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Gennadij [26K]
3 years ago
8

Consider an Erlang loss system. The average processing time is 3 minutes. The average inter-arrival time is 3 minutes. The numbe

r of servers is 3. What is r (sometimes referred to as the offered load)
Computers and Technology
1 answer:
AnnyKZ [126]3 years ago
3 0

Answer:

r = 1

Explanation:

Average processing time ( p ) = 3 minutes

Average inter-arrival time ( a ) = 3 minutes

number of servers ( m ) = 3

<u>Determine the value of r </u>

r ( offered load ) = p/a

                          = 3 / 3  = 1

∴ value of r ( offered load ) = 1

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When the email was sent as a group email 

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2 years ago
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Consider the following declaration: (1, 2) double currentBalance[91]; In this declaration, identify the following: a. The array
Rom4ik [11]

Answer:

Following are the solution to the given choices:

Explanation:

Given:

double currentBalance[91];//defining a double array  

In point a:  

The name of the array is= currentBalance.  

In point b:  

91 double values could be saved in the array. It requires 8bytes to hold a double that if the array size is 91*8 = 728

In point c:  

Each element's data type is double.

In point d:  

The array index range of values is between 0 and 90 (every array index starts from 0 and ends in N-1, here N=91).  

In point e:  

To access first element use currentBalance[0], for middle currentBalance[91/2] , for last currentBalance[90]

3 0
2 years ago
Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For f
julia-pushkina [17]

Complete Question:

Recall that with the CSMA/CD protocol, the adapter waits K. 512 bit times after a collision, where K is drawn randomly. a. For first collision, if K=100, how long does the adapter wait until sensing the channel again for a 1 Mbps broadcast channel? For a 10 Mbps broadcast channel?

Answer:

a) 51.2 msec.  b) 5.12 msec

Explanation:

If K=100, the time that the adapter must wait until sensing a channel after detecting a first collision, is given by the following expression:

  • Tw = K*512* bit time

The bit time, is just the inverse of the channel bandwidh, expressed in bits per second, so for the two instances posed by the question, we have:

a) BW  = 1 Mbps = 10⁶ bps

⇒ Tw = 100*512*(1/10⁶) bps = 51.2*10⁻³ sec. = 51.2 msec

b) BW = 10 Mbps = 10⁷ bps

⇒ Tw = 100*512*(1/10⁷) bps = 5.12*10⁻³ sec. = 5.12 msec

5 0
3 years ago
(1) Prompt the user for a string that contains two strings separated by a comma. (1 pt) Examples of strings that can be accepted
denis-greek [22]

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.

4 0
3 years ago
Which of the following is not a characteristic of a motorcycle?
MrRa [10]

Answer:

the correct answer is A

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