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Solutions of All Homework Problems, CS/EE 143, Fall 2014 Prof. Steven Low. TAs: Changhong Zhao, Jianchi Chen Last updated: 11/18/2014 1 Introduction to Internet 1.1 W&P, P1.1 2 Points. How many hosts can one have on the Internet if each one needs a distinct IP address? [Solution] Since the IP(v4) address is 4-byte, there can potentially be 232 distinct addresses. 1.2 Adapted from W&P, P1.3 8 Points. Imagine that all routers have 17 ports. There are 216 hosts to be connected. Assume you can organize the hosts and routers any way you like. Your goal is to design the network structure as well as the assignment of addresses so as to (i) minimize the number of routers required, (ii) minimize the size of the routing table required in each router. (a) 4 Points. Describe your network structure and explain why your design gives the minimum number of routers. (b) 4 Points. Describe your scheme for assigning addresses and routing. Explain why your design gives the minimum routing table size in each router. [Solution] The routers will be arranged in a tree topology, with each of 16 of the ports representing the next half-byte in the network address. The 17th port would be the link to the parent router. The root router would have a dead 17th port, and so there would be a routing table of size 16 on the first router, and 17 on every other router. This would be of the form: • (prefix, halfbyte, unmatched) routes to the port indexed by halfbyte, where prefix is the total specified address on the parent router that points to this router on its 17th port; • default route goes to port 17; • the root router is just: (halfbyte, unmatched) routes to the port indexed by halfbyte. The total number of routers we use is 212 + 28 + 24 + 20 = 4369. This is optimal because: (a) Consider all the routers and hosts as vertex and links as edge. Then in the result graph, it should be a tree, otherwise we can further remove edges and/or vertexes to simplify the graph. Assume we have N routers. For each router, we have 17 ports. For each host, we have 1 port. The total number of edges should be no less than the total number of vertex minus 1, which is (N + 216 − 1) to provide 1 connectivity. Also we have the number of edges should be no larger than half of the total ports, which is (17N + 216 )/2. Hence we have (17N + 216 )/2 ≥ (N + 216 − 1), so N ≥ 4368.9. (b) The routing table at every router except the root has only 17 entries, which is minimum with a router fanout of 17. 1.3 W&P, P1.4 3 Points. Assume that a host A in Berkeley sends packets with a bit rate of 100Mbps to a host B in Boston. Assume also that it takes 130ms for the first acknowledgment to come back after A sends the first packet. Say that A sends one packet of 1Kbyte and then waits for an acknowledgment before sending the next packet, and so on. What is the long-term average bit rate of the connection? Assume now that A sends N packets before it waits for the first acknowledgment. Express the long-term average bit rate of the connection as a function of N. [Note: 1 Mbps = 106 bits per second; 1 ms = 1 millisecond = 10−3 second.] [Solution] It takes 1KB/100Mbps = 0.08ms to transmit one packet. When N = 1, the connection transfers 1 packet per RTT, i.e., 1KB per 130.08ms ∼ 130ms. This represents a throughput of 1KB/0.13s = 61.5 kbps. (If you use 1KB = 1,024 bits instead of 1,000 bits, then the throughput will be 63 kbps.) It is OK to ignore the packet transmission time (0.08ms) since it is order of magnitude smaller than the propagation delay of 130ms. When N is small enough (see below), the sender sends N packets, and then stops and waits for the first ACK which arrives after one RTT of 130.08ms ∼ 130ms. Therefore the throughput is N KB/130ms = 61.5N kbps. 1 KB ≤ 130.08 ms, i.e., K = 1, 626. When N ≥ K, the Let K be the largest integer such that K 100Mbps sender would have already received the first ACK when it finishes transmitting N packets, and therefore can immediately transmit the N + 1st packet. In this case, the throughput is 100Mbps. In summary, the throughput is: ( 61.5N kbps throughput = 100 Mbps 1.4 if N < 1, 626 if N ≥ 1, 626 = min {61.5N kbps, 100 Mbps} B&G, 2.9: Horizontal and vertical parity checks 4 Points. A horizontal and vertical parity check of size K by L takes (K − 1) binary sequences each of length (L − 1) bits and outputs K binary sequences each of length L bits such that if these sequences are arranged into an K by L matrix, then each row and each column will have an even number of 1’s. The parity check works as follows. To send (K − 1) data sequences each of length (L − 1) bits, the sender sends the K by L matrix. The receiver checks the parity of each row and each column. If there is any row or column that has odd parity, an error is detected. (a) 2 Points. Give an example of a pattern of six errors that cannot be detected by horizontal and vertical parity checks. (b) 2 Points. Find the number of different patterns of four errors that will not be detected by horizontal and vertical parity checks. [Solution] (a) Any 6 bit errors in the following entry of the K × L matrix will not be detected: (i1 , j1 ), (i1 , j2 ), (i2 , j2 ), (i2 , j3 ), (i3 , j3 ), (i3 , j1 ), because every row and every column has zero or 2 bit errors which maintains valid parity of the matrix. (b) Any 4 bit errors distributed such that every row and every column has zero or 2 bit errors will not 2 be detected. There are 1 K L = KL(K − 1)(L − 1) 2 2 4 such configurations. 1.5 Adapted from B&G, P3.1 2 Points. Customers arrive at a fast food restaurant at a rate of five per minute and wait to receive their order for an average of 10 minutes. Customers eat in the restaurant with probability 1/2 and carry out their order without eating with probability 1/2. A meal requires an average of 20 minutes. What is the average number of customers in the restaurant? [Solution] For customers who carry out their orders, the average time spent in the restaurant is 10 minutes; for those who eat in, the average time spent is 30 minutes. Therefore, the average time spent in the restaurant over all customers equals 20 minutes. Applying Littles law, this means the average number of customers in the restaurant equals 20 × 5 = 100. 1.6 Capacity constrained network design 3 Points. Two nodes A and B need to be connected via a communication link. On average, it is estimated that 1000 packets will arrive at A destined for B each second, each packet having an average size of 1 Kbyte. There are two design choices: (i) you build a single link with rate 10Mbps between A and B, (ii) you build two parallel 5Mbps links, and send a packet arriving at A randomly via either link. Assume each link is equipped with an infinite sized buffer. Assuming the M/M/1 formula for the delay on a link, compute the average packet delay in each case. Which design choice is better? Can you explain why? You may ignore packet propagation times. [Solution] In case (i), the link service rate equals 10Mbps/8Kb = 1250 pkts/sec. Therefore, the mean delay equals 1/(1250-1000) = 4 ms. In case (ii), each link has a service rate of 625 pkts/sec, and sees an arrival rate of 500 pkts/sec. The mean delay therefore is 1/(625-500) = 8ms. The first design choice is better since if you double the arrival rate as well as the service rate in an M/M/1 queue, the average waiting time decreases by a factor of 1/2. 