Floating Point Multiplication

Question

I got this problem I have to solve where I have to multiply to floating point numbers (16 bit), but I have no way of double checking it. Any help is immensely appreciated.

Floating Point A: 0 11000 0111000000 Floating Point B: 0 11010 1011000000

I calculate the exponents:

A: 24-15=9 B: 26-15=11

Calculate mantissas (a & b):

(2^9*b) * (2^11*b) = 2^9+11 * (a*b) + 2^20 * (a*b)

Overflow, so I increase exponent of A to sane as B(11).

Then I shift mantissa of A in accordance with the calculation:

1.0111 > 0.10111 > 0.010111.

Then I multiply to get mantissa.

                    0.010111
                  * 1.101100
                     0000000
                   0000000
                 0010111
               0010111
             0000000
           0010111
         0010111_____
       0.100110110100

I shift again.

0.100110110100 < 1.00110110100

Decrease exponent by 1, so it's 10.

Sign is 0, so it's positive number.

Answer: 0 01010 00110110100.

Is it correct?

Thanks in advance!

Answer 1

Looks like binary16

Is it correct?: No.

Both "mantissas" or fraction need to include an implied bit as its not the minimum exponent. So .1011000000 becomes 1.1011000000.

Instead 1.01 1100 0000 * 1.10 1100 0000 = 10.0110 1101 0000 0000 0000

The change in exponent is something to consider after the multiplication, in addition to rounding.

Shift right the product 1.0011 0110 1000 0000 0000 0 and +1 to the unbiased exponents 9 + 11 + 1 --> 21.

Round the product to 10 fraction bits
1.0011 0110 1000 0000 0000 0
1.0011 0110 10

Rebuild result
sign 0^0 = 0
biased exponent 21 + 15 = 36 = 100100, which overflows. @Yves Daoust
fraction = 0011 0110 10.

Overflow is typically set to INF 0 11111 0000000000 = infinity

Have not doubled check my work yet.

Answer 2

I see two problems:

• you don't account for the implicit 1 as the leftmost bit of the mantissas: 1.0111000000 * 1.1011000000,
• the exponent overflows and the result just cannot be represented [had the exponent underflown, you could have denormalized].