Question

Consider the bandwidth of the line is 1.2 Mbps, packet size is 1 KB and round-trip time
is 50 msec. What is the link utilization in stop and wait?​

Answer 1

If the bandwidth of the line is 1.2 Mbps, packet size is 1 KB and RTT  is 50 msec, the link utilization in stop and wait is 0.12 or 12%.

Step-by-step explanation:

It is given,

The bandwidth of the line = 1.2 Mbps = 1.2 * 10⁶ bits per sec

Packet size = 1 KB = 2¹⁰ * 8 bits

Round Trip Time, RTT = 50 msec

The required formula for finding the link utilisation or efficiency in stop and wait:

ɳ = $\frac{1}{1+2\frac{Tp}{Tt}}$ …… (i)

Step 1:

Firstly we will find the value for Transmission delay “Tt”.

Therefore,  

Transmission Delay, Tt  

= [Packet size]/[Bandwidth]

= [2¹⁰ * 8] / [1.2 * 10⁶]

= 6.82 * 10⁻³ sec

= 6.82 msec

Step 2:

Secondly, we will find the value for Propagation Delay “Tp”.

Therefore,  

Propagation Delay, Tp

= [Round trip time] / 2

= 50 / 2

= 25 msec

Step 3:

Now, substituting the value of Tt & Tp in the formula in eq. (i) and, we will find the efficiency or the link utilisation in stop and wait.

Therefore,

ɳ = $\frac{1}{1+2\frac{25}{6.82}}$ = $\frac{1}{1+2*3.66}$ = $\frac{1}{8.32}$ = 0.12 or 12%

Thus, the efficiency or the link utilization in stop and wait is 0.12 or 12%.

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Answer 2

and wait is 0.12 or 12%.

Step-by-step explanation:

It is given,

The bandwidth of the line = 1.2 Mbps = 1.2 * 10⁶ bits per sec

Packet size = 1 KB = 2¹⁰ * 8 bits

Round Trip Time, RTT = 50 msec

The required formula for finding the link utilisation or efficiency in stop and wait:

ɳ = \frac{1}{1+2\frac{Tp}{Tt}}

1+2

Tt

Tp

1

…… (i)

Step 1:

Firstly we will find the value for Transmission delay “Tt”.

Therefore,

Transmission Delay, Tt

= [Packet size]/[Bandwidth]

= [2¹⁰ * 8] / [1.2 * 10⁶]

= 6.82 * 10⁻³ sec

= 6.82 msec

Step 2:

Secondly, we will find the value for Propagation Delay “Tp”.

Therefore,

Propagation Delay, Tp

= [Round trip time] / 2

= 50 / 2

= 25 msec

Step 3:

Now, substituting the value of Tt & Tp in the formula in eq. (i) and, we will find the efficiency or the link utilisation in stop and wait.

Therefore,

ɳ = \frac{1}{1+2\frac{25}{6.82}}

1+2

6.82

25

1

= \frac{1}{1+2*3.66}

1+2∗3.66

1

= \frac{1}{8.32}

8.32

1

= 0.12 or 12%

Thus, the efficiency or the link utilization in stop and wait is 0.12 or 12%.

----------------------------------------------------------------------------------

Also View:

How does TCP determine the round trip time and time out interval?Discuss the algorithm.

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An IP packet with 2500 bytes of data (plus header) passes through an IP network with MTU = 500 bytes. How many additional bytes will be delivered at the destination?

https://brainly.in/question/11782992