1.7 W&P: P2.6 3 Points. Consider a router in the backbone of the Internet. Assume that the router has 24 ports, each attached to a 1Gbps link. Say that each link is used at 15% of its capacity by connections that have an average throughput of 200kbps. How many connections go through the router at any given time? Say that the connections last an average of 1 minute. How many new connections are set up that go through the router in any given second, on average? [Solution] Aggregate average throughput per link is 1Gbps × 15% = 150Mbps. Hence average number of connections per link is 150Mbps/200kbps = 750, and average number of connections at the router is 750 × 24 = 18, 000. Then Littles Law implies the number of new connections set up at the router: λ = N/T = 18, 000/60 = 300 connections/sec. 3 CS/EE143 Networking WF: 10:30 – 11:55am Fall 2011 Professor Steven Low Aggregate average throughput per link = 1Gbps x 15% = 150Mbps. Hence, average #connections per link = 150Mbps / 200kbps = 750, and average #connections/router = 750 x 24 = 18,000. Little’s Law then implies: = N / T = 18,000 / 60 = 300 connections/sec. 8. W&P, P2.7 [4 points] 1.8 W&P, We would like to transfer 20 KB (1 Byte=8 bits) file across a network from node A to node F. P2.7 Packets have a length of 1 KB (neglecting header). The path from node A to node F passes through 5 links, and 4 intermediate nodes. Each of the links is a 10 Km optical fiber with a rate 4 Points. We would like to transfer 20 KB (1 Byte=8 bits) file across a network from node A to node of 10 Mbps (assume speed of light in optical fiber is 2.0x108m/s). The 4 intermediate nodes are F. Packets have store‐and‐forward a length of 1 KB (neglecting header). The path from node A to node F passes through devices, and each intermediate node must perform a 50 s routing table 5 links, and 4 intermediate nodes. Each of the links is a 10 Km optical fiber with a rate of 10 Mbps look up after receiving a packet before it can begin sending it on the outgoing link. How long (assume speed ofdoes it take to send the entire file across the network? light in optical fiber is 2.0 × 108 m/s). The 4 intermediate nodes are store-and-forward devices, and each intermediate node must perform a 50 µs routing table look up after receiving a packet before it can begin sending it on the outgoing link. How long does it take to send the entire file across [Solution] the network? Timing diagram: [Solution] Look at the time diagram Figure 1. A B 1 2 C D T1 T2 T3 T1 1 20 T2 T3 2 1 20 2 Let: link propagation delay = T1 = 10 km / (2x108 m/s) = 0.05 ms packet transmission time = T2 = 1 kB/10Mbps = 0.8 ms Figure 1: Time diagram of packets transferred through 5 links and 4 intermediate nodes. 5 Let: link propagation delay = T1 = 10 km / (2 × 108 m/s) = 0.05 ms packet transmission time = T2 = 1kB/10Mbps = 0.8 ms node processing time = T3 = 0.05 ms. Assume A starts transmission at time 0, then: The time at which B receives and finishes processing first packet is TB = T1 + T2 + T3 . The time at which C receives and finishes processing first packet is TC = TB + T1 + T2 + T3 = 2TB . The time at which D receives and finishes processing first packet is TD = 3TB . The time at which F receives the first packet is TF = 4TB + T1 + T2 since F does not need to look up the routing table. Therefore, the time at which F receives all 20 packets is TF + 19T2 = 4(T1 + T2 + T3 ) + T1 + 20T2 = 5T1 + 24T2 + 4T3 = 5 × 0.05 + 24 × 0.8 + 4 × 0.05 = 19.65 ms. 4 1.9 W&P, P2.9 3 Points. Consider the case of GSM cell phones. GSM operates at 270.88 Kbps and uses a spectrum spanning 200 KHz. What is the theoretical SNR (in dB) that these phones need for operation? In reality, the phones use a SNR of 10 dB. Use Shannons theorem to calculate the theoretical capacity of the channel, under this signal-to-noise ratio. How does the utilized capacity compare with the theoretical capacity? [Solution] (a) By Shannon’s formula C = W log2 (1 + SNR), we have 270.88kbps = 200kHz log2 (1 + SNR). Hence, required SNR is 2 270.88 200 − 1 = 1.557 = 10 log10 (1.557) dB = 1.9 dB. (b) With SNR = 10 dB, Shannon capacity = 200kHz log2 (1 + 10) = 692 kbps. Therefore, the utilization is 270.88/692 = 39%. 1.10 IP addresses 2 Points. Consider an IPv4 subnet with private IP address space 166.111.8.0/24. If each IP interface in the subnet needs a distinct IP address, then how many IP interfaces can there be in the subnet? [Hint: to obtain the right answer, please look up what is a broadcast address online.] [Solution] There is a total of 28 = 256 IP addresses in this subnet. One of these IP addresses (the one whose last 8 digits are all 1) should be reserved for broadcasting, leaving 255 IP addresses for physical interfaces. Hence, a maximum number of 255 IP interfaces is allowed in the subnet. 1.11 Network operations 6 Points. Assume that host A sends a file of 20 KByte (1 Byte=8 bits) to host F. Host A segments the file into 22 packets, each of 1 KB (why 22 packets instead of 20?). The path from A to F passes through 5 links and 4 routers. Each link is a 10km optical fiber with 10 Mbps capacity (the speed of light in an optical fiber is 2.0 × 108 m/s). East router performs a 50µs routing table look up after receiving a whole packet, before forwarding the packet to an output port. How long does it take for the file to reach F? (Assume that a router can look up the routing table for multiple packets simultaneously.) [Solution] The 20KByte file is segmented into 22 packets of 1KB instead of 20, since each packet has header (so that the data in a packet is less than 1KB) and redundant packets may be introduced for forward error correction. Let B, C, D, E denote the four intermediate routers on the path from A to F, and let tp , ts , tl denote the propagation time of a link, transmission time of a packet, and table lookup time respectively. The timeline of the overall data transmission is shown in Figure 19. It is clear from the figure that the overall transmission time is T = 4(tp + ts + tl ) + tp + 22ts = 5tp + 26ts + 4tl 10km 1KB + 26 × + 4 × 50µs = 5× 8 2 × 10 m/s 10Mbps = 5 × 0.05ms + 26 × 0.8ms + 4 × 0.05ms = 21.25ms. 5 A B C 1 tp 2 ts tl 1 2 D E F tp ts tl 1 tp 2 ts tl 1 2 tp ts tl 1 tp 2 ts ts Figure 2: Timeline of the data transfer from A to F. Here B, C, D, E represent the four routers on the path from A to F, and tp , ts , tl represent the propagation time of a link, transmission time of a packet, and table lookup time, respectively. 