1. consider the following program. interpret the program line by line. Get an idea what the program.

TOS EQU 0FEFEH
CPU 8086
ORG 0400H

SECTION CODE:
MAIN:
MOV SP, TOS
CALL INITIALIZATION
MOV BX, PATTERN
MOV AL, [BX]
JMP $

INITIALIZATION:
MOV WORD[03FCH], COUNTERISR
MOV WORD[03FCH], 0
MOV AX, 0
STI
RET

COUNTERISR:
OUT 0, AL
IRET

SECTION DATA:
PATTERN DB 255, 128, 170, 15, 240

Do the following:
Part a:
a) by following the program as above, determine what would be the pattern that should be displayed on the LEDs.
Answer:

b) change the special instruction in the above [BX+1]. Determine what would be the pattern that should be displayed on the LEDs.
Answer:

c) change the special instruction in the above [BX+2]. Determine what would be the pattern that should be displayed on the LEDs.
Answer:

d) change the special instruction in the above [BX+3]. Determine what would be the pattern that should be displayed on the LEDs.
Answer:

e) change the special instruction in the above [BX+4]. Determine what would be the pattern that should be displayed on the LEDs.
Answer:

Since the program does not provide the specific instructions for controlling the LEDs, it is not possible to determine the pattern that would be displayed on the LEDs accurately. The program only shows the initialization process and the structure of the program, but it does not contain the relevant instructions for controlling or changing the patterns on the LEDs.

To interpret the program line by line:

1. TOS EQU 0FEFEH - This line defines a constant called TOS and assigns it the value 0FEFEH (a hexadecimal value).

2. CPU 8086 - This line specifies that the program is written for an 8086 CPU.

3. ORG 0400H - This line sets the origin address of the program to 0400H (another hexadecimal value).

4. SECTION CODE - This line starts a new section called CODE.

5. MAIN - This line labels the following code as the main program.

6. MOV SP, TOS - This line moves the value of TOS (0FEFEH) into the stack pointer (SP) register.

7. CALL INITIALIZATION - This line calls the INITIALIZATION subroutine.

8. MOV BX, PATTERN - This line moves the offset of the label PATTERN into the BX register.

9. MOV AL, [BX] - This line moves the value at the memory location specified by BX (the first element of the PATTERN array) into the AL register.

10. JMP $ - This line jumps to the current instruction (an infinite loop).

11. INITIALIZATION subroutine - This subroutine is called to initialize certain values.

12. MOV WORD[03FCH], COUNTERISR - This line moves the offset of the COUNTERISR label into the memory location specified by 03FCH.

13. MOV WORD[03FCH], 0 - This line moves the value 0 into the memory location specified by 03FCH.

14. MOV AX, 0 - This line moves the value 0 into the AX register.

15. STI - This line enables interrupts.

16. RET - This line returns from the subroutine.

17. COUNTERISR - This is an interrupt service routine for the counter.

18. OUT 0, AL - This line sends the value in the AL register to output device 0.

19. IRET - This line returns from the interrupt.

20. SECTION DATA - This line starts a new section called DATA.

21. PATTERN DB 255, 128, 170, 15, 240 - This line defines an array called PATTERN with these specific values.

Part a:
a) The pattern that would be displayed on the LEDs is 255, 128, 170, 15, 240.

b) Changing the special instruction [BX+1] would modify the pattern displayed on the LEDs. Since BX points to the PATTERN array, [BX+1] refers to the second element. The new pattern would depend on the value stored at the memory location specified by [BX+1].

c) Similarly, changing [BX+2] would modify the pattern displayed on the LEDs. The new pattern would depend on the value stored at the memory location specified by [BX+2].

d) Changing [BX+3] would modify the pattern displayed on the LEDs. The new pattern would depend on the value stored at the memory location specified by [BX+3].

e) Changing [BX+4] would modify the pattern displayed on the LEDs. The new pattern would depend on the value stored at the memory location specified by [BX+4].