1 Communicatienetwerken Oefeningen 3 : TCP & IP Woensdag 28 november 2007

2 VRAAG 1 one way delay = 25 msec MSS = 1 kB application at sender bitrate = 1000 kByte/s = 1000 MSS/sec Receiver buffer = 4 kB = 4 MSS accumulated acknowledgments (after 20 msec), but when 2 segments are received : immediate ack. sender sends every 5 msec one MSS (if allowed) receiver processing takes 15 msec slow start (congestion control) segment has to wait 25 msec extra in a router Make a detailed timing diagram (show Ack, Seq number, Rec. window, send window and congestion window) Step 1: till segment is sent Step 2: from segment

3 0=>2=>1, CW=9 0=>1=>0, CW=1 0=>2=>1, CW=2 1=>0 0=>3=>2, CW=3 2=>1 1=>0 0=>2=>1, CW=4 1=>0 0=>2=>1, CW=5 1=>0 0=>1=>0, CW=6 0=>1=>0, CW=8 Delayed ack (20 msec) To application zal niet verwerkt worden Want TCP wacht nog op wordt pas hier verwerkt 1=>0 VRAAG 1 (a) Note: this is and not because the sender is allowed to send segments 9,10,11 and 12 ! (because segment went already to the application).

4 0=>2=>1, CW=9 1=>0 0=>3=>2, CW=9 2=>1 1=>0 0=>2=>1, CW=10 1=>0 0=>2=>1, CW=11 1=>0 0=>1=>0, CW=12 0=>2=>1, CW=13 1=>0 0=>2=>1, CW=14 1=>0 0=>1=>0, CW=9 Regimetoestand : 36 x 5 msec = 180 msec voor 10 MSS = 10 kBYTE of 55.5 kBYTE/s of kbit/s Theorie (propagatietijd en receiver window) : 4 MSS per RTT (=50 msec) of 80 kBYTE/s of 640 kbit/s Theorie (verwerking bij receiver) 1 MSS per 15 msec of 66.6 kBYTE/s of kbit/s [dit is dus de eenvoudigste bovenlimiet] 1 periode VRAAG 1 (b)

A B VRAAG 2 - Ken aan alle routerinterfaces IP adressen toe (eerst kleinste subnetten bepalen). Neem IP adressen zo hoog mogelijk binnen het subnet. - Geef de vrije subnetten weer - routeringstabel A en B ? (zo klein mogelijk, gebruik OSPF)

6 VRAAG 2 : Split subnetwork.122

/30.FC.116/30.FC.120/29.F8.96/28.F0.224/28.F0.240/28.F0.48/29.F8.56/30.FC.60/30.FC.0/27.E FF.FF.FF.00 A B VRAAG 2 : subnetten en interfaces (Hex en CIDR)

8 VRAAG 2 : Split subnetwork.122

/30.FC.116/30.FC.120/29.F8.96/28.F0.224/28.F0.240/28.F0.48/29.F8.56/30.FC.60/30.FC.0/27.E Overblijvende subnetten :.32 /28.F014 IP adr.64 /27.E030 IP adr.128 /26.C062 IP adr.192 /27.E030 IP adr FF.FF.FF.00 A B VRAAG 2 : subnetten en interfaces (Hex en CIDR)

10 VRAAG 2 : Split subnetwork.122

/30.FC.116/30.FC.120/29.F8.96/28.F0.224/28.F0.240/28.F0.48/29.F8.56/30.FC.60/30.FC.0/27.E Overblijvende subnetten :.32 /28.F014 IP adr.64 /27.E030 IP adr.128 /26.C062 IP adr.192 /27.E030 IP adr FF.FF.FF.00 A B VRAAG 2 : subnetten en interfaces (Hex en CIDR).112 /28.96 /27 DEFAULT

12 Aggregatie van subnet.112,.116 en.120 !!! Aggregatie van subnet.0,.60,.224,.240 !!! in default VRAAG 2 : routeringstabel A & B (Hex en CIDR).56 /30.FC /28.F /29.F /28.F / A Aggregatie van subnet.96,.112,.116 en.120 !!! Aggregatie van subnet.0,.60,.224,.240 !!! in default.56 /30.FC /29.F /27.F / B

Overblijvende subnetten : IP adr IP adr IP adr IP adr A B VRAAG 2 : subnetten en interfaces (decimaal)

14 Oplossing routing table A Aggregatie van subnet.112,.116 en.120 !!! Aggregatie van subnet.0,.60,.224,.240 !!! in default Oplossing routing table B Aggregatie van subnet.96,.112,.116 en.120 !!! Aggregatie van subnet.0,.60,.224,.240 !!! in default VRAAG 2 : routeringstabel A en B (decimaal)