6 2 Ethernet 2.1 W&P, P3.2 3 Points. Consider the Slotted ALOHA MAC protocol. There are N nodes sharing a medium, and time is divided into slots. Each packet takes up a single slot. If a node has a packet to send, it attempts transmission with a certain probability. The transmission succeeds if no other node attempts transmission in that slot. Now, suppose that we want to give differentiated services to these nodes, i.e., we want different nodes to get a different share of the medium. The scheme we choose works as follows: If node i has a packet to send, it will try to send the packet with probability pi . Assume that every node has a packet to send all times. In such a situation, will the nodes indeed get a share of the medium in the ratio of their probability of access? [Solution] Consider a given slot. Since every node has a packet to send and transmits with probability pi , the probability of success for any node i is Y pi β pi (1 − pj ) = 1 − pi j6=i where β := QN j=1 (1 − pj ) is independent of any user j. Therefore, the share of the media any user i gets is proportional to 2.2 pi 1−pi , not to pi . W&P, P3.4 4 Points. Consider a commercial 10 Mbps Ethernet configuration with one hub (i.e., all end stations are in a single collision domain). (a) 2 Point. Find the Ethernet efficiency for transporting 512 byte packets (including Ethernet overhead) assuming that the propagation delay between the communicating end stations is always 25.6 µs, and that there are many pairs of end stations trying to communicate. (b) 2 Points. Recall that the maximum efficiency of Slotted Aloha is 1/e. Find the threshold for the frame size (including Ethernet overhead) such that Ethernet is more efficient than Slotted Aloha if the fixed frame size is larger than this threshold. Explain why Ethernet becomes less efficient as the frame size becomes smaller. [Solution] (a) Following the derivation of efficiency of the text, we have η= 1 1 + 3.4A A := PROP . TRANS where Now TRANS = packet transmission time = Therefore A = 25.6 µs 409.6 µs 512 × 8 bits = 409.6 µs. 10 Mbps = 0.0625, and hence efficiency η= 1 = 82%. 1 + 3.4 × 0.0625 7 (b) Let the Ethernet frame size be s bits, then TRANS = and A= PROP 10PROP = . TRANS s η= 1 1 + 3.4 × We want s such that which implies s≥ s bits s = µs 10 Mbps 10 10PROP s ≥ 1 e 34 PROP 34 × 25.6 = = 506.6 bits. e−1 2.718 − 1 A smaller frame size implies a larger A, which implies a smaller efficiency. Note that on average, an interval of length 2(e − 1)PROP gets wasted between successful transmissions. A smaller frame size implies therefore that the fraction of time spent transmitting data successfully is smaller. 2.3 W&P, P3.5 4 Point. Ethernet standards require a minimum frame size of 512 bits in order to ensure that a node can detect any possible collision while it is still transmitting. This corresponds to the number of bits that can be transmitted at 10 Mbps in one roundtrip time. It only takes one propagation delay, however, for the first bit of an Ethernet frame to traverse the entire length of the network, and during this time, 256 bits are transmitted. Why, then, is it necessary that the minimum frame size be 512 bits instead of 256? [Solution] A station detects collision by listening and comparing what it hears on the wire with what it transmits. Therefore, if it transmits for a round-trip time, it can compare the signal it detects on the wire with what it just transmitted onto the wire, and a collision is detected if they are different. If it transmits only for 12 RTT, it would have stopped transmitting before the signal from the other end reaches it, and hence miss the detection. This is illustrated in Figure 3. 2.4 Switch vs Hub, W&P, P3.7 6 Points. In the network shown in Figure 4, all of the devices want to transmit at an average rate of R Mbps, with equal amounts of traffic going to every other node. Assume that all of the links are half-duplex and operate at 100Mbps and that the medium access control protocol is perfectly efficient. Thus, each link can only be used in one direction at a time, at 100Mbps. There is no delay to switch from one direction to the other. (a) 3 Points. What is the maximum value of R? (b) 3 Points. The hub is now replaced with another switch. What is the maximum value of R now? [Solution] Denote the flows on links as in Figure 5. (a) There are two potential bottlenecks: the switch-hub link or the hub. Switch-hub link bottleneck: Each node generates 7 flows, each of rate R/7, to each of the other nodes. Therefore the rate crossing the link from switch to link is R 16 X= ×4×4= R. 7 7 8 CS/EE143 Networking WF: 10:30 – 11:55am Fall 2011 Professor Steven Low CS/EE143 Networking WF: 10:30 – 11:55am Fall 2011 Professor Steven Low from the root; the rest are all blocked ports), no LAN segment will use Bridge 8 to send or receive Ethernet frames. Therefore, removing Bridge 8 has no effect on the spanning tree, except that it has one fewer leaf node now. 5. W&P, P3.7 [4 points.] In the network shown below, all of the devices want to transmit at an average rate of R Mbps, with equal amounts of traffic going to every other node. Assume that all of the links are half‐ 4. duplex and operate at 100 Mbps and that the media access control protocol is perfectly W&P, P3.6 [4 points] efficient. Thus, each link can only be used in one direction at a time, at 100 Mbps. There is no Consider the corporate Ethernet shown in the figure below. Each switch is labeled with its ID. Diagram demonstrating why the minimum frame should be transmitted for RTT in delay to switch from one direction to the other. Figure 3: successfully detect collision. order to (a) What is the maximum value of R? [2 points] (a) Determine which links get deactivated after the Spanning Tree Algorithm runs, and indicate (b) The hub is now replaced with another switch. What is the maximum value of R now? [2 them on the diagram by putting a small X through the deactivated links. [3 points] points] Figure 4: An(b) A disgruntled employee wishes to disrupt the network, so she plans on unplugging central ethernet. Each circle represents a host, that wants to send an aggregate of R Mbps traffic, [Solution] evenly to other hosts. Bridge 8. How does this affect the spanning tree and the paths that Ethernet frames follow? [1 point] [Solution] 10 12 Figure 5: Flows in the hub-switch ethernet. 9 The rate crossing the link from hub to switch is Y = 4R since the hub repeats all input flows on all links except from they come from. Then we have X + Y ≤ 100 Mbps =⇒ R ≤ 15.9 Mbps. Hub bottleneck: Take any link, say, from hub to node A. The total rate of traffic on this link is the sum of all input traffic to the link except its own: SA = 3R + X = (3 + 16 37 )R = R. 7 7 This traffic shares the link with the traffic from A to hub, which has the rate R. Therefore, 37 + 1)R ≤ 100 Mbps =⇒ R ≤ 15.9 Mbps. 7 Hence, both bottlenecks impose the same upper bound for R. SA + R ≤ 100 Mbps =⇒ ( (b) When the hub is replace with a switch, then X = Y and the switch-switch link is the only bottleneck. 2X = 2.5 32 R ≤ 100 Mbps =⇒ R ≤ 21.875 Mbps. 7 Spanning Tree Protocol, W&P, P3.6 6 Points. Consider the network topology shown in Figure 6, where 1, 2, . . . , 8 denote 8 switches interconnecting 9 Ethernets. Figure 6: Each circle represents a switch, interconnecting 9 Ethernets. (a) 3 Points. Determine which links get deactivated after the Spanning Tree protocol runs, and indicate them on the diagram by putting a small X through the deactivated links. (b) 3 Points. A disgruntled employee wishes to disrupt the network, so she plans on unplugging central bridge switch 8. How does this affect the spanning tree and the paths that Ethernet packets follow? [Solution] (a) The STP operates as follows: 10 CS/EE143 Networking WF: 10:30 – 11:55am Fall 2011 Professor Steven Low (a) The STP operates as follows: 1. The bridge with the smallest ID will be the root (Bridge 1) 1. The2.bridge with the smallest ID will be the root (Bridge 1). Each bridge finds a shortest path to the root, where link cost is simply hop count (# bridges en route to root). The port on a bridge through which it roots is called a root 2. Each bridge finds a shortest path to the root, where link cost is simply hop count (# bridges en port (RP). route to root). The port on a bridge through which it roots is called a root port (RP). 