Why would you activate more than one nic on a pc?
1 answer:
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Answer:
the average arrival rate \lambda in units of packets/second is 15.24 kbps
the average number of packets w waiting to be serviced in the buffer is 762 bits
Explanation:
Given that:
A single channel with a capacity of 64 kbps.
Average packet waiting time in the buffer = 0.05 second
Average number of packets in residence = 1 packet
Average packet length r = 1000 bits
What are the average arrival rate \lambda in units of packets/second and the average number of packets w waiting to be serviced in the buffer?
The Conservation of Time and Messages ;
E(R) = E(W) + ρ
r = w + ρ
Using Little law ;
r = λ × T_r
w = λ × T_w
r / λ = w / λ + ρ / λ
T_r =T_w + 1 / μ
T_r = T_w +T_s
where ;
ρ = utilisation fraction of time facility
r = mean number of item in the system waiting to be served
w = mean number of packet waiting to be served
λ = mean number of arrival per second
T_r =mean time an item spent in the system
T_w = mean waiting time
μ = traffic intensity
T_s = mean service time for each arrival
the average arrival rate \lambda in units of packets/second; we have the following.
First let's determine the serving time T_s
the serving time T_s
= 0.015625
now; the mean time an item spent in the system T_r = T_w +T_s
where;
T_w = 0.05 (i.e the average packet waiting time)
T_s = 0.015625
T_r = 0.05 + 0.015625
T_r = 0.065625
However; the mean number of arrival per second λ is;
r = λ × T_r
λ = r / T_r
λ = 1000 / 0.065625
λ = 15238.09524 bps
λ ≅ 15.24 kbps
Thus; the average arrival rate \lambda in units of packets/second is 15.24 kbps
b) Determine the average number of packets w waiting to be serviced in the buffer.
mean number of packets w waiting to be served is calculated using the formula
w = λ × T_w
where;
T_w = 0.05
w = 15238.09524 × 0.05
w = 761.904762
w ≅ 762 bits
Thus; the average number of packets w waiting to be serviced in the buffer is 762 bits
Answer:
Check the explanation
Explanation:
To solve the question above, we will be doing some arithmetic and bit wise operations on enum values. in addition to that, we are asserting the array with INTCOUNT and also printing the size of array i.e 12 and we are adding BIG_COUNT enum value to 5 and the result is 25.
#include<stdio.h>
enum EnumBits
{
ONE = 1,
TWO = 2,
FOUR = 4,
EIGHT = 8
};
enum Randoms
{
BIG_COUNT = 20,
INTCOUNT = 3
};
int main()
{
// Basic Mathimatical operations
printf("%d\n",(ONE + TWO)); //addition
printf("%d\n",(FOUR - TWO)); //subtraction
printf("%d\n",(TWO * EIGHT)); //multiplication
printf("%d\n",(EIGHT / TWO)); //division
// Some bitwise operations
printf("%d\n",(ONE | TWO)); //bitwise OR
printf("%d\n",(TWO & FOUR)); //bitwise AND
printf("%d\n",(TWO ^ EIGHT)); //bitwise XOR
printf("%d\n",(EIGHT << 1)); //left shift operator
printf("%d\n",(EIGHT >> 1)); //right shift operator
// Initialize an array based upon an enum value
int intArray[INTCOUNT]; // declaring array
// Have a value initialized be initialized to a static value plus
int someVal = 5 + BIG_COUNT; //adding constant and BIG_COUNT variable
printf("%d\n",someVal); //value will be 25
printf("%d",sizeof(intArray)); //value will be 12
return 0;
}
Kindly check the attached image below for the Code and Output Screenshots:
