GATE IT 2004 | Question: 83

Question

A $20$ $\text{Kbps}$ satellite link has a propagation delay of $400$ $\text{ms}$. The transmitter employs the "go back $n$ $ARQ$" scheme with $n$ set to $10$. Assuming that each frame is $100$ $\text{byte}$ long, what is the maximum data rate possible?

• A. $5$ $\text{Kbps}$
• B. $10$ $\text{Kbps}$​​​​​​​
• C. $15$ $\text{Kbps}$​​​​​​​
• D. $20$ $\text{Kbps}$​​​​​​​

Comments

here, max data rate = throughput

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Tt = 40 msec  ;  Tp = 400 msec
Effec. =  Window Size  /  [1 + (2 * Tp / Tt)]
Effec. =  10 / [1 + (2 * 400 / 40)]  =  10 / 21
Throughput = effec * BW = (10 / 21)  *  20 Kbps  =  9.52 Kbps

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$$\textbf{If we find the efficiency, we can find how much of the available bandwidth we are actually able to use.}$$

 

Answer 1

Answer: B
Transmission TIme $=\dfrac{100\times 8\text{bits}}{20\text{ Kbps}}=40\ ms$

Propagation Time $=400\ ms$

Efficiency $=\dfrac{\text{Window Size$\times$ Transmission Time}}{\text{(Transmission Time + 2$\times$ Propagation Time)}}$

$=\dfrac{10\times 40}{(40+2\times 400)}=0.476$

Maximum Data Rate$=0.476\times 20\text{ Kbps}=9.52\text{ Kbps}$

which is close to option B.

Comments

Data rate = throughput

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A satellite link has a propagation delay of 400 ms

 Look at hear what hapen actually we send a data from source to satellit and get back ACK so only 2*propagation dealy requir sources to .satellite(destination) hear RTT = 2* prpgation dealy

when A 1 mbps satellite link connects two ground stations then

 RTT is 4* propgation dealy
I think now all of you clear...
 

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Understanding Satellite Propagation Delay in Networking Problems

Scenario A: Question Gives the Distance (Altitude)

Example

"A satellite is at an altitude of 36,000 km..."

Meaning

The examiner is giving only the physical distance, so you must calculate the propagation delay yourself.

In satellite communication, the signal travels:

Ground Station → Satellite → Ground Station

This is one complete one-way communication path.

For acknowledgements, the signal again travels:

Ground Station → Satellite → Ground Station

Therefore, the total round-trip communication involves:

4 × (satellite altitude)

Hence, in such problems, the Round Trip Time (RTT) effectively becomes:

[
RTT = 4T_p
]

where:

[
T_p = \frac{\text{Satellite altitude}}{\text{Signal speed}}
]

Key Idea

When only altitude/distance is given, you must manually account for:

• upward propagation

• downward propagation

• ACK return upward

• ACK return downward

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Scenario B: Question Gives the Propagation Delay ((T_p))

Example

"A satellite link has a propagation delay of 400 ms..."

Meaning

Here, the examiner has already calculated the one-way propagation delay for the entire satellite link.

In networking terminology, the propagation delay of a link represents the complete one-way travel time from sender to receiver.

So the upward and downward satellite travel have already been included in the given (T_p).

Therefore, you should NOT multiply by 4 again.

You simply use the standard formula:

[
RTT = 2T_p
]

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Quick Summary

Given in Question	What it Means	RTT Used
Satellite altitude/distance	You calculate propagation manually	(4T_p)
Propagation delay (T_p) already given	Full one-way delay already included	(2T_p)

---

Easy Memory Trick

• Distance given → think physically → UP + DOWN + ACK UP + ACK DOWN → (4T_p)

• Propagation delay given directly → one-way already complete → just use (2T_p)

Answer 2

Option(B) is correct!

There are 3 ways to do  this question.
1.
Tp = 400 ms,  R.T.T = 2*400 ms = 800 ms
Tt = 40 ms
In one round trip, we only  send 10 packets = 10*100 = 1000 Bytes = 8000 bits.

In 800 ms we send 8000 bits, throughput = In 1 second, how many bits can be sent? 
1 ms = $\frac{8000}{800}$ = 10 bits, and in 1 second = 10 k bits 
2.
  $$\text{Throughput = efficiency*B.W}$$
efficiency = $\frac{n}{(1+2*a)}$ where $a=\frac{tp}{tt}$
efficiency = $\frac{10}{1+2*400/40}$ = $\frac{10}{21}$
Throughout = (10/21)*20Kbps = 10 Kbps (almost)

3.
R.T.T = 2*TP = 800 ms
Tt = 40 ms
in 1 R.T.T we can send maximum $\frac{800ms}{40ms}$ = 20 packets

But as mentioned in the question we are sending only 10 packets, it means efficiency is 50% (or) we are utilizing the bandwidth only 50%, so effective bandwidth or throughout will be 50% of total bandwidth i.e. 10 kbps.

