INSTITUT FÜR KOMMUNIKATIONSNETZE UND RECHNERSYSTEME Universität Stuttgart Prof. Dr.-Ing. Andreas Kirstädter Sample Solution Communication Networks I Examiner: Professor Dr.-Ing. Andreas Kirstädter Date: March 5, 2010 Duration: 120 Minutes Required: All Problems Permitted Resources: All Problem 1 Link Layer ARQ Protocol Question 1 2 Points TD = d / c = 50 µs TS = lF / r = 10 µs TA = lA / r = 0.2 µs Question 2 a) The receiver has to differentiate between original and retransmitted frames. b) The feedback is always referring to the last transmitted frame. c) The SN alternates between 0 and 1. No pipelining is used. 3 Points Question 3 a) 10 Points R S TS 2TD F1 ACK TA F0 ACK F1 ACK b) R S TS F1 2TD ACK TA F0 NACK F0 ACK F1 ACK c) R S TS 2TD F1 ACK TA F0 NACK F0 ACK F1 ACK Question 4 4 Points Problem 1 In case of acorrupted ACK S will be triggered to retransmit although R may have already received the frame correctly. R will then discard this frame and will Page 2 not forward the packet included to the network layer. However, S is waiting for a ACK from R to see that this transmission was ok and thus to be able to continue. R S F1 ACK F1 ACK Question 5 a) 9 Points B Wait for feedback Sender SN=1 T Feedback A Frame Wait for packet Reset T Set (now+T0, T) Check feedback Packet Including Sequence Number SN (unchanged) B N Generate frame FB ok? Y Frame Frame Packet confirm Set (now+T0, T) Set (now+T0, T) A B B Modulo-2 SN++ b) Receiver D ESN=1 Extract RSN C N Wait for frame RSN == ESN? Sequence Number in arrived frame Y Extract packet Frame (unchanged) Packet Check frame N Frame ok ? C ESN++ Modulo-2 Y D Feedback = ACK Feedback (unchanged) Question 6 6 Points Problem 1 C a) TS, TD, TA b) Too large values of T0 slow down protocol operation ⇒ T0 should be chosen as small as possible Page 3 T0 ≥ TS+2TD+TA = 110.2 µs in order to receive feedback within T0 ⇒ T0 = 110.2 µs (+ a small "delta") c) R S F1 ACK F0 T0 ? F0 ACK F1 ACK Question 7 a) Especially, at low bit error rates the probability of an errored ACK/NACK is much smaller than that of an errored frame due to the small size of the ACKs/NACKs. b) py = Prob{"y link layer transmissions needed"} p1 = 1-E p2 = E (1-E) py = Ey-1 (1-E) for y > 0 c) Number of frame transmissions = y Expectation: n = 1p1+2p2+3p3+...= (1 - E)-1 d) Duration of a transmit cycle: TS+2TD+TA Cycles per packet transmission: n Resulting throughput: T1 = lF/[n(TS+2TD+TA)]=lF(1-E)/[lF/r+2d/c+lA/r] a) A timer for TS has to be introduced together with a second waiting state "Wait for TS" that is entered after the transmission of the first frame and left after the timer expires. Only then, the second frame is sent and S enters the state "Wait for feedback". b) S collects both the feedback for FA and FB. If neither feedback is an uncorrupted ACK, S will retransmit FA and FB. 11 Points Question 8 9 Points Problem 1 Page 4 c) B Wait for feedback Sender SN=1 Feedback G FB_cnt++ ACK_cnt=0 FB_cnt=0 A Wait for packet Packet Generate frame Frame Check feedback N Y FB ok? Set (now+TS, T) N ACK? Y ACK_cnt++ Wait TS G N T FB_cnt ==2? B Y N ACK_cnt Y Frame G >0? SN++ Modulo-2 B Packet confirm A Question 9 a) Probability for at least one uncorrupted frame out of two: p1*=1-E2 Cycles per packet transmission: n* = 1/p1*= (1-E2)-1 Duration of a transmit cycle: 2TS+2TD+TA Resulting throughput: T2 = lF/[n(2TS+2TD+TA)]=lF(1-E2)/[2lF/r+2d/c+lA/r] b) Calculation of break even point E0: 9 Points T1 = T2 2 lF ⋅ ( 1 – E0 ) lF ⋅ ( 1 – E0 ) lF ⋅ ( 1 – E0 ) ⋅ ( 1 + E0 ) ----------------------------- = -------------------------------- = ----------------------------------------------------lF lF lF d lA d lA d lA ---- + 2 --- + ----2 ---- + 2 --- + ----2 ---- + 2 --- + ----r c r r c r r c r lF E 0 = ------------------------------dr l F + 2 ----- + l A c T2 > T1 for E > E0 Problem 1 Page 5 Problem 2 Ethernet & IP Question 1 flooding ⇒ congested network spanning tree ⇒ inefficient resource usage flat addresses ⇒ impractical large forwarding tables no separation of (sub-)networks possible ⇒ security and privacy concerns 3 Points Question 2 7 Points a) address resolution protocol (ARP) b) local subnetwork host broadcasts an ARP request message to all hosts within subnetwork; this message requests the MAC address belonging to the IP of the destination host; the according destination host sends an ARP response message containing its IP and MAC address another subnetwork host broadcasts an ARP request message to all hosts within subnetwork; this message requests the MAC address belonging to the IP of the router responsible for the destination host (the routers IP is derived by the hosts routing