3. A tie is broken by choosing the port / path where the next hop (bridge) has a smallest 3. A tie isID. broken by choosing the port / path where the next hop (bridge) has a smallest ID. 4. Each LAN segment will choose a bridge towards the root; the port on the bridge that’s 4. Each LAN segment will choose a bridge towards the root; the port on the bridge thats chosen by chosen by the LAN segment is called a designated port (DP). the 5. LAN segment is called a designated port (DP). A tie is broken by the LAN choosing a bridge with the smallest ID. 5. A tie is broken by the LAN choosing a bridge with the smallest ID. For the following figure, In Figure 7: BP: blocked port BP: blocked port DP: designated port (incoming chosen by LAN) DP: designated port (incoming chosen by LAN) RP: rootRP: root port (outgoing chosen by bridge) port (outgoing chosen by bridge). Note that LAN chooses Bridge 2 over 8 become of smaller bridge ID. (b) Since Bridge 8 has no DP (it only has 1 RP which will be used by Bridge 8 to send / receive Figure 7: Spanning Tree Protocol. 11 Note that LAN chooses Bridge 2 over 8 because of smaller bridge ID. (b) Since Bridge 8 has no DP (it only has 1 RP which will be used by Bridge 8 to send / receive from the root; the rest are all blocked ports), no LAN segment will use Bridge 8 to send or receive Ethernet frames. Therefore, removing Bridge 8 has no effect on the spanning tree, except that it has one fewer leaf node now. 11 2.6 Aloha (a) (Equal Share). Assume that n hosts share a medium using the slotted ALOHA protocol: at every time slot, each host attempts to send a packet with probability p. A host succeeds to send a packet at a given time slot if and only if it is the only host that sends a packet at that time slot. 1. 2 Points. What is the probability that a host sends a packet successfully at a give time slot? 2. 2 Points. What is the probability P{a packet sent} that a packet be sent at a given time slot? 3. 4 Points. What choice of p maximizes the probability P{a packet sent}? How does this maximum probability behave as n → ∞? (b) (Unequal Share). Assume that n hosts share a medium using the slotted ALOHA protocol, but at every time slot, each host attempts to send a packet with a different probability. More specifically, let N := {1, . . . , n} denote the collection of hosts and assume that host i attempts to send a packet with probability pi ∈ (0, 1) at every time slot for i = 1, . . . , n. 1. 2 Points. What is the probability Pi := P{i sends a packet successfully} that host i ∈ N sends a packet successfully at a given time slot? 2. 2 Points. What is the ratio P1 : P2 : P3 : · · · : Pn ? This ratio characterizes the share of medium among the hosts. Is the share of medium proportional to the probabilities pi that hosts attempt to send packets, i.e., is the ratio P1 : P2 : P3 : · · · : Pn equal to p1 : p2 : p3 : · · · : pn ? Pn Pn 3. (*) 5 Points. Assume i=1 pi = 1 and let P := i=1 Pi denote the probability that a packet gets successfully transmitted at a given time slot. Prove that (p1 , p2 , p3 , . . . , pn ) = (1/n, 1/n, . . . , 1/n) minimizes P , i.e., (1/n, 1/n, . . . , 1/n) is the solution to min n X i=1 pi Y (1 − pj ) j6=i s.t. 0 ≤ pi ≤ 1, n X pi = 1. i = 1, . . . , n; i=1 It means that equal share of the medium minimizes the throughput. [Solution] (a) Answers given in the lecture/textbook. (b) 1–2 same answer as Problem 2.1 in this set; 3. Consider the nontrivial case where n ≥ 3. Let P (p) := p = (p1 , . . . , pn ) 6= ( n1 , . . . , n1 ) which satisfies n X pi = 1 Pn i=1 pi Q j6=i (1 − pj ). Consider any and i=1 0 ≤ pi ≤ 1, ∀i Let M := maxi=1,...,n pi , and m := mini=1,...,n pi . Hence we have M > n1 > m. Without loss of M +m generality, let p1 = M and p2 = m, then p = (M, m, . . . , pn ). Let p0 = ( M +m 2 , 2 , . . . , pn ). It is 12 sufficient to show P (p) > P (p0 ). We have n n Y X Y P (p) − P (p0 ) = [M (1 − m) + (1 − M )m] (1 − pj ) + (1 − m)(1 − M ) pi (1 − pj ) j=3 i=3 j6=1,2,i n n Y M +m Y M +m 2X M +m − 2× (1 − ) ) pi (1 − pj ) + (1 − (1 − pj ) 2 2 2 i=3 j=3 j6=1,2,i n n Y X Y (M + m)2 − M m 2 = (1 − pj ) − pi (1 − pj ) 4 j=3 i=3 j6=1,2,i n n n X Y X Y (M + m)2 − M m 2 pi (1 − pj ) − > pi (1 − pj ) (1) 4 i=3 j=3 i=3 j6=1,2,i n n X Y X Y (M + m)2 − M m 2 pi (1 − pi ) (1 − pj ) − pi (1 − pj ) = 4 i=3 i=3 j6=1,2,i j6=1,2,i n X Y (M + m)2 = (1 − pj ) − M m (1 − 2pi )pi 4 i=3 j6=1,2,i ≥ 0 (2) Pn where the inequality in (1) is because (M + m)2 − 4M m > 0 for M 6= m and i=3 pi < 1, and the inequality in (2) is because pi ≤ 21 since otherwise pi > M , which leads to a contradiction. 13 3 3.1 Routing Longest prefix matching (exercise) 1 Point. Consider the following routing table. IP 166.111.8.0/24 166.111.0.0/16 output port 1 2 Which outport should a packet be forwarded to, if its destination IP address is 166.111.8.28? [Solution] The IP address 166.111.8.28 matches the first 24 bits of the first entry, and only 16 bits for the second entry. Hence, the packet should be forwarded to port 1 according to longest prefix matching. 3.2 Static routing, W&P, P5.1 Consider the network topology depicted in Figure 8. Each link is marked with its weight/cost. Figure 8: Network topology with link weights. (a) 3 points. Run Dijkstras algorithm on the above network to determine the routing table for node 3. (b) 3 points. Repeat (a) using Bellman-Ford algorithm. [Solution] (a) The steps of Dijkatra’s algorithm are: steps 0 1 2 3 4 5 6 F 3 3,2 3,2,1 3,2,1,7 3,2,1,7,5 3,2,1,7,5,4 3,2,1,7,5,4,6 D1, pred inf, – 3, 2 D2, pred 1, 3 D4, pred 8, 3 8, 3 8, 3 8, 3 8, 3 D5, pred inf, – inf, – 7, 1 7, 1 D6, pred inf, – inf, – inf, – 10, 7 9, 5 9,5 D7, pred inf, – inf, – 6, 1 where F is the set of points for which the shortest distance and path to node 3 have been determined, Di 14 denotes the current-step shortest distance from node i to node 3, and “pred” stands for the predecessor of node i on its shortest path to node 3. D3 = 0 at initialization. Hence the routing table for node 3 is: destination 1 2 4 5 6 7 next node 2 2 4 2 2 2 (b) The steps of Bellman-Ford algorithm are: steps 0 1 2 3 4 D1, pred inf, – 3, 2 3, 2 3, 2 3, 2 D2, pred 1, 3 1, 3 1, 3 1, 3 1, 3 D4, pred 8, 3 8, 3 8, 3 8, 3 8, 3 D5, pred inf, – 9, 4 7, 1 7, 1 7, 1 D6, pred inf, – inf, – 11, 5 9, 5 9, 5 D7, pred inf, – inf, – 6, 1 6, 1 6, 1 where D3 = 0 at initialization. The resulting routing table of node 3 is the same as (a). 