Computer Networks DRDO-2008

Question

Two ground stations are connected by a 10Mbps satellite link. The altitude of the satellite is 36,000km and the speed of the signal is 3x108 m/sec. What should be the packet size for channel utilization of 50% using GBN sliding window protocol. Window size is 100. Assume that the acknowledgement packets are negligible in size and there are no errors during communcations.

• A. 1.5 Kbytes
• B. 3 Kbytes
• C. 4.5 Kbytes
• D. 6 Kbytes

Comments

@srestha I am getting 2412.060 bits... Plz help me where I am going wrong.

Answer 1

Here the altitude of the satellite is 36,000km, i.e. $d=36000 \times 1000$ m

speed of the signal  $(v) = 3\times 10^8 m/s$

Propagation delay $T_p =\frac{2 \times 36000 \times 1000 }{3 \times 10^8} =0.24 s.$

(Packet must propagate to the satellite and then back)

Let packet size = L

bandwidth of channel B= 10Mbps

then transmission time $T_t =L/(10\times 10^6)$

$\text{Efficiency} = \frac{\text{Data transmitted}}{\text{Maximum data that could be transmitted}}$

In Go-Back-N, before first ACK comes, $N$ packets can be sent. ACK comes back after $T_t + T_p + T_p +T_{ack}$. $T_{ack} = 0$ as per question. So, we consider this time and during $T_t$ a packet of size $L$ can be sent. So,

$\text{Efficiency} = \frac{N \times L} { L + 2T_p \times B}$

$0.5 = \frac{100L}{L + 2\times 0.24 \times 10 \times 10^6}$

$0.5L + 2400000 = 100L$

$L = 2400000/99.5 = 24120.6 \text{ bits} \approx 3 \text{ Kbytes}$

Comments

Even if u multiply by 2 in Tt , you will get nearly the same answer just little difference in decimals.

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I think transmission time is time initially taken by the packet to get on the transmission link. And as soon as the bits of first packet reaches satellite from the source station. it's not accumulated to form a packet on the satellite, it will be trasmitted as soon as it is received by the satellite.
Only in this scenerio, we can ignore transmission time taken by the satellite to send packet to the destination station.

Correct me if i am wrong.

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@srestha mam, Why are we taking L in the denominator of efficiency.. See arjun sir's comment here https://gateoverflow.in/15759/throughput-calculation

Answer 2

=> (Efficiency) $\eta = \frac{N}{1 + \frac{4 * T_{P}}{T_{t}}}$

=> $\frac{1}{2} = \frac{100}{1 + \frac{4 * 12 * 10^{-2} * 10 * 10^{6}}{L}}$

=> L = $\frac{48 * 10^{5}}{199}$

=> L = 24120.6 bits = 3015 bytes = 3.015Kbytes ≌ 3Kbytes

Comments

then we would definitely consider TT at router, see its all acc to my perspective, just trying to convey what i understood.

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so if we had a router instead of a satellite, can we say the following???:

 useful time (u) = (2*transmission time)
total time (t)= (2*transmission time + 4*propagation time)

Efficiency = (N * u)/t, where N = 100

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useful time(u) will be only "transmission time",  not 2 * transmission time, (routers transmission time will not be taken into consideration in useful time however it is considered in total time)

coz we are calculating the efficiency of sender.