table); the according router sends an ARP response message containing its IP and MAC address Question 3 a) 24 bit subnetmask ⇒ 28 = 256 IP addresses within subnet network address 10.0.1.0 not usable for a PC broadcast address 10.0.2.255 not usable for a PC ⇒ 256 - 2 = 254 addresses usable for devices b) occurrence of shared segments (e.g. with repeaters, hubs) no collisions of frames in switched Ethernet a) addresses of supernet completely covered by subnetworks all subnetworks reachable over same interface b) smallest supernet including both subnetworks: 10.0.4.0/22 10.0.4.0/22 includes also 10.0.5.0/24 and 10.0.7.0/24 ⇒ gaps in address range Under normal conditions it would be not possible to build a supernet and have a common routing entry. However, in this special scenario neither 10.0.5.0/24 nor 10.0.7.0/24 is used. ⇒ in principle it would work with entry 10.0.4.0/22 3 Points Question 4 4 Points Question 5 a) 2 Points MAC(B eth 1)MAC(A eth 3) IP(PC 1)IP(PC 3) MAC DA IP SA Problem 2 MAC SA IP DA Page 6 Question 6 a) Router A 8 Points subnetwork 10.0.1.0/24 10.0.2.0/24 next hop IP(C eth 1) direct interface eth 2 eth 1 10.0.4.0/24 10.0.6.0/24 IP(B eth 1) IP(B eth 1) eth 3 eth 3 subnetwork next hop interface 10.0.1.0/24 10.0.2.0/24 10.0.4.0/24 IP(C eth 3) IP(A eth 3) direct eth 2 eth 1 eth 4 10.0.6.0/24 direct eth 3 subnetwork next hop interface 10.0.1.0/24 10.0.2.0/24 10.0.4.0/24 10.0.6.0/24 direct IP(A eth 2) IP(B eth 2) eth 2 eth 1 eth 3 IP(B eth 2) eth 3 Router B Router C (If the routing table of router A is fixed, the routing table of C may remain unchanged for correct operation and vice versa.) Question 7 5 Points Question 8 b) Traffic for 10.0.6.0/24 loops between A and C. TTL of IP packets is reduced until they are discarded. PC 1 and PC 4 cannot reach PC 3. 10.0.4.0/24 has wrong interface at B. Packets for this subnet get lost within 10.0.6.0/24 (more precisely, the according MAC addresses cannot be resolved). PC 2 ist not reachable by any other PC. a) IP addresses of interfaces, direct connected subnetworks and network mask b) OSPF c) each node has complete view on topology of network (link-state) node knows only the next hop for each destination (distance-vector) a) a node receives a routing information update from each other node ⇒ (n-1) updates b) a node reiceives a routing information update from each neighboring node ⇒ m updates 4 Points Problem 2 Page 7 Question 9 a) a node receives from each other node n-1 updates ⇒ (n-1)2 updates thereof (n-1) updates are not duplicates ⇒ (n-1) * (n-2) duplicates b) the number of received messages does not scale very well (n2) with increasing number of nodes 5 Points Question 10 a) single processor + pure delay system ⇒ offered traffic needs to be smaller than 1 Erlang 1000 / (60s) * h + 60000 / (5*60s) * 0.001 s < 1 Erlang ⇒ h < 48 ms the processing time of a non-duplicate needs to be below 48 ms 4 Points b) Yes. In case of a link-state protocol there are not significantly more updates in case if a network change (in contrast to a distance-vector protocol). Question 11 a) network topology + propagation delays + line rates + processing times determine time needed for information spread in the network used protocol determines the required number of updates until there is again a consistent view (e.g. count-to-infinity problem) update time interval 7 Points b) although the links itself might be still alive, this does not guarantee that the routing engine is still working; cost of links might have been changed Question 12 Router A 6 Points subnetwork next hop cost S S S S B C D 0 2 3 5 subnetwork S 1 S 2 S 3 next hop A C cost 2 0 2 S 4 C 5 1 2 3 4 Router B Problem 2 Page 8 Router C subnetwork S 1 S 2 next hop A B cost 3 2 S 3 S 4 D 0 3 subnetwork next hop cost S 1 S 2 S 3 A C C 5 5 3 S 4 - 0 subnetwork next hop cost S S S S B B B 0 2 2 2 subnetwork S 1 S 2 S 3 next hop B B - cost 2 2 0 S 4 B 2 subnetwork S 1 S 2 next hop C C cost 5 5 S 3 S 4 C - 3 0 Router D Question 13 a) Router A 5 Points 1 2 3 4 Router C Router D b) Problem 2 Subnet 4 is not reachable anymore from A, B and C as with A and C the only routers directly connected to D route the traffic to B. Page 9 Question 14 a) 7 Points In order to solve the problem of Question 13 either A or C has to forward the traffic for subnet 4 to D and not to B. Example to achieve this: Update sent to A subnetwork S 1 cost 0 S 2 S 3 S 4 0 0 0 Update sent to C b) Problem 2 subnetwork S 1 S 2 cost 0 0 S 3 0 S 4 2 (or greater) No, traffic exchanged between subnetworks 4 and 3 does not go over B. Also traffic form subnetwork 4 to 1 does not go over B. All other traffic goes over B. Page 10