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

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

2 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 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 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 444.4. 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 533.3 kbit/s [dit is dus de eenvoudigste bovenlimiet] 1 periode VRAAG 1 (b)

5 5 157.193.122.0 255.255.255.00.117.114.125.232.231.50.57.61.1.16.242.251.105.98 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 6 VRAAG 2 : Split subnetwork.122

7 7.117.114.125.232.231.50.57.61.1.16.242.251.105.98.112/30.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.E0.126.124.118.113.110.109.58.53.54.238.237.254.30.62 157.193.122.0 FF.FF.FF.00 A B VRAAG 2 : subnetten en interfaces (Hex en CIDR)

8 8 VRAAG 2 : Split subnetwork.122

9 9.117.114.125.232.231.50.57.61.1.16.242.251.105.98.112/30.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.E0.126.124.118.113.110.109.58.53.54.238.237.254.30.62 Overblijvende subnetten :.32 /28.F014 IP adr.64 /27.E030 IP adr.128 /26.C062 IP adr.192 /27.E030 IP adr 157.193.122.0 FF.FF.FF.00 A B VRAAG 2 : subnetten en interfaces (Hex en CIDR)

10 10 VRAAG 2 : Split subnetwork.122

11 11.117.114.125.232.231.50.57.61.1.16.242.251.105.98.112/30.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.E0.126.124.118.113.110.109.58.53.54.238.237.254.30.62 Overblijvende subnetten :.32 /28.F014 IP adr.64 /27.E030 IP adr.128 /26.C062 IP adr.192 /27.E030 IP adr 157.193.122.0 FF.FF.FF.00 A B VRAAG 2 : subnetten en interfaces (Hex en CIDR).112 /28.96 /27 DEFAULT

12 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.58.96 /28.F0.109.48 /29.F8.54.112 /28.F0.110.109.00 /0.00.53.54 A Aggregatie van subnet.96,.112,.116 en.120 !!! Aggregatie van subnet.0,.60,.224,.240 !!! in default.56 /30.FC.54.50.48 /29.F8.50.96 /27.F0.54.50.00 /0.00.53.50 B

13 13.117.114.125.232.231.50.57.61.1.16.242.251.105.98.112.252.116.252.120.248.96.240.224.240.48.248.56.252.60.252.0.224.126.124.118.113.110.109.58.53.54.238.237.254.30.62 Overblijvende subnetten :.32.24014 IP adr.64.22430 IP adr.128.19262 IP adr.192.22430 IP adr A B VRAAG 2 : subnetten en interfaces (decimaal)

14 14 Oplossing routing table A.56.252.58.96.240.109.48.248.54.112.240.110.109.00.53.54 Aggregatie van subnet.112,.116 en.120 !!! Aggregatie van subnet.0,.60,.224,.240 !!! in default Oplossing routing table B.56.252.54.50.48.248.50.96.224.54.50.00.53.50 Aggregatie van subnet.96,.112,.116 en.120 !!! Aggregatie van subnet.0,.60,.224,.240 !!! in default VRAAG 2 : routeringstabel A en B (decimaal)