Education for ALL: Kathford College

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Access Control List (ACL): Differences between Standard ACL and Extended ACL

It is a Layer 3 security which controls the flow of traffic from one router to another.
It is also called as Packet Filtering Firewall.

Types of ACL

Standard ACL
Extended ACL
Named ACL

Standard Access List

The access-list number lies between 1 - 99.
Can block a Network, Host and Subnet.
Two-way communication is stopped.
All services are blocked.
Implemented closest to the destination.

Extended Access List

The access-list number lies between 100 - 199.
Can block a Network, Host, Subnet and Service.
One-way communication is stopped.
Selected services can be blocked.
Implemented closest to the source.

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C Program for Least Square Method (Regression Analysis)

#include<stdio.h>

#include<conio.h>

void main()

{

float x[100],y[100],sumx=0,sumx2=0,sumy=0,sumyx=0,b,a;

int i,n;

printf("Enter n: ");

scanf("%d",&n);

printf("Enter the values:\n");

for(i=0;i<n;i++)

{

printf("Enter x[%d] and y[%d]: ",i,i);

scanf("%f%f",&x[i],&y[i]);

}

for(i=0;i<n;i++)

{

sumx=sumx+x[i];

sumx2=sumx2+x[i]*x[i];

sumy=sumy+y[i];

sumyx=sumyx+y[i]*x[i];

}

//for y=ax+b

b=(sumx2*sumy-sumyx*sumx)/(n*sumx2-sumx*sumx);

a=(n*sumyx-sumx*sumy)/(n*sumx2-sumx*sumx);

printf("\nHence, the required eqn is y = %fx + %f",a,b);

getch();

}

OUTPUT:

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How to construct a Bezier curve?

ALGORITHM:

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Liang-Barsky Line Clipping Algorithm

For given window with (XWmin, YWmin) and (XWmax, YWmax), the given line has end-points (X1, Y1) and (X2, Y2).
Calculate
Calculate pk, qk and rk for k = 1, 2, 3, 4 as:
If pk=0 for some k, then the line is parallel to clipping boundary. Now test qk:

If one qk < 0 for these k, then the line is outside.

If all qk ≥ 0 for these k, then some portion of the line is inside.

For all pk < 0 (i.e. line proceeds from outside to inside the boundary), calculate

to determine intersection point with the possibly extended clipping boundary k and obtain a new starting point for the line at u1.
For all pk > 0 (i.e. line proceeds from inside to outside the boundary), calculate

to determine intersection point with the possibly extended clipping boundary k and obtain a new end point for the line at u2.
If

discard the line.
Else, the line is now between
Hence,
Updated points are

8085 ALP: ASCENDING ORDER in a block of DATA

Suppose, there is an array of ten 8-bit data stored in memory addresses from 2030h to 2039h.
Starting from 2030h, we compare the data of the first location with that of the second location.
If the data of the first location is greater than that of the second location, they are swapped.
Running external loop for 9 times, 8 times, 7 times and so on, we run every internal loop as 9 times, 8 times, 7 times and so on for every corresponding external loop. For e.g. for external loop 9, we run respective internal loop 9 times.
Doing so, after every external loop, the largest number will be at the last memory location.

MVI C, 09h

AGAIN: MOV D, C

LXI H, 2030h

NEXT: MOV A, M

INX H

CMP M

JC SKIP

MOV B, M

MOV M, A

DCX H

MOV M, B

INX H

SKIP: DCR D

JNZ NEXT

DCR C

JNZ AGAIN

HLT

Computer Networks (CN) LAB - Application Layer Services

Setup Application Layer Services w/ Servers in Packet Tracer - Part 1

Setup Application Layer Services w/ Servers in Packet Tracer - Part 2

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C Program to Sort Elements in Lexicographical Order (Dictionary Order) with Input

#include<stdio.h>

#include<conio.h>

#include<string.h>

void main()

{

char word[12][50],temp[50];

int i,j;

printf("Enter 12 words:\n");

for(i=0;i<12;++i)

gets(word[i]);

for(i=0;i<11;i++)

for(j=i+1;j<12;j++)

{

if(strcmp(word[i],word[j])>0)

{

strcpy(temp,word[i]);

strcpy(word[i],word[j]);

strcpy(word[j],temp);

}

printf("\nIn lexicographical order: \n");

for(i=0;i<12;++i)

puts(word[i]);

getch();

}

OUTPUT:

C Program to Store Information (name, roll and marks) of a Student Using Structure >>

Differences between 8086 and 8088 Microprocessor

8086

8088

The instruction Queue is 6 byte long.

The instruction Queue is 4 byte long.

In 8086, memory divides into two banks

-even or lower bank

-odd or higher bank

The memory in 8088 does not divide into two banks.

The data bus of 8086 is 16-bit wide

The data bus of 8088 is 8-bit wide.

It has BHE (bar) signal on pin no. 34 & there is no SSO (bar) signal.

It does not have BHE (bar) signal on pin no. 34 & has only SSO (bar) signal. It has no S7 pin.

Control pin in 8086 is M/IO (bar).

Control pin in 8088 is IO/M (bar).

It needs one machine cycle to R/W signal if it is at even location otherwise it needs two.

In 8086, all address & data Buses are multiplexed.

In 8088, address bus; AD₇- AD₀ buses are multiplexed.

It needs two IC 74343 for de-multiplexing AD₀-AD₁₉.

It needs one IC 74343 for de-multiplexing AD₀-AD₇.

Maximum supply current 360mA.

Maximum supply current 340mA.

Three clock speed: 5, 8, 10 MHz

Two clock speed: 5, 8 MHz

PROGRAM TO COUNT NUMBER OF 1s IN REGISTER A

MVI C,00H ;COUNTER FOR 1

MVI B, 8 ;DOWN COUNTER

;FOR 8-BIT REGISTER A

UP: RAL

JNC PASS

INR C

PASS: DCR B

JNZ UP

HLT

Final Year Project Report

It's the final year project report in 8th semester "Wireless Weather Monitoring System" using Arduino.

8085 Program - Transferring block of data with "Condition"

Q. There are two tables of data stored at 40A1H and 40B1H having 10 data each. Write a program to store the data in the first table to third table starting from address 40C1H if the corresponding data in the first table is greater than the second and table else store FF in the third table.

Program:

LXI B,40A1H ;assume A1 as the starting number

LXI H,40B1H

LXI D,40C1H

START: LDAX B

CMP M

JC NEXT

STAX D

INX B

INX H

INX D

MOV A,C