What is the effect on line utilization, if we increase the number of frames for a constant message size?

Question

a) lower line efficiency

b) higher line efficiency

c) no change in line efficiency

d) no relation between line efficiency and frame size

Comments

If a constant message gets divided into multiple frames, each frame will receive its own header at the data link layer. So, effectively the total header size should increase than before where we only had 1 frame. This should in turn increase our transmission time because we are now transmitting more header than before, which should be increasing our line utilization.

---

Hey,

Wouldn’t dividing a message into multiple frames will increase the total data transmitted over the line (because of Link layer's header and trailer)?
I guess Yes, therefore link utilization should decrease.

Answer 1

No change in line efficiency.

 

Line efficiency will increase , only when we have some hops between sender and receiver.

For this question we will consider that there is no hop between sender and receiver so , dividing a constant message in k frames or increase the number of frames to ( k+i),will not effect link utilization.

Comments

I think it will decrease the efficiency of a link,the reason is for this is given above by @Sohini.

Moreover if we break our message into smaller frames then for each frames overhead of header will increase and thus lower line efficiency.

---

line utilization will remain same. Line utilization will be affected if we are sending more than one frame at a time through the same link and in that case it will surely decrease

---

the line utilization will decrease in terms of overhead bytes sent like the bytes of header consumed and acks recieved per frame

Answer 2

Efficiency should decrease,

efficiency= 1/ 1+ 2a where a = (d/v) * (B/ L) so when L increase B/L decrease that means (d/v) * (B/ L) is also decreasing.

d- distance, v- velocity, B- bandwidth, L - frame length.

Increase number of frames increase frame length L.

So , option A is correct.

Comments

Why do you assume that stop&wait  protocol is used, why not SR or GBN?

---

@Bikram your solution is incorrect you said that as a decreases efficiency decreases but that is not the case coz efficiency = 1/1+2a.

---

But message size is constant hereR

---

@Bikram if L increases then efficiency should increase.. how can it decrease? pls clear

Answer 3

yes i agree with you @Bikram if it was a normal case but keeping the message size constant 

let L is divided into n packets of lenght l then L = n*l 

Transmission time T = n*l / B which is equivalent to L / B

So no change in transmission time but there will be change in propogation time (P)

IN STOP AND WAIT – 

the propagation time will become = n*P So there will be low utilization 

IN sliding window protocol –

there will be no change in efficiency as it does not waste propogation time so n*P will be utilized anyway