Size of Sequence Number Field

Question

Consider a Go Back N sliding window protocol that uses a frame size of $2$KB to send data on a $10$KBPS link with a Round Trip Time of $100$ milli seconds.To achieve a link utilization of $50\%$, the minimum number of bits required to represent the sequence number field is  _____ .

---

Here per frame transmission time is 0.2 seconds & given RTT is 0.1 second.

How it is possible?

Am I misinterpreting the question?

Comments

before attempting the question , lets clear out somethings

1. its not 10KBPS(Bytes per second) , its 10Kbps(bits per second ),here K = 10^3
   ​​​​​​​
2. Packet size is 2KB = Kilo Bytes means 2*8 Kilo bits ,here K = 2^10

Answer 1

Transmission time= 2KB / 10KBPS

                            =2*2¹⁰ / 10 * 2¹⁰ sec

                               =1/5 sec= 0.2 sec=200ms

RTT= 100m =0.1 sec

then, propagation delay = 100/2=50 ms

So, 2a = 2t_p / t_t = 2 *50/200 = 1/2 

Min no. of bits in sequence no of field    ceil(log₂(1+2a))

                                                              =ceil(log(3/2)) = 1

Comments

@Mari Ganesh Kumar

"size of sender's windows must be equal to 1" That's a typo, thanks for pointing that out. It should be 

"size of receiver's windows must be equal to 1"  .I'll change it.

Yes, here Sender's window size is coming out to be 1, so it will be Stop & Wait.

But Stop & Wait is a specific case of GBn, it is GB1.

& even if the Sender's Window Size is 1, we will require 2 sequence numbers(or 1 bit sequence number field) to handle delayed ACKs. 

So Stop & wait do require sequence numbers.

---

@Anurag Pandey
I agree with you.
Stop & wait  is an algorithm designed for noiseless channel and hence, we don't require any sequence no.( Since there will be no frame loss)
GoBackN is designed for noisy channel hence, for GBn with sender window size 1 we require two sequence no.
 

---

Hello,

Please clear my doubt.see  efficiency=w/1+2a

so when efficeincy was given why didn't we consider it in getting value for w? why did we consider 100%.

Answer 2

Efficiency of GBN = N / 1+2a             so 0.5 = N / 1+2a where a= Tp / Tt, 

so Here N=0.5(1+2a)

2a = 100*10^-3 *10* 10^3 / 2*1024*8 = 0.061

N= 0.5 (1+0.061)=0.53 IN GBN  N+1 <= ASN implies 1.53<=ASN ,

ASN =2 so min no of bits =1

Answer 3

when using GO Back N sliding window protocol receiver window size will be 1

propogation delay is 50ms

tansmission time is 2kb/10kbps is 200ms

and sender window size if using 50% utilization

sender window size>=1/2 *(1+2a)

so a=transmission delay/transmission time

a=50ms/200ms

a=1/4

so (1+2a) =3/2

so sender window size is ceil(3/4)=1

available sequence no >=sender window size + receiver window size

available sequence no >=1+1

ASN>=2

bits require to represent sequence no is ceil(log₂2) is 1