INSTITUT FÜR KOMMUNIKATIONSNETZE UND RECHNERSYSTEME Universität Stuttgart Prof. Dr.-Ing. Andreas Kirstädter Sample Solution Communication Networks II Examiner: Professor Dr.-Ing. Andreas Kirstädter Date: September 29th, 2010 Duration: 90 Minutes Required: All Problems Permitted Resources: All [56 P.] Problem 1 Traffic Engineering Part 1 Shortest-Path Routing Question 1 a) eAB, eGB, eGE, eED b) Shortest path tree: 15 Points C B G E D A F c) H Forwarding table of node A: Destination Node Outgoing Interface (edge) B eAB C eAB D eAB E eAB F eAF G eAB H eAF d) Question 2 10 Points Forwarding table of node B (attention: new shortest path tree has to be derived!): Destination Node Outgoing Interface (edge) A eBA C eBC D eBG E eBG F eBG G eBG H eBH a) 24 Mbit/s b) 5 Mbit/s c) In case of IP routing, the edge eED along the shortest path is limiting the capacity to 5 Mbit/s. No, IP routing provides only a single path from source to destination. Therefore, only up to 10 Mbit/s can be reached by modifying the link cost values. d) Network operators should select link costs inversly proportional to the transmission capacity of the link. In Figure 1 the ratio of cost to transmission capacity varies between 0.06 and 1.5. The high capacity links, i.e. the 50 MBit/s links, have low cost. For the 10 MBit/s links the cost values vary between 7 and 15. The 5 MBit/s link is comparable cheap and must have other advantages. e) Other criteria • Delay: Low delay links are required by some applications • Monetary Cost: The usage of some links might be charged by the owner of the link • MTU: Usage of links with a small MTU might lead to packet fragmentation • Load: In dynamic scenarios the link load might be of interest. That means low loaded links should be preferred to high loaded links. Part 2 MPLS Switching Question 3 a) The LSR reads the label of the arriving MPLS packet and switches the packet according to the information for this label and input link found in the forwarding table. Also, the label is changed according to the entry in forwarding. b) The table entries can be set by 5 Points Problem 1 Page 2 Question 4 • Label distribution protocol: In case of dynamic MPLS path establishment a so called label distribution protocol is used to inform the LSR about the used labels. LSP A-B-C-D involves the edges eAB, eBC, and eCD. The other paths provide less transmission capacity b) Node B: I-label = 1, O-interface = eBE, O-label = 2, swapping Node C: I-label = 2, O-interface = eEH, O-label = 3, swapping c) Node A operates as a Label Edge Router: It reads the IP address of the arriving packets and pushes a label to them according to the Forward Equivalence Class (FEC) they belong to and according to the LSP that they have to follow in the MPLS network. Node D also operates as a LER: It pops the label from the arrving packet. a) A TE link from the IP perspective are recognized as ordinary links to transport packets. They are realized in so called Multi-layer networks, e.g. by MPLS LSP. That means a path on the lower layer (e.g. MPLS) appears as a single link on the higher layer (e.g. IP). (In many cases TE links use reserved resources, e.g. a certain amount of bandwith.) b) The routing of OSPF can be influenced, that means specific traffic can be directed via these links without affecting other traffic in the network. a) The MPLS-based TE links are considered during the IP routing like physical links. Virtual interfaces are created and added as outgoing interfaces. That means the forwarding engine uses physical as well as virtual interfaces for packet transport. b) The Ethernet link-layer frames contain a type field that indicates the type of the packet contained in the arriving frame. Ethernet frame type for IP packets: 0x0800 Ethernet frame type for MPLS: 0x8847 c) The routers need to implement a priority scheduling at their output ports. 4 Points Question 6 Network management: Via a network management interface the network operator is able to set MPLS paths manually. Such MPLS paths are static. a) 6 Points Question 5 • 6 Points Part 3 ECMP Routing Question 7 a) If several paths with the same minimum cost exist between the source and destination node, the traffic will be evenly split among them at the current node (source). b) By default packets belonging to the same flow (IP src/dst. address, transport protocol, src/dst. port) are routed on the same path to avoid reordering of packets. If ECMP wouldn’t be aware of flows reordering might take place, because packets on different paths experience different delays. In case of TCP, reordering of packets would lead to a significant goodput/throughput decrease. 