Arithmetic Operations

• Multiply 3.2x10⁴ * 5.1x10⁶

     5.1
     3.2
     --
    102
   153
   ----
   16.32 x 104 + 6
   = 1.632x1011
     

• Notice, we did the following
  • Normalized the operands (both in the for a.bx10^e, where a in 1-9
  • Add the exponents
  • Multiply the significands
  • Normalize the result.
• Let's try this in IEEE 754-DB, a standard where the significands have only 10 bits.
• But let's just do 5.1 x 3.2
  • 5.1 is 101.00011001100
    • Normalize this, 1.010001100110 x 2²
    • We need to represent e (2) in excess 127, so 127+2 = 129 = 10000001
    • The sign is +, so the sign bit is 0
    • The final representation is
    • 0 10000001 0100011001
    • This is actually 5.097656
  • Do the same thing with 3.2
    • 3.2 = 11.00110011001
    • Normalized 1.100110011001 x 2¹
    • 127 + 1 = 128 = 10000000
    • Sign = 0
    • 0 10000000 1001100110
    • This is really 3.199219
  • Multiply the two
    • They are normalized
    • Add the exponents

      •  10000001
         10000000
         --------
        100000001
        	    

      • But this is excess 127 so we must subtract 127 from the result.

      • 100000001
         01111111
         --------
         10000010
        	    

      • This is really 128 + 2 = 130, 130-127 = 3
    • Multiply the significands
      • Remember to add the leading 1.

        	     10100011001
        	     11001100110
        	     -----------
        	    101000110010
        	   1010001100100
        	   -------------
        	   1111010010110
        	10100011001
        	----------------
        	1100000110110110
               10100011001
               -----------------
              100000011111110110
            10100011001
            --------------------
            11000011100111110110
           10100011001
           ---------------------
          1000001001110111110110
        

        10.0001001110111110110 x 2⁰
    • Normalize this 1.00001001110111110110 x 2¹
  • This means adjusting the exponent by 1.
  • e = 10000011
  • The signs were the same so they stay a 0.
  • The reesult needs to be represented in 10 bits (but remember to delete the leading 1)

      1000001001110111110110
       0000010011
         

  • The final product is 0 10000011 0000010011
• We should double check this
  • Sign = 0 (+)
  • e = 10000011 = 127+4
  • f= 1.0000010011
  • 10000.010011 = 16.296875
  • Not the 16.32 we wanted
  • But it is the product of the two numbers (rounded)
  • In this case, we experienced round off error.
• 

• We can do an addition too.
  • 3.2x10⁴ + 5.1x10⁶
  • Obtain a common exponent (in terms of the larger)

    .032x106
       

  • Add the significands

    5.1
     .032
    5.132
       

  • Normalize the result
  • answer 5.132x10⁶
• Again, we can do this in IEEE 754-DB
  • 5.1 = 0 10000001 0100011001
  • 3.2 = 0 10000000 1001100110
  • Represent in common exponent

    3.2 = 0 10000001 1100110011  (remember add the 1)
    	

  • Add the bases

     10100011001
     01100110011
     -----------
    100001001100
    	

  • Normalize the results

    10.0001001100 x 20
    1.00001001100 x 21
    so e = 10000010
    	

  • So the answer is 0 10000010 0000100110
  • Which really is 1.0000100110x 2³
  • Which is 1000.0100110
  • This is 8.296875, which is the sum of the two numbers given.