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2 answers:
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Answer:
a.) k² - 3k + 3
b.) 1/(1 - k)²
c.)
Explanation:
a.) A packet can make 1,2 or 3 hops
probability of 1 hop = k ...(1)
probability of 2 hops = k(1-k) ...(2)
probability of 3 hops = (1-k)²...(3)
Average number of probabilities = (1 x prob. of 1 hop) + (2 x prob. of 2 hops) + (3 x prob. of 3 hops)
= (1 × k) + (2 × k × (1 - k)) + (3 × (1-k)²)
= k + 2k - 2k² + 3(1 + k² - 2k)
∴mean number of hops = k² - 3k + 3
b.) from (a) above, the mean number of hops when transmitting a packet is k² - 3k + 3
if k = 0 then number of hops is 3
if k = 1 then number of hops is (1 - 3 + 3) = 1
multiple transmissions can be needed if K is between 0 and 1
The probability of successful transmissions through the entire path is (1 - k)²
for one transmission, the probility of success is (1 - k)²
for two transmissions, the probility of success is 2(1 - k)²(1 - (1-k)²)
for three transmissions, the probility of success is 3(1 - k)²(1 - (1-k)²)² and so on
∴ for transmitting a single packet, it makes:
∞ n-1
T = ∑ n(1 - k)²(1 - (1 - k)²)
n-1
= 1/(1 - k)²
c.) Mean number of required packet = ( mean number of hops when transmitting a packet × mean number of transmissions by a packet)
from (a) above, mean number of hops when transmitting a packet = k² - 3k + 3
from (b) above, mean number of transmissions by a packet = 1/(1 - k)²
substituting: mean number of required packet =
Answer:
Here is the C++ program:
#include<iostream> // to use input output functions
using namespace std; //to identify objects like cin cout
void StartChar(string str) { // function that takes a word string as parameter and returns the first letter of that word in capital
int i;
char c;
do //start of do while loop
{
for (int i = 0; i < str.length(); i++) { //iterates through each character of the word str
if(str.length()>25){ //checks if the length of input word is greater than 25
cout<<"limit exceeded"; } //displays this message if a word is more than 25 characters long
c = str.at(i); // returns the character at position i
if (! ( ( c >= 'a' && c <= 'z' ) || ( c >= 'A' && c <= 'Z' ) ) ) { //checks if the input word is contains characters other than the alphabet
cout<<str<<" is not a word!"<<endl; break;} //displays this message if user enters characters other than the alphabet
if (i == 0) { //check the first character of the input word
str[i]=toupper(str[i]); //converts the first character of the input word to uppercase
cout<<str<<" starts with letter "<<str[i]<<endl; } // prints the letter that the word starts with in capital letter
} cout<<"Enter a word: "; //prompts user to enter a word
cin>>str; //reads input word from user
}while(str!="#"); } //keeps prompting user to enter a word until the user enters #
int main() //start of the main() function body
{ string str; //declares a variable to hold a word
cout<<"Enter a word: "; //prompts user to enter a word
cin>>str; //reads input word from user
StartChar(str); } //calls function passing the word to it
Explanation:
The program prompts the user to enter a word and then calls StartChar(str) method by passing the word to the function.
The function StartChar() takes that word string as argument.
do while loop starts. The for loop inside do while loop iterates through each character of the word string.
First if condition checks if the length of the input word is greater than 25 using length() method which returns the length of the str. For example user inputs "aaaaaaaaaaaaaaaaaaaaaaaaaaa". Then the message limit exceeded is printed on the screen if this if condition evaluates to true.
second if condition checks if the user enters anything except the alphabet character to form a word. For example user $500. If this condition evaluates to true then the message $500 is not a word! is printed on the screen.
Third if condition checks if the first character of input word, convert that character to capital using toupper() method and prints the first letter of the str word in capital. For example if input word is "Computer" then this prints the message: Computer starts with the letter C in the output screen.
The program keeps prompting the user to enter the word until the user enters a hash # to end the program.
The output of the program is attached.
