UDP checksum is calculated as follows:

$1.$ At the sender$,$ the checksum field is set to all $0$s when computing the checksum.

$2.$ Checksum is generated by adding the $16-$bit words successively. When two $16-$bit values are

added, if there is a carry$-$out at the leftmost bit, then add $1$ to the result.

$3.$ Take the $1$s complement of the final result.

$4.$ At the receiver, the checksum is calculated by adding all the received $16-$bit words.

$5.$ If the received UDP datagram contains no error, the resulting sum must be all $1$s$.$

$3.$a Assume the UDP datagram contains $4$ words shown in the table. Compute the

checksum at the source and checksum at the destination if the received datagram is error free $3.$b Explain why the checksum at the destination is all one$$s if the received UDP

datagram contains no error?