3.3 Dynamic routing Consider 5 stations connected in a bi-directional ring, as shown in Figure 9. Suppose station 0 is Figure 9: A bi-directional ring topology. the only sender, and it sends packets to all other stations 1, 2, 3, 4 at rates 4, 3, 2, 1 packets/sec, respectively. Note that these are end-to-end traffic rates between source 0 and all destinations, not the link flow rates which depend on the routing. These end-to-end source-to-destination rates and the routing decision jointly induce a traffic pattern on the network and hence flow rates on the links. (a) 3 points. Table 1 shows the routing tables at each station. For each station, the first column is D (destination) and the second column is NN (next node). Indicate in a diagram the routes from station 0 to stations 1, 2, 3, 4 as implemented by the routing table. 15 These end‐to‐end source‐to‐destination rates and the routing decision jointly induce a traffic pattern on the network and hence flow rates on the links. (a) The routing tables at each station is as shown below (for each station, the first column is D (destination) and the second column is NN (next node)). Table 1: Routing tables of stations Station 0 Station 1 Station 2 Station 0 Station 1 Station 2 StationStation 3 3 Station Station 4 4 D NN D NN D NN D NN D NN D1 1 NN D0 2 NN D 0 3 NN D 0 4 NN D NN 0 0 2 1 2 2 1 3 1 4 1 1 0 2 0 3 0 4 0 1 0 0 3 1 3 2 3 3 2 4 2 0 2 1 24 2 2 1 4 3 3 1 4 4 4 1 3 0 0 4 1 Indicate in 3a diagram the 1 3 routes 2 from 3 station 3 0 to 2stations 4 1, 2, 2 3, 4 0 as implemented by the routing table. 4 1 4 2 4 3 4 4 3 0 (b) Suppose the flow rate on a link is used as the cost of that link (10 of them). Use the Dijkstra algorithm to compute the shortest paths from station 0 to all other stations. (b) 3 points. Suppose the flow rate on a link is used as the cost of that link (there are 10 links in total). Use the Dijkstra algorithm to compute the (new) shortest paths from station 0 to all other stations. (c) Compute the new flow rates and then the shortest paths from station 0 to all other stations, (c) 3 points. using the Bellman‐Ford algorithm. Will the routing ever converge? Compute the new flow rates and then the shortest paths from station 0 to all other stations, using the Bellman-Ford algorithm. If this procedure is repeated, will the routing ever converge? [Solution] (a) The shortest-path from station 0 is such that all traffic clockwise, asclockwise, shown inas Fig. (a) The tree shortest‐path tree implemented by that the routing table goes is: all traffic goes shown: 10. 0 4 1 3 2 [Solution] 18 Figure 10: [Solution] Shortest-path tree in P3.3 (a). (b) The costs (flow rates) on the links induced by the routing table are as shown in Fig. 11. Note that all links in the counter-clockwise direction have zero costs. The steps of Dijkstra’s algorithm are shown in the table below: steps 0 1 2 3 4 F 0 0,4 0,4,3 0,4,3,2 0,4,3,2,1 D1, pred 10, 0 10, 0 10, 0 0, 2 D2, pred inf, – inf, – 0, 3 D3, pred inf, – 0, 4 D4, pred 0, 0 Therefore, all traffic is routed in the counter-clockwise direction in the shortest-path tree, exactly opposite to the routing used in (a), as shown in Fig. 12. (c) The costs (flow rates) on the links induced by the routing in (b) are as shown in Fig. 13. Note that all links in the clockwise direction have zero costs. The steps of Bellman-Ford algorithm are: 16 that all links in the counter‐clockwise direction have zero costs. 0 0 10 CS/EE143 Networking Fall 2011 0 WF: 10:30 – 11:55am Professor Steven Low 0 4 1 0 0 6 1 0 CS/EE143 Networking Fall 2011 3 2 0 WF: 10:30 – 11:55am Professor Steven Low 3 Dijkstra computation: 4 1 Figure 11: [Solution] Network topology with link costs (flow rates) in P3.3 (b). t F D1, pred D2, pred D3, pred D4, pred 0 0 10, 0 Inf, ‐ Inf, ‐ 0, 0 0 1 0, 4 10, 0 Inf, ‐ 0, 3 2 4 2 0, 4, 3 0, 3 4 10, 0 1 3 0, 4, 3, 2 0, 2 4 0, 4, 3, 2, 1 (c) The costs (flow rates) on the links induced by the routing table are as shown below. Note that all links in the clockwise direction now have zero cost. 3 2 Therefore, all traffic is routed in the counter‐clockwise direction in the shortest‐path tree, exactly opposite to the routing used in the previous period, as shown: Figure 12: [Solution] New shortest path tree calculated in P3.3 (b). (c) The costs (flow rates) on the links induced by the routing table are as shown below. Note 19 0 0 that all links in the clockwise direction now have zero cost. 0 0 10 4 1 9 4 0 0 7 0 0 0 3 2 10 0 0 4 1 Bellman‐Ford computation: Figure 13: [Solution] Network link 9 topology 4 costs (flow rates) in P3.3 (c). t D1, pred D2, pred D3, predwithD4, pred 0 0 7 0 0, 0 Inf, ‐ ‐ 10, 0 steps D1, pred Inf, D2, pred D3, pred D4, pred 3 2 0 0, 0 inf, – inf, – 10, 0 1 0, 0 0, 0 1 19, 4 10, 0 2 0, 0 0, 1 0, 2 10, 0 Bellman‐Ford computation: 3 0, 0 0, 1 0, 3 20 0, 2 4 0, 0 0, 1 0, 2 0, 3 t D1, pred D2, pred D3, pred D4, pred 0 0, 0 Inf, ‐ Inf, ‐ 17 10, 0 Therefore, all traffic is routed in the clockwise direction in the shortest-path tree, exactly the same as the routing used in (a), as shown in Fig. 10. Indeed, the routing updates will continue to oscillate across routing update periods, between the results of (a) and (b), and will never converge. 3.4 Dynamic routing (exercise) Consider the case where H1 sends 2Mbps traffic to H2 via one of two links as in Figure 14, either through R1 or through R2. Consider the dynamic routing case where the routing table is updated every 3 minutes. When the routing table is updated, the link weight at a link is computed by the following equation: weight = 1Mbps + average throughput over the past 3 minutes . capacity Assume that at t = 0, the routing table is updated, and at t = 1ms, H1 starts sending traffic to H2. 2Mbps R1 2Mbps H1 H2 4Mbps R2 4Mbps Figure 14: A sends packets of 1KB to B via a 1Mbps link with 20KB buffer. Give the traffic throughput through Routers R1 and R2 at t = 1, 4, 7, . . . minutes. [Solution] At t = 0, the traffic is 0 on both paths (H1→R1→H2 and H1→R2→H2). Therefore the link weights are t=0 R1 R2 H1 1/2 1/4 H2 1/2 1/4 After running the shortest algorithm, H1 decides to choose the path via R2, during 0∼3 minutes. At t = 3, the traffic is 0 on the path via R1 and 2Mbps on the path via R2. Therefore the link weights are t=3 R1 R2 H1 1/2 3/4 H2 1/2 3/4 After running the shortest algorithm, H1 decides to choose the path via R1, during 3∼6 minutes. At t = 6, the traffic is 2Mbps on the path via R1 and 0 on the path via R2. Therefore the link weights are t=6 R1 R2 H1 3/2 1/4 18 H2 3/2 1/4 After running the shortest algorithm, H1 decides to choose the path via R2, during 6∼9 minutes. Then at t = 9, the situation is the same as t = 3, and routing starts oscillating every 6 minutes. As a result, the traffic throughput is shown below. t R1 R2 3.5 1 0 2Mbps 4 2Mbps 0 7 0 2Mbps 10 2Mbps 0 13 0 2Mbps ··· ··· ··· Routing on a continuum of nodes (exercise) Consider the network given in Figure 15. Each point represents a router, connected to its neighbors Figure 15: Network topology for problem 3.5. via links of capacity 1. The links form the ring. Lable the routers by x ∈ [0, 1), and give router 0 two labels: 0 and 1. Assume that all traffic have the same destination: router 0. Let r(x) denote the source rate at x for x ∈ [0, 1), and assume r(x) = 2x. (a) (Static Routing). Consider the following static routing strategy: pick a y ∈ (0, 1), let every router x ∈ (0, y) forwards packets clockwise, and every router x ∈ (y, 1) forwards packets counter-clockwise. • What is the traffic throughput f − (x, y) at link x for x ∈ (0, y), and what is the traffic throughput