How to determine the critical crack size for a plate
1 answer:
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Answer:
Check the explanation
Explanation:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define MAX_LIMIT 50
int checkComma(char *input)
{
int flag = 0;
for(int i = 0; i < strlen(input); i++)
{
if(input[i] == ',')
{
flag = 1;
break;
}
}
return flag;
}
int main(void)
{
char input[MAX_LIMIT];
char *words[2];
char delim[] = ", ";
printf("\n");
do
{
printf("Enter input string: ");
fgets(input, MAX_LIMIT, stdin);
size_t ln = strlen(input) - 1;
if (*input && input[ln] == '\n')
input[ln] = '\0';
if(strcmp(input, "q") == 0)
{
printf("Thank you...Exiting\n\n");
exit(1);
}
else
{
if(checkComma(input) == 0)
{
printf("No comma in string.\n\n");
}
else
{
char *ptr = strtok(input, delim);
int count = 0;
while(ptr != NULL)
{
words[count++] = ptr;
ptr = strtok(NULL, delim);
}
printf("First word: %s\n", words[0]);
printf("Second word: %s\n\n", words[1]);
}
}
}while(strcmp(input, "q") != 0);
return 0;
}
Kindly check the attached image below for the output.
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 =