Question

How many ATM cells (described in Example 28) can be

transmitted in 10 seconds over a link operating at the following

rates?

a)128 kilobits per second (1 kilobit=1000 bits)

b)300 kilobits per second

c)1 megabit per second (1 megabit = 1,000,000 bits)

Short answer

a) 3018

b) 7076

c) 23585

Step-by-step solution

Step 1:

We have to find number of ATM cells which can be transmitted in 10 second at 128 kilobits per second (1 kilobit = 1000 bits) rate.

Now we ATM transmitting 128 kilobits per second

= 1281000

=128000 bits per second

We give each cell is of 53 byte long, which means that it is

= 538 = 424 bits long (since 1 byte = 8 bits)

So, the number of cells that can be transmitted in 10 seconds

=$\frac{128000\times 10}{424}=3018$

Step 2:

We have to find number of ATM cells which can be transmitted in 10 second at 300 kilobits per second (1 kilobit = 1000 bits) rate.

Now we ATM transmitting 128 kilobits per second

= 3001000

=300000 bits per second

We are given each cell is of 53 byte long, which means that it is

= 538 = 424 bits long (since 1 byte = 8 bits)

So, the number of cells that can be transmitted in 10 seconds

=$\frac{300000\times 10}{424}=7076$

Step 3:

We have to find number of ATM cells which can be transmitted in 10 second at 1 megabits per second (1 kilobit = 1000 bits) rate.

Now we ATM transmitting 1 megabits per second

= 1000000

=1000000 bits per second

We are given each cell is of 53 byte long, which means that it is

= 538 = 424 bits long (since 1 byte = 8 bits)

So, the number of cells that can be transmitted in 10 seconds

=$\frac{1000000\times 10}{424}=23585$