f + (x, y) at link x for x ∈ (y, 1)? • Use the expression of the queueing delay for the M/M/1 queue. What is the queueing delay + d− s (x, y) at link x for x ∈ (0, y), and what is the queueing delay ds (x, y) at link x for x ∈ (y, 1)? − − Let ds (y) := supx∈(0,y) ds (x, y) denote the maximum queueing delay over links x ∈ (0, y), i.e., + over the links that forward packets clockwise. And let d+ s (y) := supx∈(y,1) ds (x, y) denote the maximum queueing delay over links x ∈ (y, 1), i.e., over the links that forward packets counter− clockwise. What is d+ s (y) and ds (y)? • Assume that the propagation delay d− i (x) from x to 0 (clockwise) is x, and that the propagation delay d+ i (x) from x to 1 (counter-clockwise) is 1 − x. Each router x has two paths, clockwise or counter-clockwise, to forward packets to the destination—router 0(1). Label the clockwise path by superscript - and the counter-clockwise path by superscript +, and define costs D+ (x, y) + := d+ s (y) + di (x), D− (x, y) − := d− s (y) + di (x) 19 for the two paths. For what values of x is D+ (x, y) equal to D− (x, y)? • Let x(y) denote the x where D+ (x, y) = D− (x, y). Use matlab (or other tools) to draw x(y) as y increases from 0 to 1, when does the line intersect z(y) = y? • Let y ∗ denote the y ∈ (0, 1) where x(y) intersects z(y). Show that 0<x<y ⇒ D− (x, y) < D+ (x, y), y<x<1 ⇒ D− (x, y) > D+ (x, y) when y = y ∗ . That is, when x < y ∗ , the left path has smaller cost, and the right path has bigger cost. This is considered “stationary.” Give an interpretation of why this is called “stationary”. (b) (Dynamic Routing). Let’s extend (a) to the dynamic routing case where routing y is updated over time. Let y k denote the routing strategy at time k = 0, 1, 2 . . . and assume y k+1 = x(y k ) for k = 0, 1, 2, . . . For what initial values of y0 does the sequence {yk }y≥0 converge? [Solution] (a) • Z − y f (x, y) = f + (x, y = Z xx r(s)ds = y 2 − x2 , x ∈ (0, y) r(s)ds = x2 − y 2 , x ∈ (y, 1) y . • d− s (x, y) = 1 Hence we have = 1 , 1 − y 2 + x2 1 1 = , d+ s (x, y) = 1 − f + (x, y) 1 − x2 + y 2 1− f − (x, y) x ∈ (0, y) x ∈ (y, 1) 1 1 − y2 1 + d+ s (y) = ds (1, y) = 2 y − d− s (y) = ds (0, y) = . • Let D− (x, y) = 1 1 + x = 2 + 1 − x = D+ (x, y) 1 − y2 y , and hence we have x(y) = . 1 1 1 (1 + 2 − ) 2 y 1 − y2 • See Fig. 16. The unique solution of x(y) = y on [0, 1] is y ∗ = 0.6756. • When 0 < x < y ∗ we have D− (x, y ∗ ) < D− (y ∗ , y ∗ ) = D− (x(y ∗ ), y ∗ ) = D+ (x(y ∗ ), y ∗ ) = D+ (y ∗ , y ∗ ) < D+ (x, y ∗ ) We can show D− (x, y ∗ ) > D+ (x, y ∗ ) when y ∗ < x < 1, in a similar way. When y = y ∗ , the routing protocol happens to guarantee that all the nodes select the shorter path, and there is no motivation for them to deviate to any other routing. (b) Let y := sup{y ∈ [0, 1]|x(y) = 1} and y := inf{y ∈ [0, 1]|x(y) = 0}. From Fig. 16 we see that |x0 (y)| > 1 for y ∈ [y, y], which implies that y k+1 = x(y k ) will not converge, unless y 0 = y ∗ . 20 1 0.9 0.8 X: 0.68 Y: 0.68 0.7 x 0.6 0.5 0.4 0.3 0.2 0.1 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 y Figure 16: [Solution] x(y) as a function of y. 3.6 Forward error correction code The forward error correction code discussed in the lecture (and textbook) can be represented in the matrix form as in Figure 17, i.e., C = P A, where C is a k ×m matrix that represents m encoded packets C1 C2 C3 k P1 P2 P3 Cm = Pn A Figure 17: Matrix representation of forward error correction. each of k bits (the m columns of C), P is a k × n matrix that represents n original packets each of k bits (the n column of P ), and A is an n × m 0-1 matrix that represents the coding Aij ∈ {0, 1}. For example, the jth column of C is: n X Cj = Pi Aij i=1 for j = 1, . . . , m. 1 1 0 1 Let A = 0 0 1 0 Here the summation is elementwise XOR. 0 1 0 1 0 0 1 1 1 1 0 1 1 1 0 1 (a) 3 points. If packets C1 = [1 0 0 1]T , C2 = [0 0 1 1]T , C5 = [0 0 1 0]T , C6 = [1 1 1 1]T are received, find the original packets P1 , P2 , P3 , P4 . (b) 3 points. If another set of packets C1 = [1 0 0 1]T , C3 = [0 0 1 0]T , C4 = [1 0 1 0]T , C6 = [0 1 1 0]T are received, find the original packets P1 , P2 , P3 , P4 . [Solution] 21 (a) The corresponding matrices 1 0 0 1 are 0 0 1 1 0 0 1 0 1 1 = C 0 = P A0 = P 1 1 1 0 0 1 1 1 0 0 0 1 0 0 1 1 1 1 where P is the 4 × 4 matrix whose j-th column is the original packet Pj . It amounts to inverting A0 to obtain P , and can be done as follows. Note that P2 = C30 = [0 0 1 0]T , then C20 = P1 + P2 and hence P1 = C20 + P2 = [0 0 1 1]T + [0 0 1 0]T = [0 0 0 1]T . Similarly, C10 = P1 + P4 and hence P4 = C10 + P1 = C10 + C20 + P2 = [1 0 0 1]T + [0 0 1 1]T + [0 0 1 0]T = [1 0 0 0]T . Finally, C40 = P1 + P2 + P3 + P4 and hence P3 = C40 + P1 + P2 + P4 = [0 1 0 0]T . Hence 0 0 0 1 0 0 1 0 P = 0 1 0 0 . 1 0 0 0 (b) Similarly, the corresponding matrices are 1 0 1 0 0 0 0 1 00 00 0 1 1 1 = C = P A = P 1 0 0 0 1 0 0 1 0 0 1 1 1 0 1 1 1 1 1 1 Note that C100 = P1 + P4 and C300 = P1 + P3 + P4 , and hence P3 = C100 + C30 = [0 0 1 1]T . Then from C200 = P3 + P4 we have P4 = C200 + P3 = [0 0 0 1]T . Then P1 = C100 + P4 = [1 0 0 0]T and, since C400 = P1 + P2 + P3 + P4 , we have P2 = C400 + P1 + P3 + P4 = [1 1 0 0]T . Hence 1 1 0 0 0 1 0 0 P = 0 0 1 0 . 0 0 1 1 3.7 Network coding, W&P, P5.4 Consider a wireless network with nodes X and Y exchanging packets via an access point Z. For simplicity, we assume that there are no link-layer acknowledgments. Suppose that X sends packets to Y at rate 2R packets/sec and Y sends packets to X at rate R packets/sec; all the packets are of the maximum size allowed. The access point uses network coding. That is, whenever it can, it sends the “exclusive or” of a packet from X and a packet from Y instead of sending the two packets separately. (a) 2 points. What is the total rate of packet transmissions by the three nodes without network coding? (b) 2 points. What is the total rate of packet transmissions by the three nodes with network coding? [Solution] (a) Without network coding: • X transmits at rate 2R to Z. • Y transmits at rate R to Z. 22 • Z transmits at rate 3R to X, Y Hence, total rate required = 6R. (b) With network coding, in every 1 second, on average there is 0.5 second when Z broadcasts R packets x+y to both X and Y. In the other 0.5 second, Z relays R packets from X to Y. Hence, total rate required = 2R + R + (R+ R)/(0.5+0.5) = 5R. 23 4 4.1 Internetworking W&P, P6.1 (a) 2 points. How many IP addresses need to be leased from an ISP to support a DHCP server (with L ports) that uses NAT to service N clients, if every client uses at most P ports? (b) 2 points. If M unique clients request an IP address every day from the above mentioned DHCP server, what is the maximum lease time allowable to prevent new clients from being denied access assuming that requests are uniformly spaced throughout the day, and that the addressing scheme used supports a max of N clients? [Solution] (a) Consider the worst case that all the P ports of all the N clients are connected, i.e., there are N P connections. To use NAT, the DHCP server must maintain an address translation table in which every entry takes the form IPa , TCPb ↔ IPb , TCPm , where IPa is the IP address actually assigned by ISP, TCPb is an port at the NAT server, IPb is the (virtual) IP of a client that is known only by the NAT server, and TCPm is the actual port used by the client for a connection. Hence the maximum number of entries in this table is restricted by the number L of ports of the NAT server multiplied by the number S of assigned IPs, and must not be smaller than the number of connections, i.e., SL ≥ #entries in the table ≥ N P , which implies S ≥ NLP . Therefore there should at least be d NLP e IP addressed assigned by the ISP. 24 (b) Suppose all the IP addresses are available at the beginning of the day (time 0). Every M hours 24N there is a request of IP address from a new client, and therefore at time M hours a maximum number of N clients are supported and there is no more IP address available. In that case the first client must release the IP address it has been occupying to satisfy any new request. Hence the maximum lease time is 24N M hours. 