
• Department of Computer Science
• Ohio Northern University

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Floating Point

• many applications requires non integer numbers
• one non-integer type is called floating point
• two parts -- exponent --- significand
• 1.52(-3) 0.1875 exponent -3, significand1.5
• IEEE standard

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IEEE standard

• when rounding a halfwayresult to the nearest
  floating point number it picks the one that is
  even 6.10.53.05 ( if this need it to be rounded
  2 digits 3.0 not 3.1)
• It includes the special values NaN(Not a Number),
  ? and -?
• divide by zero (will give ? )or square root of
  negative number(NaN)

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Continue

• It uses denormal number that represent less than
  1.02(Emin)
• xy ltgtx-y 0
• x1.25610Emin, y1.23410Emin
• x-y0.2210Emin which flushes to 0
• even though x not equal to y, x-y0
• It rounds to nearest by default, but it also has
  three other rounding modes
• It has sophisticated facilities for handling
  exceptions

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representation of floating point

• single precision stored in 32 bitgt1 for sign, 8
  for exponent, 23 for fraction
• Ex 1 10000001 010.0
• exponent is 129 making the value 129-1272
• (-1)S(1significand)2(exponent-Bias)
• (-1)(10.25)22-5 ( significand
  12(-2)1/40.25)
• left to right (significand bit)s1,s2,s3, gt
  s12(-1)s22(-2)
• significand 1/4gt0.25

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floating point multiplication

• xs2s
• (s12(e1))(s12(e1))(s1s2)(2(e1e2))
• 1 10000010 00000 -123
• 0 10000011 000000 124
• When unpacked, the significance are both 1.0
  their product is 1.0 so the result is is of the
  form 1 ?????? 000000
• To compute the exponent use this formula

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conti

• biased exp(e1e2)biased exp(e1)biased
  exp(e2)-bias
• exp bias 127 01111111 so in 2s complement
  -12710000001
• 10000010
• 10000011
• 10000001
• -------------
• 10000110 since this is 134 , 134-1277 as exponent

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rounding

• the interest part of floating point is rounding
• example
• 1.236.788.3394 8.34
• 2.834.4712.65011.27101
• 1.287.8109.99681.00101
• round digit gt 5 add 1 as in first case
• in last case when adding one there may be
  carry-out

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continue

• if the round digit is exactly 5 as in second case
  then additional bit must be examined to decide
  between truncation or incrementing
• in the following algorithm it will recorded in
  sticky bit
• there is a straight forward method of handling
  rounding using the multiplier together with an
  extra sticky bit
• p number of pit in significand
• A,B, P registers should be p bit wide

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• multiply the significand, obtain 2p bits (P,A)
  registers
• during the multiplication the first p-2 times a
  bit is shifted into A or into sticky bit
• let s represent sticky bit, g the most
  significant bit of A, r the second most
  significant bit of A, there are two cases
• 1. Higher order bit of P is 0,Shift P left 1 bit,
  shifting in the g bit from A. Shifting the rest
  of A is not necessary
• 2. Higher order bit 1, set ssVr, and rg, and
  add 1 to the exponent

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two cases of floating point multiply algorithm
x0,x1,x2, x3,x4,x5
g, r, s, s ,s,s
sticky
x1,x2,x3,x4,x5,g
Case1x00 shift needed
sticky
x0,x1,x2,x3,x4,x5
x01 increment needed
Adjust binary point add 1 to exponent to
components
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Floating point addition

• floating point operations take two inputs with p
  bits of precision and return a p bit result
• Example
• sum of binary 6-bit number 1.10011 and
  1.100012(-5)
• 1.10011
• .0000110001
• -----------------
• using 6-bit adder gives

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• 1.10011
• .00001
• ----------
• 1.10100
• the first discard bit is 1
• this isnt enough to decide whether to round up
• the rest of the discarded bits 0001
• so sticky is 1
• the final sum is 1.10101

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• Here is another example
• 1.11011
• .0101001
• ----------
• A 6 bit adder gives
• 1.11011
• .01010
• ----------
• 10.00101

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• Because the carry out on the left, the round bit
  is isnt first discarded bit rather, it is the
  low order bit of the sum (1)
• the discard bit is 01
• because the round and sticky are both 1
• the final answer is 10.0011

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• Here is the subtraction
• 1.00000
• -.00000101111
• ------------------
• convert negative by 2s complement and add and
  add
• 100000
• 1.11111010001
• ----------------
• 0.11111

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Floating point addition algorithm

• a1 and a2 be two number to be added
• ei exponent
• sisignificand
• there are 8 steps
• we will go for each step with this example
• a1-1.0012(-2)
• a2-1.11120
• s11.001 e1-2 s21.111 e20

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step1

• if e1lte2 swap the operands. This ensures that the
  difference of the exponents satisfies de1-e2gt0
  tentatively set the exponent of the result to e1
• e1lte2 so swap d2 tentative exp0

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step 2

• if the signs of a1 and a2 differs replace s2 by
  its 2s complement
• here the signs of the both are negative dont
  negate s2

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step3

• place s2 in a p-bit register and shift it
  dee1-e2 places to the right.From the bits
  shifted out, set g to be the most significant
  bit, r to the next most sig, and set sticky hit
  the OR of the rest
• Here shift s2(1.001after swap) right by 2, giving
  s2.010, g0, r1., s0

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step 4

• compute a preliminary significand Ss1s2, by
  adding s1 to the p-bit registers containing s2,If
  the signs of a1, a2 are different the most
  significant bit of S is 1, and there was no
  carry-out, the S is negative, Replace S with its
  twos complement. this can only happen when d0
• 1.111
• .010
• -------
• (1)0.001 S0.001 with a carry out

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step 5

• Shift S as follows. If the signs of a1 and a2 are
  the same and there was a carry out in step 4,
  shift S right by one, filling in the high order
  position with 1. Otherwise shift it left until it
  is normalized. When left shifting on the first,
  shift fill in the low-order position with the g
  bit. After that shift,in zeros. Adjust the
  exponent of the result accordingly
• Carry-out, so shift S right , S1.000, expexp1,
  exp1

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step 6

• adjust r and s. if S was shifted right in step 5,
  set rlow-order bit of S before shifting and sg
  OR s. If there was no shift se rg, sr OR s If
  there was a single left shift, dont change r and
  s, If there were two or more left shifts,
  r0,s0,
• rlow-order bit of sum1, sg v r v s0v1v01(v
  denotes OR)

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step7

• Round S by using the follows. if the entry is non
  empty and 1 to the low-order bit of S. If
  rounding causes carry-out, shift S right and
  adjust the exponent. this the signiciand of the
  result
• r AND strue so round up , SS1, S1.001

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Table

• swap compl sign(a1)
  sign(a2)sign(result)
• yes
  - -
• yes -
  
• no no
  -
• no no -
  -
• no yes
  - -
• no yes -
  

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step 8

• compute the sign of the result. If a1 and a2 have
  the same sign, this is the sign of the result. If
  a1 and a2 have different signs, then the sign of
  the result depends on which of a1, a2 is
  negative,. whether there is swap in step 1 and
  whether S was replaced by its 2s complement ins
  step 4 use the above table
• both signs are negative so sign of result is
  negative
• Final answer -S2exp-1.00121

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Exercise

• Use the algorithm to compute the sum
  (-1.010)1.100