1 Communicatienetwerken Oefeningenreeks 1 Hoofdstuk : Transport Laag Woensdag 10 oktober 2007

2 VRAAG 1 S 100 kbit/s of 10 Mbit/s km Y X 10 Gbit/s vezel : 2000 km 100 m TP 100 m coax 10 x MSS of 100 x MSS A = FTP client, B = FTP server Upload from A to B of a large file MSS = 10 kbit Receiver buffer : 10 or 100 x MSS 1 acc. Ack. from B to A if 10 or 100 MSS received A B

3 VRAAG 1 : delay ? 64 kbit/s (transmission delay = 125 msec) 10 Gbit/s (transmission delay = 0) 100 m (propagation delay = 0) km (GEO satellite) (propagation delay = 250 msec) Packet of 1 kByte (=8 kbit) over 64 kbit/s ==> transmission delay ==> 8 (kbit) / 64 (kbit/s) = 125 msec propagation delay = delay due to wave (e.g. light) propagation speed ( m/s in air, m/s in fiber) transmission delay = time between first and last bit sent 125 msec first bit last bit use faster access line use shorter route (transatlantic fiber) Speed of light : 3 x 10 8 m/s ==> propagation delay ==> 0.7 x 10 8 (m)/3 x 10 8 m/s = s (  250 msec) See also Chapter 1 (part 1.6)

4 VRAAG 1 : voorbeeld Y X A B 10 Mbit/s vezel : 2000 km 10 Gbit/s 100 m TP 100 m coax 10 x MSS 10 x MSS gives 1 Ack => transfer of 100 kbit How long does it take ? (1) 10 MSS from A to B : A to X : propagation delay = 0, transmission delay = 100 (kb) / 10 (Mb/s) = 0.01 s X to Y : propagation delay = 2 x 10 6 / 2 x 10 8 (m/s) = 0.01 s, trans. delay = 0 Y to B : prop. delay & trans. delay = 0 (short distance, very high bitrate) (2) Ack from B to A : ONLY propagation delay important (because ACK segment is very short) Y to X gives 0.01 s Total : = 0.03 s for 100 kb gives : 100 / 0.03 kb/s = 3.33 Mb/s

5 VRAAG 1 : voorbeeld Y X A B 10 Mbit/s vezel : 2000 km 10 Gbit/s 100 m TP 100 m coax 10 x MSS Acc.Ack s

6 VRAAG 1 : uitkomst andere stukken Link AXLink XYBuffer BBitrate (Mbit/s) 100 kbit/sSatelliet10 x MSS 83 kbit/s 10 Mbit/sSatelliet10 x MSS 0.48 Mbit/s 10 Mbit/sVezel10 x MSS 3.3 Mb/s 100 kbit/sSatelliet100 x MSS 98 kbit/s 10 Mbit/sVezel100 x MSS 8.3 Mbit/s

7 VRAAG 2 one way delay = 25 msec MSS = 1 kB application at sender bitrate = 1000 kByte/s = 1000 MSS/sec Receiver buffer = 4 kB = 4 MSS immediate acknowledgments sender sends every 5 msec one MSS (if allowed) receiver OPTION 1 : immediate processing receiver OPTION 2 : 15 msec processing / kB (but first kB treated immediately, when receiver buffer empty; and : if application process is ready, it will first take a new segment before any action is taken at the TCP layer {e.g. send an ACK} ) no congestion control, no slow start, no TCP set-up -1- Make a detailed timing diagram (show Ack, Seq number, Rec. window, send window) -2- Compare throughput with simplified theoretical limit

8 VRAAG 2 : OPTION 1 5 msec steps b.v.: SW = Send Window SW 0=>1=>0 0=>1=>0 0=>1=>0 0=>1=>0 etc. NO processing time at receiver ! 4 ==> 3 3 ==> 2 2 ==> 1 1 ==> 0 Theory : 4 kB/50 msec = 640 kbit/s Detail : 4 kB/50 msec = 640 kbit/s 50 msec

9 VRAAG 2 : OPTION 2 5 msec steps b.v.: SW = Send Window 4 ==> 3 3 ==> 2 2 ==> 1 1 ==> 0 processing time at receiver :15 msec ! First process and then send ACK for ==> receive window remains 2 One ACK but also rec.window reduced by one ==> no segment can be sent ! One ACK and rec.window NOT reduced ==> segment can be sent ! 0=>1=>0 0=>1=>0 0=>2=>1 1==>0 0=>1=>0 etc. Theory : 4 kB/50 msec = 640 kbit/s Detail (steady state): 3 kB / 50 msec = 480 kbit/s 0=>2=>1 1==>0 0=>1=>0 50 msec

10 VRAAG 3 one way delay = 25 msec MSS = 1 kB application at sender bitrate = 1000 kByte/s = 1000 MSS/sec Receiver buffer = 4 kB = 4 MSS immediate acknowledgments sender sends every 5 msec one MSS (if allowed) receiver immediate processing slow start (congestion control) Make a detailed timing diagram (show Ack, Seq number, Adv. Rec. window, send window and congestion window)

11 VRAAG 3 b.v.: SW = Send Window CW = Congestion Window 1 ==> 0, CW= 1 0=>2=>1, CW= 2 0=>2=>1, CW= 3 1=>3=>2, CW= 4 2==>1, CW= 4 1==>0, CW= 4 etc. 1=>0, CW= 2 0=>1=>0, CW= 5 0=>1=>0, CW= 6 0=>1=>0, CW= 7 0=>1=>0, CW= 8