B are primary and C are secondary.
Answer:
a. Utilization = 0.00039
b. Throughput = 50Kbps
Explanation:
<u>Given Data:</u>
Packet Size = L = 1kb = 8000 bits
Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps
RTT = 20 msec
<u>To Find </u>
a. Sender Utilization = ?
b. Throughput = ?
Solution
a. Sender Utilization
<u>As Given </u>
Packet Size = L = 8000 bits
Transmission Rate = R = 1 Gbps = 1 x 10⁹ bps
Transmission Time = L/R = 8000 bits / 1 x 10⁹ bps = 8 micro-sec
Utilization = Transmission Time / RTT + Transmission Time
= 8 micro-sec/ 20 msec + 8 micro-sec
= 0.008 sec/ 20.008 sec
Utilization = 0.00039
b. Throughput
<u>As Given </u>
Packet Size = 1kb
RTT = 20ms = 20/100 sec = 0.02 sec
So,
Throughput = Packet Size/RTT = 1kb /0.02 = 50 kbps
So, the system has 50 kbps throughput over 1 Gbps Link.
The employee would access an intranet to maintain security.
Application Programming Interface.
Answer:
- def average_num_in_file(fileName):
- with open(fileName) as file:
- rows = file.readlines()
-
- sum = 0
- count = 0
- for x in rows:
- sum += float(x)
- count += 1
-
- average = sum / count
- return average
-
- print(average_num_in_file("cans.txt"))
Explanation:
The solution code is written in Python 3.
Firstly create a function that take one parameter, fileName (Line 1).
Open the file stream and use readlines method to read the data from the file (Line 2-3). Create variable sum and count to tract the total of the number from text files and number of data from the file (Line 5-6). Use a for loop to loop over each row of the read data and add the current value of each row to sum and increment the count by one (Line 7-9).
After the loop, calculate the average (Line 11) and return the result (Line 12).
At last, we can test the function by passing the cans.txt as argument (Line 14).
