A = 0 10011 0011110111

B = 1 00011 0010011000

exponent is 15, mantissa is 10 bits and first bit is implicit. Can somebody please tell me the final answer cause I am having trouble figuring out what is happening with exponent since there are 3 extra zeroes in front of mantissa multiplication. I got that exponent is 7(00111 in binary). Thanks in advance


1 Answer 1


When multiplying floating points:

  1. we add the (real) exponents, to get the output exponent
  2. we multiply the two mantissas (remember the implied 1), round and shift as necessary.

So the new exponent is $E=(19-15) + (3-15)= -8$.

The multiplication of the mantissas give $$1.0011110111_2 \times = 1.0010011000_2 = 1.01101100111010101$$

taking only 10 digits (rounding down, see here) we get $M= 0110110011$.

The sign is a minus.

So the output is $Out= 1\ 00111\ 0110110011$.

For sanity check, let's convert to decimal and see that it makes sense.

$A = (-1)^0 \times 2^{4} \times 1.2412109_{10} = 19.8593744$

$B = (-1)^1 \times 2^{-12} \times 1.1484375_{10} = -0.0002804 $


$A\cdot B = -0.0055686$

We can convert $Out$ from above to decimal, and get $Out = (-1)^1 \times 2^{-8} \times 1.4248047 = -0.0055656$ and the error is $0.000003 \approx 2^{-18}$ which makes sense (since the real exponent is $2^{-8}$ and we have a 10 digit significand)

See also https://oletus.github.io/float16-simulator.js/ for a calculator, and https://en.wikipedia.org/wiki/Half-precision_floating-point_format or https://en.wikipedia.org/wiki/Floating_point for definition and some more explanations.

  • $\begingroup$ I got 1 00111 011011 0011 by multiplying both mantises, I got 101 as guard digit, round digit and sticky bit but since the signs are different it doesn't change anything, what did I miss to get 0011 instead of 1110 ? $\endgroup$
    – vucko95
    Mar 27, 2016 at 23:21
  • $\begingroup$ No, you are correct. My answer performs the multiplication in the decimal basis, then converting back to half-point (hence, the $\approx$; and since your question was about the exponent anyways). But note that rounding the $0011...$ should probably give you $0100$, no? $\endgroup$
    – Ran G.
    Mar 27, 2016 at 23:36
  • $\begingroup$ In my notes it says if the signs of the numbers that I am multiplying is same then I add that 1 to result. When they are different just like in this case (a -0 , b-1) then the result stays the same. I don't know if that's correct but that said the professor. I hope it is correct .Thank you $\endgroup$
    – vucko95
    Mar 27, 2016 at 23:43
  • $\begingroup$ "The IEEE standard for floating point arithmetic requires that the programmer be allowed to choose 1 of 4 methods for rounding:; see for instance pages.cs.wisc.edu/~markhill/cs354/Fall2008/notes/… for a nice explanation. It seems your Prof uses method 2? $\endgroup$
    – Ran G.
    Mar 28, 2016 at 0:08
  • $\begingroup$ In my task it says, find A*B and round it with round to nearest together with round away from zero. Probably I understood question wrong. But there is this thing that if at the I end is 101 I have to look at the sign of the numbers to decide if I should change mantisa. So with round away from zero and round to nearest it is 1 00111 011011 0100? $\endgroup$
    – vucko95
    Mar 28, 2016 at 19:00

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