LDA 8051H

LXI H,8061H

MOV B,M

MVI E,00H

ADD B

JNC L1

INR E

L1:STA 8071H

LDA 8050H

LXI H,8060H

MOV B,M

ADD E

ADD B

MVI E,00H

JNC L2

INR E

L2:STA 8070H

MOV A,E

STA 8072H

HLT

****************************************************************

LHLD 4000H

: Get first I6-bit number

XCHG

: Save first I6-bit number in DE

LHLD 4002H

: Get second I6-bit number in HL

DAD D

: Add DE and HL

SHLD 4004H

: Store I6-bit result in memory locations 4004H and 4005H.

HLT

: Terminate program execution

Sample problem:

(4000H) = 15H

(4001H) = 1CH

(4002H) = B7H

(4003H) = 5AH

LDA 8051H

LXI H,8061H

MOV B,M

MVI E,00H

ADD B

JNC L1

INR E

L1:STA 8071H

LDA 8050H

LXI H,8060H

MOV B,M

MUL E

MUL B

MVI E,00H

JNC L2

INR E

L2:STA 8070H

MOV A,E

STA 8072H

HLT

****************************************************************

LHLD 4000H

: Get first I6-bit number

XCHG

: Save first I6-bit number in DE

LHLD 4002H

: Get second I6-bit number in HL

DAD D

: Add DE and HL

SHLD 4004H

: Store I6-bit result in memory locations 4004H and 4005H.

HLT

: Terminate program execution

Sample problem:

(4000H) = 15H

(4001H) = 1CH

(4002H) = B7H

(4003H) = 5AH

TITLE GETDTTM

PAGE 60, 132

;   This program retrieve the system date and time,

;   converts it to ASCII, and displays it to the screen

;   Define constants

;

CR      EQU 0DH ;define carriage return

LF      EQU 0AH ;define line feed

EOM     EQU '$' ;define end of message marker

NULL    EQU 00H ;define NULL byte

;

;   Define variables

;

JMP START

PROMPT  DB  CR, LF, "The current time is: ",EOM

PROMPT2 DB  CR, LF, "The date is: ",EOM

TIME    DB  "00:00:00", CR, LF, EOM

DATE    DB  "00/00/0000", CR, LF, EOM

;

;   Program code

;

START: 

CALL GET_TIME   ;call function to get system time

CALL GET_DATE   ;call function to get system date

LEA DX, PROMPT  ;print time prompt to screen

MOV AH, 09H

INT 21H

LEA DX, TIME    ;print time

MOV AH, 09H

INT 21H

LEA DX, PROMPT2 ;print date prompt to screen

MOV AH, 09H

INT 21H

LEA DX, DATE    ;print date

MOV AH, 09H

INT 21H

CVT_TIME:   ;converts the time to ASCII

CALL CVT_HR

CALL CVT_MIN

CALL CVT_SEC

RET

CVT_HR:

MOV BH, CH  ;copy contents of hours to BH

SHR CH,4    ;convert high char to low order bits

ADD CH, 30H ;add 30H to convert to ASCII

MOV [TIME], CH  ;save it

AND BH, 0FH ;isolate lower 4 bits

ADD BH, 30H ;convert to ASCII

MOV [TIME+1], BH    ;save it

RET

CVT_MIN:

MOV BH, CL  ;copy contents of minutes to BH

SHR CL, 4   ;convert high char to low order bits

ADD CL, 30H ;add 30H to convert to ASCII

MOV [TIME+3], CL    ;save it

AND BH, 0FH ;isolate lower 4 bits

ADD BH, 30H ; convert to ASCII

MOV[TIME+4], BH ;save it

CVT_SEC:

MOV BH, DH  ;copy contents of seconds to BH

SHR DH, 4   ;convert high char to low order bits

ADD DH, 30H ;add 30H to convert to ASCII

MOV [TIME+6], DH    ;save it

AND BH, 0FH ;isolate lower 4 bits

ADD BH, 30H ;convert to ASCII

MOV[TIME+7], BH ;save it

GET_DATE:   ;get date from the system

    MOV AH, 04H    ;BIOS function to read date

    INT 1AH        ;call to BIOS, run 04H

    CALL CVT_DATE

    RET

;CH = Century

;CL = Year

;DH = Month

;DL = Day

;CF = 0 if clock is running, otherwise 1

CVT_DATE:

    CALL CVT_MO

    CALL CVT_DAY

    CALL CVT_YR

    CALL CVT_CT

    RET

CVT_MO:     ;convert the month to ASCII

MOV BH, DH  ;copy month to BH

SHR BH, 4   ;convert high char to low order bits

ADD BH, 30H ;add 30H to convert to ASCII

MOV [DATE], BH  ;save in DATE string

MOV BH, DH  ;copy month to BH

AND BH, 0FH ;isolate lower 4 bits

ADD BH, 30H ;convert lower bits to ASCII

MOV [DATE+1], BH;save in DATE string

RET

CVT_DAY:    ;convert the day to ASCII

MOV BH, DL  ;copy days to BH

SHR BH, 4   ;convert high char to low order bits

ADD BH, 30H ;add 30H to convert to ASCII

MOV [DATE+3], BH    ;save in DATE string

MOV BH, DL  ;copy days to BH

AND BH, 0FH ;isolate lower 4 bits

ADD BH, 30H ;convert lower bits to ASCII

MOV [DATE+4], BH;save in DATE string

RET

CVT_YR:     ;convert the year to ASCII

MOV BH, CL      ;copy year to BH

SHR BH, 4       ;convert high char to low order bits

ADD BH, 30H     ;convert to ASCII

MOV [DATE+8], BH    ;save it

MOV BH, CL      ;copy year to BH

AND BH, 0FH     ;isolate low order bits

ADD BH, 30H     ;convert to ASCII

MOV [DATE+9], BH    ;save in DATE string

RET

CVT_CT:     ;convert the century to ASCII

MOV BH, CH      ;copy century to BH

SHR BH, 4       ;convert high char to low order bits

ADD BH, 30H     ;convert to ASCII

MOV [DATE+6], BH    ;save it

MOV BH, CH      ;copy century to BH

AND BH, 0FH     ;isolate low order bits

ADD BH, 30H     ;convert to ASCII

MOV [DATE+7], BH    ;save it

RET

;

;Program End

;

End

Boolean Algebra is the mathematics we use to analyse digital gates and circuits. We can use these “Laws of Boolean” to both reduce and simplify a complex Boolean expression in an attempt to reduce the number of logic gates required. Boolean Algebra is therefore a system of mathematics based on logic that has its own set of rules or laws which are used to define and reduce Boolean expressions.

The variables used in Boolean Algebra only have one of two possible values, a logic “0” and a logic “1” but an expression can have an infinite number of variables all labelled individually to represent inputs to the expression, For example, variables A, B, C etc, giving us a logical expression of A + B = C, but each variable can ONLY be a 0 or a 1.

Let A and B be the probabilities of solving the problem by George and James respectively.

Then P(A)=1/2 and P(B)=1/3.

The problem will be solved if it is solved at least by one of them also.

So, we need to find P(AUB).

By addition theorem on probability, we have

P(AUB) = P(A) + P(B)- P(A∩B).

P(AUB) = 1/2 +.1/3 – 1/2 * 1/3 = 1/2 +1/3-1/6 = (3+2-1)/6 = 4/6 = 2/3