Frames are generated at node A
sent to node C through node B.
Determine the minimum transmission rate required between nodes B and C
so that the buffers in node B will not flooded, based on the following
figures:

A<---4000--->B<-1000->C

 (I) the data rate between A and B is 100 Kbps. (ii) the propagation
delay is 10 microsec/mile for both lines, (iii) there are full duplex
lines between the nodes, (iv) all data frames are 1000 bits long; ACK
frames are separate frames of negligible length, (v) between A and B,
a sliding window protocol with a window size of 3 is used, (vi)
between B and C, stop and wait is used, (vii) there are no errors.

http://66.102.9.104/search?q=cache:DlHyHBvtIwQJ:www.ignoutnf.org/mca/09jun99.htm

Hi rndgroup!!


I assume that the numbers 4,000 and 1,000 are the distance between nodes in miles.

In order to not flood buffers in node B we must balance what comes in
from node A with what goes out to node C.
With the above statement in mind we can start to solve the problem:

Since the size of an ACK is negligible, we assume that
Time for ACK = Propagation time

From A to B:
Propagation Time = 4,000 miles * 10 microsec/mile =
                 = 4,000 miles * 0.01 milisec/mile = 
                 = 40 msec (milisec)

Transmission time per frame = 1,000 bits/100 kbps =
                            = 1,000 bits/100,000 bits/sec =
                            = 0.01 sec =
                            = 10 msec

From B to C:
Propagation time = 1,000 miles × 10 microsec/mile = 
                 = = 1,000 miles * 0.01 milisec/mile = 
                 = 10 msec (milisec)

Transmission time per frame = X = 1,000 bits/R ,
where R = data rate between B and C (unknown)

Due (v) assumption node A A can then send three frames to B and must
then stop in order for an ACK to be received before transmitting
additional frames. The first frame takes 10 msec to transmit, the last
bit of the first frame arrives at node B 40 msec after it was
transmitted, and therefore 50
msec after the frame transmission began. It will take an additional 40
msec for node B's acknowledgment to return to node A. This means that
A can transmit 3 frames in 90 msec.
Node B can transmit one frame to node C at a time using stop and wait,
and it takes for each frame to be received at node C:
Propagation time + Transmition times = (10 + X) msec  
and the acknowledgement will be back after another 10 msec. Thus, node
B can transmit one frame every (20 + X) msec, or 3 frames every:
(60 + 3*X) msec.

Recalling the initial statement (we must balance what comes in from
node A with what goes out to node C), so we have that the time that
takes node A to transmit 3 frames to node B must be the same time that
node B takes to transmit 3 frames to node C, in other words:
(60 + 3*X) msec = 90 msec

Then:
X = (90 msec - 60 msec)/3 =
  = 10 msec 

Finally:
R = 1,000 bits/X msec = 
  = 1 Kbit/10 msec =
  = 1 Kbit/0.010 sec =
  = 100 Kbps


I hope that this helps you. Feel free to request for a clarification
if you need it.

Best regards.
livioflores-ga