4.2 Insufficient IP addresses (exercise): A variant of P4.1 (a) Consider a type D subnet (there are 256 IP addresses in a type D network, and one of these IP addresses is used for broadcasting). Assume that 15 IP addresses are assigned to servers, then how many hosts can this subnet support if there is no DHCP service nor NAT service? (b) Assume that there is a DHCP server in the subnet, and that each host connects to the Internet 8 hours a day. How many hosts can this subnet support? (c) Assume that the DHCP server also runs the NAT protocol, and can use up to 2000 TCP ports for each dynamic IP address. Also assume that each host needs 20 TCP connections. How many hosts can this subnet support? [Solution] (a) There is a total of 256-1-15=240 IP addresses that can be assigned to hosts. If there is no DHCP service nor NAT service, then 1 IP address can be assigned to at most one host. Hence, this subnet can only support 240 hosts. (b) If the subnet runs DHCP, then 1 IP address can be assigned to different hosts if the hosts get online at different times of a day. If each host connects to the Internet 8 hours a day, then 3 hosts can share a single IP address. Hence, this subnet can support 240*3=720 hosts if DHCP service is enabled. (c) If NAT protocol is run, then multiple hosts can use the same IP address simultaneously, as long as they are assigned different port numbers. If 2000 TCP ports can be used for NAT and each host has 20 TCP connections, then 1 IP address can be shared by 2000/20=100 hosts. Hence, this subnet can 24 support 720*100=72000 hosts. 25 5 5.1 Transport W&P, P7.1 (a) 2 points. Suppose you and two friends named Alice and Bob share a 200 Kbps DSL connection to the Internet. You need to download a 100 MB file using FTP. Bob is also starting a 100 MB file transfer, while Alice is watching a 150 Kbps streaming video using UDP. You have the opportunity to unplug either Alice or Bob’s computer at the router, but you cannot unplug both. To minimize the transfer time of your file, whose computer should you unplug and why? Assume that the DSL connection is the only bottleneck link, and that your connection and Bob’s connection have a similar round trip time. (b) 2 points. What if the rate of your DSL connection were 500 Kbps? Again, assuming that the DSL connection were the only bottleneck link, which computer should you unplug? [Solution] If I share with Bob, then each of us will get half of the link rate because we both use the same TCP algorithm for file transfer and both have the same RTT. If I share with Alice, then Alice will get 150kbps since she is using UDP without congestion control, and I will get the remaining link rate. Therefore, to maximize my own throughput: (a) I should unplug Alice, so I will get 100kbps. (b) I should unplug Bob, so I will get 500 − 150 = 350 kbps, as opposed to 250kbps if sharing with Bob. 5.2 W&P, P7.2 As shown in Figure 18, station A has an unlimited amount of data to transfer to Station E. Station A uses a sliding window transport protocol with a fixed window size. Thus, station A begins a new packet transmission whenever the number of unacknowledged packets is less than W and any previous packet being sent from A has finished transmitting. The size of the packets is 10000 bits (neglect headers). So for example if W > 2, station A would start sending packet 1 at time t = 0, and then would send packet 2 as soon as packet 1 finished transmission, at time t = 0.33 ms. Assume that the speed of light is 3 × 108 meters/sec. (a) 4 points. Suppose station B is silent, and that there is no congestion along the acknowledgement path from C to A. (The only delay acknowledgements face is the propagation delay to and from the satellite.) Plot the average throughput as a function of window size W . What is the minimum window size that A should choose to achieve a throughput of 30 Mbps? Call this value W ∗ . With this choice of window size, what is the average packet delay (time from leaving A to arriving at E)? (b) 4 points. Suppose now that station B also has an unlimited amount of data to send to E, and that station B and station A both use the window size W ∗ . What throughput would A and B get for their flows? How much average delay do packets of both flows incur? (c) 4 points. What average throughput and delays would A and B get for their flows if A and B both used window size 0.5W ∗ ? What would be the average throughput and delay for each flow if A used a window size of W ∗ and B used a window size of 0.5W ∗ ? [Solution] (a) Recalling that throughput is inverse proportional to the round-trip time, we start with calculating the round-trip time of data transfer from A to E. The timeline of data transfer from A to E is shown in Figure 19, where ts , tq , and tp denote the transmission time of a packet, queueing delay at the buffer of C, and propagation delay on a space-to-ground link respectively. The transmission time of a packet is ts = 0.33 ms. 26 Figure 18: Transmitting data from stations A, B to E. A C 1 D E ts 2 tq 1 tp 2 ts 1 2 tp ts tp 1 2 tp Figure 19: [Solution] Timeline of data transfer from A to E. When B does not transmit, the buffer at C is empty (since incoming traffic rate is limited by the capacity of the link from A to C, which is no greater than the capacity of the output link from C to D), and 27 therefore the queueing delay at the buffer of C is tq = 0. The propagation delay on a space-to-ground link is 5 ∗ 104 km tp = = 166.7ms. 3 ∗ 108 m/s Hence, the round-trip time is RTT = 3ts + tq + 4tp = 667.7ms. The throughput is given by x= W ∗ pkt size W ∗ 10, 000bits 30W = = Mbps RTT 667.7ms 2003 when 30W/2003 ≤ 30, i.e., W ≤ 2003. When W > 2003, the throughput is 30Mbps. Average throughput as a function of the window size W is shown in Figure 20. The minimum window size to throguhput (Mbps) 30 25 20 15 10 5 0 0 2003 window size (pkts) Figure 20: [Solution P5.2] Throughput as a function of the window size. achieve a throughput of 30Mbps is W ∗ = 2003 ≈ 2000. The average packet delay is (average packet delay) = 3ts + tq + 2tp = 334.3ms. (b) When A and B both use a window size of W ∗ , the buffer at C will build up, leading to a longer round-trip time for both flows. This increased round-trip time will slow down the throughput of both A and B. Eventually, the system comes to a steady state, where the round-trip time