 

Comments

@Manu Thakur @Puja Mishra

What puts in in trouble is why we did not taken RTT = 4 x PT. It is a satellite link right so 2 x PT will be the time needs to be taken from source to destination. Same from destination to source as well.

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@Manu Thakur  in your first approach, why haven't you included transmission time for sending 8000 bits?

Tt+2Tp=840 ms----------------can send 8000 bits,

1 ms------------------------------8000/840 =9.52 bits

1 s----------------------------------9.52 K bits =10 K bits(almost)

Answer 3

The definition of round trip time in Stop and wait protocol or Sliding window protocols is author dependent. Let us go through them one by one.

Definition 1(More popular):

The total time from the moment we start transmission of a frame till we receive the acknowledgment of the respective frame.

Round Trip Time(RTT) = Transmission Time + 2*(Propagation Time)

Resources that follow this definition:

Tanenbaum : https://csc-knu.github.io/sys-prog/books/Andrew%20S.%20Tanenbaum%20-%20Computer%20Networks.pdf

(Page 233)

Definition 2:

The total time from the moment we place a frame completely into the link till we receive the acknowledgment of the respective frame.

Round Trip Time(RTT) = 2*(Propagation Time)

Resources that follow this definition:

1 . Foruzan : https://dpvipracollege.in/wp-content/uploads/2023/01/Data-Communications-and-Networking-By-Behrouz-A.Forouzan.pdf   

(Page 322 | Example 11.4)

2. Washington University:  https://courses.cs.washington.edu/courses/cse461/17au/lectures/reliable_data_transfer.pdf 

(Page 9)

Conclusion:

So either or not we have to include the Transmission time in the Round trip time is completely author dependent. Although the definition 1 is more commonly used , some questions may require us to follow the definition 2 also. According to the following MIT question it is always better to check both.

Link : https://web.mit.edu/6.02/www/s2011/handouts/tutprobs/transport.html

 

Efficiency $=\dfrac{\text{Window Size$\times$ Transmission Time}}{\text{Round Trip Time(RTT)}}$

Transmission Time  $=\dfrac{100\times 8\text{bits}}{20\text{ Kbps}}=40\ ms$

Propagation Time $=400\ ms$

Using definition 1:

Efficiency $=\dfrac{\text{Window Size$\times$ Transmission Time}}{\text{(Transmission Time + 2$\times$ Propagation Time)}}$

$=\dfrac{10\times 40}{(40+2\times 400)}=0.476$

Maximum Data Rate $=0.476\times 20\text{ Kbps}=9.52\text{ Kbps}$  (None of the options matches)

 

Using definition 2:

Efficiency $=\dfrac{\text{Window Size$\times$ Transmission Time}}{\text{2$\times$ Propagation Time}}$

$=\dfrac{10\times 40}{2\times 400}=0.5$

Maximum Data Rate $=0.5\times 20\text{ Kbps}=10\text{ Kbps}$  (Option B matches)

This question requires us to use definition 2.

Comments

Thank you @akashkumarsen4 

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Good work brother

Answer 4

calculate transmission time = (100*8)/20 =40ms
throughput is 100*8*10/(40+2*400)= 9.52 Kbps = 10Kbps

Comments

but why 8 ??

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to convert byte into bit

Answer 5

Here is the answer

Thanks to Shiva Sagar Rao

Answer 6

Answer  is B

Given that;

Bandwidth = 20Kbps 

\(T_p\) = 400 mS

Window Size = \(N\) = 10

Frame Size = 100 Bytes

\[
\text{So, }T_t = \frac{100 \times 8}{20 \times 10^3} = \frac{800}{20000}  = 40\ \text{ms}
\]

RTT can be calculated in two ways (or we can say that, "RTT has 2 definitions varying from Author to Author") :

Way 1 : We INCLUDE \(T_t\) in RTT (the Popular one)

\[\text{RTT} = T_t + 2\times T_p\]

In this case, \[\text{Efficiency} = \frac{N \times T_t}{T_t + 2 \times T_p} = \frac{10 \times 40}{40 + 2 \times 400} = \frac{400}{840} = 0.476\] 

So, maximum data rate possible (or maximum bandwidth consumed) = \[0.476 \times 20 Kbps = 9.52 Kbps\]

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Way 2 : We DO NOT INCLUDE \(T_t\) in RTT 

\[\text{RTT} =  2\times T_p\]

In this case, \[\text{Efficiency} = \frac{N \times T_t}{2 \times T_p} = \frac{10 \times 40}{2 \times 400} = \frac{400}{800} = 0.5\] 

So, maximum data rate possible (or maximum bandwidth consumed) = \[0.5 \times 20 Kbps = 10 Kbps\]

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In the above question there is no any option which equals \(9.52 Kbps\) but there is an option which equals \(10 Kbps\).

So, the question was framed on the assumption that RTT doesn't include \(T_t\).

Hence, B is the correct option.
 

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There are some GATE PYQs which follow the assumption that RTT includes \(T_t\) like this one GATE CSE 2003 | Question : 84.

Solve these two questions and you will get full clarity on \(T_t\) being included and excluded from RTT.