4 Points Problem 1 Page 3 Question 8 a) The traffic has to use all three links leading to the destination node: eCD, eHD, and eED. As B shall be the only splitting node, the following three branches have to be used: A-B-C-D, A-B-H-D, and A-B-G-E-D. Therefore we have to set: cBC = 5, cBH = 3. b) Modified link costs potentially change the routing of other traffic in the network. In case of traffic from H to C, the route is changed. Originally, the traffic has been routed by node D. After the change, traffic is routed via node B. 6 Points Problem 1 Page 4 [49 P.] Problem 2 Performance Evaluation of Communication Networks Part 1 Modelling Question 1 a) Max. frame size 1518 bytes = 12144 bits 10e9 bits/s / 12144 bits/frame ≈ 823452 frames/s b) x/100 * 823452 frames/s c) 1/λ = x/100 * 823452 6 Points fX ( y ) = λ ⋅ e – λy d) Question 2 a) 16 Points The link delays the packets by the transmission time and by the propagation delay 1518bytes ⁄ packet ⋅ 8bits ⁄ byte Transmission Time: -------------------------------------------------------------------------------- ≈ 1.2μs 10e9 bits/s 200m Propagation Delay: ------------------------ = 1μs 8 2 ⋅ 10 m/s Total delay: t del = 1.2μs + 1μs = 2.2μs Problem 2 b) Because the link delay is shorter than the transmission time of a packet, a packet does not fit completely on the link.This means that 2 packets can be simultaneously on the link at maximum. c) 3 states are needed • Wire idle • 1 frames on the wire • 2 frames on the wire Page 5 d) e) Time indicating the complete reception of a frame • during the tranmission of the previous frame: t2=T1+ttrans • after the previous frame has left the sender completly: t2=NOW+ttrans+tprop Question 3 4 Points Problem 2 " " ! Page 6 Part 2 Simulation Methodology Question 4 Transient/Warm-Up Phase: 2 Points • System/buffers get loaded with events • Reach the stationary system state Evaluation Phase: • Collect measured values The differentiation is necessary to avoid bias w.r.t. the stationary state of the system. Question 5 5 Points Mean: ( 10 + 7 + 8 + 15 + 5 + 10 + 9 + 6 + 10 + 5 )ms x = ---------------------------------------------------------------------------------------------------------------- = 8.5ms 10 Sample variance: S = 1 2 2 --- ( ( 10 – 8.5 ) + ( 7 – 8.5 ) + … )ms ≈ 3.028ms 9 1 + 0.99 Var 3.028ms Half length: l = t 9 ⎛ ⎛ -------------------⎞ ⋅ ----------⎞ ≈ 3.25 ⋅ --------------------- ≈ 3.11ms ⎝⎝ 2 ⎠ 10⎠ 10 Confidence interval: [5.39 ms , 11.61 ms] Question 6 a) Those parts, with small confidence intervals, i.e. the first 3 values are significant. The last 3 values show trend with a more or less acceptable conficende interval. b) The forth simulation value has a low significance, since the confidence interval is very high. That means the mean waiting time can be in area of the confidence interval, which might lead to very different results (compare waiting time at the upper bound vs. lower bound). Moreover there are not much simulation values around 4th value. In order to improve the significane it is required to add additional simulation values below and above the 4th simulation values. In addition it is required to simulated longer in order improve the significance of the 4th value itself. c) The number of samples has a quadratic influence on the length of the confidence interval. That means that we have to simulate four times as much to halve the confidence intervals. 7 Points Problem 2 Page 7 Part 3 Capacity Extension Question 7 It is required that the switch deals with the two physical ports as one virtual port in order to jointly learn MAC addresses. Manual configuration is needed out of two reasons 3 Points • Avoid infinite cyling of IP packets: If the port aggregation feature is not configured on both ports, frames might cycle for infinity. Basically it would be possible to use the spanning tree protocol with the drawback of having only 1 GBit/s between the switches. • Port Aggregation provides a couple of different configuration options that have to be the same on both switches. This has to be configured manually. (Not expected as an answer in the exam) Question 8 6 Points Problem 2 Round Robin Random Hash Function Implementation Effort Low Low Depends on hash function State Yes No No Reordering of Pak- Possible kets within Flow Possible Not possible Link Overload Pos- No sible Yes Yes Equal Link Utilization Not Guaranteed Depends on Hash Function Yes Page 8