$3.$ Analyze the following program. Comment on the duty cycle.

MOV TMOD, $\backslash$#$01$ ; Timer $0,$ mode $1(16-$bit mode$)$

HERE: MOV TLO, $\backslash$# $\backslash (0\backslash$mathrm$\{$~F$\}2\backslash$mathrm$\{$H$\}$ ; $\backslash$mathrm$\{$TL$\}0=\backslash$mathrm$\{$F$\}2\backslash$mathrm$\{$H$\}\backslash ),$ the low byte

MOV THO, $\backslash$#$0$FFH ; THO $\backslash (=\backslash )$ FFH$,$ the high byte

CPL P$1\mathrm{.}5$ ; toggle P$1\mathrm{.}5$

ACALL DELAY

SJMP HERE

; load TH$,$ TL again

;$-$ delay using Timer $0$

DELAY:

SETB TRO ; start Timer $0$

AGAIN: JNB TF$0,$ AGAIN ; monitor Timer $0$ flag until it rolls over

CLR TRO ; stop Timer $0$

CLR TF$0$ ; clear Timer $0$ flag

RET