Exception Arithmetic Overflow

Code: Problem 4
org 100h
section .data
prompt1 db 0dh, 0ah, 0dh, 0ah, "Please input a signed base-10 integer: $"
prompt2 db 0dh, 0ah, "Your number in binary is: $"
prompt3 db 0dh, 0ah, "Pretty sure that wasn't a number. Please enter a number value. $"
prompt4 db 0dh, 0ah, "Your number in octal is: $"
section .text
START:
mov ah, 9 ;Display input prompt
mov dx, prompt1
int 21h
mov bx, 0 ;Get input
mov ah, 1
int 21h
DEC_IN:
cmp al, 0dh ;compare input to carriage return; check if user is finished
je DEC_OUT ;if yes, go display the prompt
cmp al, '0' ;compare to '0'
jg IS_DIGIT ;jump to IS_DIGIT to confirm that it is a number
jl NAN_ERROR ;if below 0, print error prompt and start over
IS_DIGIT:
cmp al, '9' ;confirms digit value
jl BIN_CONV ;if digit, convert to binary
jg NAN_ERROR ;if not, print 'not a number' error message
BIN_CONV:
and ax,000Fh ; convert from ASCII to base 10 value
push ax ; and save it on stack
mov ax,10 ; set up to multiply bx by 10
imul bx ; dx:ax = bx*10
pop bx ; saved value in bx
add bx,ax ; bx = old bx*10 + new digit
mov ah,1 ; input char function
int 21h ; read next character
jmp DEC_IN ; loop until done
NAN_ERROR:
mov ah, 9 ;display error message
mov dx, prompt3
int 21h

jmp START ;Go back to beginning
DEC_OUT:
mov ah, 9 ;Display the signed decimal value
mov dx, prompt2
int 21h
mov cx, 16
TOP: rol bx,1 ; rotate msb into CF
jc ONE ; CF = 1?
mov dl,'0' ; no, set up to print a 0
jmp print ; now print
ONE: mov dl,'1' ; printing a 1
print: mov ah,2 ; print char fcn
int 21h ; print it
loop TOP ; loop until done
mov ah, 2 ;Display decimcal value in bx
mov dl, bl
int 21h
jmp EXIT
EXIT:
mov ah,04ch ;DOS function: exit
mov al,0 ;exit code
int 21h ;call DOS, exit
Code: Problem 5
.stack 100h
.data
base10_string dw "-32799$"
biggest_16bits_signed equ 7FFFh
special_16bits_signed equ 8000h
special_16bits_signed_str dw "1000000000000000$"
error_overflow dw "Arithmetic overflow encountered. Abort$"
error_not_a_number dw "Illigal Character. Abort$"
base10 equ 10
tmp dw 0
is_negative dw 0
.code
main proc
;Initialization
mov ax,@data
mov ds,ax
;;;part 1: convert string to value
mov ax,0 ;number = 0;
lea si,base10_string
;check if positive or negative