The subroutine delay sub uses a simple loop to implement a delay. The

assembly listing of this subroutine is

$000002$f$4$ :

$2$f$4$: $004080$ mov.w count, w$0$

$2$f$6$: f$1$ ff $2$f mov.w #$0$xffff$,$ w$1$

$000002$f$8$ :

$2$f$8$: $008040$ add.w w$1,$ w$0,$ w$0$

$2$fa: $000000$ nop

$2$fc: fd ff $3$a bra NZ$,0$x$2$f$8$

$2$fe: $000006$ return

Assume an instruction frequency of $16$ MHz for an an instruction cycle time of $62\mathrm{.}5$ ns$.$

$($a$)$ How many bytes of program memory does this subroutine, delay sub, occupy? How many

bytes of data memory are used?

Program memory $=$ bytes, Data memory $=$ bytes.

$($b$)$ What value should be used for the variable count if the delay is to be as close as possible to$,$

but no greater than, $10$ ms$?$ Complete the following to initialize count with this value.

$.$data

count: $.$word

$($c$)$ To make this delay exactly $10$ ms we may need to add nop instruction$($s$).$ Where in the code

and how many nop instructions, if any, should be added