Ask for Some Help regarding PIC16 series

Hi there,
I am new to the forum. I am currently doing an electrical engineering course which has some electronic parts. One part is about explaining the test result of a program. I find this quite difficult as I have never done this before and I am never going to use it in my job so I would just need help understand the changes happen in the program if the delay loop is changed from 0X32 to 0X02 in order to speed up the process. How the values stored in SFR and variable registers after the change can be explained?
This is the program:
 
; blinks LEDs on outputs in a rotating pattern, with input option to reverse direction
;Set microprocessor as 16F84
INCLUDE <P16F84.INC>
; Setup processor configuration
__config _RC_OSC & _WDT_OFF & _PWRTE_ON
; label 2 symbols as memory addresses for counting loops
J equ 1F
K equ 1E
 
SETUP:
org 0 ;
BANKSEL TRISB;
clrf TRISB;
bcf STATUS , RP0 ;
bcf STATUS , RP1 ;
movlw 0x01 ;
movwf PORTB ;
bcf STATUS,C ;
 
ROTATE:
btfss PORTA,0 ;
goto RR
rlf PORTB,f ;
goto DELAY
RR: rrf PORTB,f ;
 
DeLAY:
movlw 0x32 ;
movwf J ;
jloop: movwf K ;
kloop: decfsz K,f ;
goto kloop
decfsz J,f ;
goto jloop
 
DIRECTION:
movlw 0x02 ;
xorwf PORTA,0;
goto ROTATE ;
end ;
 
Thanks for your help.

DeLAY:
        movlw   0x4             ; Total loops for K > 1 & J > 1: (K-1)+(J-1)*(K)
        movwf   J               ; Total loops for K = 1 & J > 1: ( 0 )+(J-1)*(1)
jloop:  movwf   K               ; Total loops for K = 0 & J > 1: (255)+(J-1)*(256)
kloop:  decfsz  K,f             ; Total loops for K > 1 & J = 1: (K-1)+(   0   )
        goto    kloop           ; Total loops for K = 1 & J = 1: ( 0 )+(   0   )
        decfsz  J,f             ; Total loops for K = 0 & J = 1: (255)+(   0   )
        goto    jloop           ; Total loops for K > 1 & J = 0: (K-1)+(255*K)
                                ; Total loops for K = 1 & J = 0: ( 0 )+(256)-1
                                ; Total loops for K = 0 & J = 0: (256)*(256)-1

<EDIT>
Verified in simulation.

1and0
That delay takes
 
Instruction Cycles = 2 + {[1 + (3 * WREG - 1) + 3] * WREG - 1}

Oh! Just go and give him the correct answer while the old guy struggles for an hour. :)

1and0

dan1138
Oh! Just go and give him the correct answer while the old guy struggles for an hour. :)

I haven't derived such calculations in years -- need to make sure my gray matter still working. ;)  Normally I'd just use the Simulator Stopwatch.

May have an issue if WREG is ever zero.

For completeness ;)

DeLAY:
        movlw   K

 
N = K & 0xFF;
if (N == 0) N = 256;
Tcy = 3*N*N + 3*N + 1;