Question

IEEE 754-2008 contains a half precision that is only 16 bits wide. The left most bit is still the sign bit, the exponent is 5 bits wide and has a bias of 15, and the mantissa is 10 bits long. A hidden 1 is assumed. Write down the bit pattern to represent $\mathbf{-}\mathbf{1}·\mathbf{5625}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{1}}$assuming a version of this format, which uses an excess-16 format to store the exponent. Comment on how the range and accuracy of this 16-bit floating point format compares to the single precision IEEE 754 standard.

Short answer

Bit pattern to represent $-1.5625\times {10}^{-1}$

1

01100

0100000000

Step-by-step solution

Convert into binary

We can writeas$-1.5625\times {10}^{-1}$as$-015625$

Now, we convert it into binary representation

So, we write it

${\left(-0.15625\right)}_{10}={\left(-0.00101\right)}_2$

Step 2: Normalize the binary representation

Since a hidden 1 is assumed, the normalized form is

${\left(-0.00101\right)}_2={\left(-1.01\times 2^{-3}\right)}_2$

Find sign, exponent, and fraction

The number is negative so Sign =1

, 15 being bias so the exponent =12

Fraction = 01

Binaryrepresentation

1

01100

0100000000

The range is much smaller than the range of single-precision IEEE754 standard as the exponent is much smaller. And this half-precision standard is not as accurate as the fraction is also much smaller.