$1.(20)$ MIPS Assembly Language Programming: Write MIPS assembly code $($implement it as a

function$)$ to reverse the bits in a register. Use as few instructions as possible. Assume the register of

interest is $t$3.$

$2.(10)$ Floating Point Computation: $1).$ Write down the binary representation of the decimal

number $63\mathrm{.}25$ assuming the IEEE $754$ single precision format.

$2).$ Write down the binary representation of the decimal number $63\mathrm{.}25$ assuming the IEEE $754$

double precision format.

$3.(30)$ Floating Point Computation: $1).$ 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 $-$

$1\mathrm{.}5625\backslash$times $10-1$ 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.

$2).$ Calculate the sum of $2\mathrm{.}6125\backslash$times $101$ and $4\mathrm{.}150390625\backslash$times $10-1$ by hand, assuming A and B are

stored in the $16-$bit half precision described in previous question. Assume $1$ guard, $1$ round bit, and

$1$ sticky bit, and round to the nearest even. Show all the steps