GATE CSE 2015 Set 3 | Question: 28

Question

Consider a network connecting two systems located $8000$ $\text{Km}$ apart. The bandwidth of the network is $500 \times 10^6$ $\text{bits}$ per second. The propagation speed of the media is $4 \times 10^6$ $\text{meters}$ per second. It needs to design a Go-Back-$N$ sliding window protocol for this network. The average packet size is $10^7$ $\text{bits}$. The network is to be used to its full capacity. Assume that processing delays at nodes are negligible. Then, the minimum size in bits of the sequence number field has to be ______.

Comments

full capacity means 100% efficiency

---

👍

---

GO to the Root of Concept: Video Explanation with timestamp: Sliding Window PYQs| GATE 2009 Q57, 58| 2006 Q46, 64| 2003, 2014, 2015, 2004, 2016, 2007| With NOTES

Answer 1

”The network is to be used to its full capacity”, Therefore, we can assume efficiency (E) = 1.

Transmission Time (Tt)= Packet Size/Bandwidth = 10^7 bits/(500 * 10^6)

Progation Time (Tp)= Distance/Velocity = 8000KM/4 * 10^6 meter per second = (8000 * 10^3)/(4 * 10^6) = 2s

Channel Capacity (a) = Tp/Tt = 2/(10^7 bits/(500 * 10^6)) = 100 bits

Efficiency = Windows Size/ (1 + 2a)

1 = Ws/(1 + 200)     Ws = 201

 total number of sequence numbers required : Sender WIndows SIze (201) + Receiver Windows SIze (1)

sequence Numbers = 202

Sequence bit required = Logbase2(202)

Answer = 8 bits

 

Answer 2

Answer is 8 bits.

Comments

how it is 8

---

In order to achieve full utilization, sender has to keep on sending frames till the acknowledgement arrives for the first frame.

Time taken for acknowledgement to arrive is 2 times propagation delay.

(we can ignore transmission delay at receiver as it is negligible)

One way propagation time = 8000x10^3 / (4x10^6)

= 2 secs

 

Time taken to transmit one frame = 10^7/(500x10^6)

=0.02 secs

 

No of frames that can be transmitted in 4 secs  = 4/0.02

=200

 

Hence minimum number of bits required for sequence numbers till 200 is 8 (as 2^8 = 256)

---

So, total number of frames that can be sent = 200 + 1 =201.

Please confirm.