does not change. Now compute the steady state. Let tq denote the steady-state queueing delay at the buffer of C, and note that A and B have the same steady-state round-trip time RTT = 3ts + tq + 4tp . Then the steady state throughputs of A and B are xA = W ∗ × pkt size = xB . RTT and therefore xA = xB = 15Mbps. We have RTT = W ∗ × pkt size 2000 ∗ 10, 000bits = = 1.333s. xA 15Mbps It follows that tq = RTT − 3ts − 4tp = 1333.3 − 3 × 0.33 − 4 × 166.7 ≈ 665.7ms. 28 CS/EE143 Networking WF: 10:30 – 11:55am Fall 2011 Professor Steven Low Therefore the average packet delay is (average packet delay) = 3ts + tq + 2tp = 1000ms . (c) When each has window size 0.5W ∗ , the two flows have combined window sizes that are sufficient to keep the pipe full. Their combined throughput will be 30 Mbps, so by symmetry each will get 15 Mbps. The round-trip delay, which is the same for A and B, can be found by: RTT = 0.5 × 2000 × 10, 000bits = 666.7ms 15Mbps Subtracting the delay on the reverse path, i.e., 2tp = 333.3ms, we get a forward path delay of 333.3ms, which is exactly the propagation delay 2tp , i.e., the queueing delay is tq = 0 (we ignored the transmission time 3ts from the RTT for simplicity). When A’s window is W ∗ and B’s is 0.5W ∗ , the two flows have combined window sizes that are sufficient to keep the pipe full. Their combined throughput will be 30Mbps. A will have twice as many packets in the pipeline as B, so their throughputs will have a ratio of 2:1. Thus A gets 20 Mbps and B gets 10 Mbps. The roud-trip time for A and B is the same, which is found by: 0.5 × 2000 × 10, 000bits 1 × 2000 × 10, 000bits = = 1000ms RTT = (a) Suppose station B is silent, and that there is no congestion along the acknowledgement path 20Mbps 10Mbps from C to A. (The only delay acknowledgements face is the propagation delay to and from the Subtracting the delay on the reverse path, i.e., p = 333.3ms, we get a forward path delay of 666.7ms satellite.) Plot the average throughput as a 2t function of window size W. What is the minimum for both A and B. Their queueing delay is t = 666.7 − 333.3 = 333.3ms (we ignored the transmission q window size that A should choose to achieve a throughput of 30 Mbps? Call this value W*. With time 3ts from the RTT for simplicity). this choice of window size, what is the average packet delay (time from leaving A to arriving at E)? Suppose now that station B also has an unlimited amount of data to send to E, and that 5.3 (b) W&P, P7.3 station B and station A both use the window size W*. What throughput would A and B get for As shown Figure 21, flows 1 and 2 sharepackets of both a link with capacity C = 120 Kbps. There is no other theirinflows? How much average delay do flows incur? bottleneck. The round trip time of flow 1 is 0.1 sec and that of flow is 0.2 sec.and B both used Let x1 and x2 denote (c) What average throughput and delays would A and B get for their2 flows if A the rates obtained by the two flows, respectively. The hosts use AIMD to regulate their flows. That window size 0.5W*? What would be the average throughput and delay for each flow if A used a is, as long as x1 + x2 < C, the rates increase linearly over time: the window of a flow increases by one window size of W* and B used a window size of 0.5W*? packet every round trip time. Rates are estimated as the window size divided by the round-trip time. Assume that as soon as x1 + x2 > C, the hosts divide their rates x1 and x2 by the factor α = 1.1. 4. W&P: P7.3 [4 points] As shown in the figure, flows 1 and 2 share a link with capacity C = 120 Kbps. There is no other bottleneck. The round trip time of flow 1 is 0.1 sec and that of flow 2 is 0.2 sec. Let x1 and x2 Figure 21: Two flows sharing a link. 10 (a) 3 points. Draw the evolution of the vector (x1 , x2 ) over time. (b) 3 points. What is the approximate limiting value for the vector? [Solution] (a) Since flow 2’s RTT is twice that of flow 1’s RTT, and flow 1’s window increases at a pace that is twice that of flow 2’s, then flow 1’s rate (=window/RTT) increases at a pace that is 4 times that of flow 2’s. Hence on the plot, the rates increase along a line that has a slope of 1/4. When x1 + x2 = C, 29 [Solution] (a) Since B’s RTT is twice that of A’s RTT, A’s window increases at a pace that’s twice that of B’s, and A’s rate (=window/RTT) increases at a pace that’s 4 times that of B’s. Hence on the ( x1 x2 ) plot, the rates increase along a line that has a slope of ¼. When x1 x2 C, both divide their rates by 1.1 immediately without feedback dealy, i.e., their new rates are 90% of those before the multiplicative decrease. Therefore, the evolution of their rates is as shown. x2 C slope = 1/4 0.9C steady state slope = 1/4 x1 x2 C 0.9C C x1 In the steady state, both rates increase at the rate of s/Ti2, where s = packet size in bits (b) of flow i, a and decrease (assuming use the same constant packet size and flows Ti = RTT Figure both 22: [Solution] Rates change ofs) two sharing link. multiplicatively by 10%. Therefore the rates oscillate along the double‐arrowed line segment indicated above, between the lines x1 x2 0.9C and x1 x2 C. To calculate the both divide their rates by 1.1 immediately, their new rates are (about) 90% of those before the approximate steady‐state rates xi.e., 1 , x2 , assume that x1 4x2 and x1 x2 C. Hence, multiplicative decrease. Therefore, the evolution of x 4C / 5 96kbps and x C / 5 24kbps. their rates is as shown in Fig. 22. 1 2 steady state, (b) In the the rates oscillate along the double-arrowed line segment indicated above, between the lines x1 + x2 = 0.9C and x1 + x2 = C. To calculate the approximate steady-state rates 26 x1 , x2 , assume that x1 + x2 = C. Since we also have x1 = 4x2 , we get x1 = 4C/5 = 96kbps and x2 = C/5 = 24kbps. 5.4 W&P, P7.4 Consider a TCP connection between a client C and a server S. (a) 3 points. Sketch a diagram of the window size of the server S as a function of time. (b) 2 points. Using the diagram, argue that the time to transmit N packets from S to C is approximately equal to a + bN for large N . (c) 3 points. Explain the key factors that determine the value of b in that expression for an Internet connection. [Solution] Here all reasonable answers are accepted. (a) The graph depends on the protocol. But basically it looks like Figure 7.10 of the textbook, which is also shown in Fig. 23. (b) The total number of packets transmitted equals to the number of packets transmitted in the slow start stage and the congestion avoidance stage. For the slow start stage, you are transferring lower than the maximum rate possible, which cause a constant part a. In the congestion avoidance stage, you are transferring at the maximum rate possible, say 1/b, which results in a linear part bN . (c) Possible factors can be the maximum rate of Ethernet connection, efficiency of the Ethernet, RTT, routing, etc. 5.5 Window size control (exercise) Assume that a host A in Los Angeles sends packets, each of 1KByte, to a host B in San Francisco, through a connection of 100Mbps capacity. Also assume that the round-trip time of each packet is a 30 (c) Explain the key factors that determine the value of b in that expression for an Internet connection. [Solution] Here all reasonable answers are accepted. (a) The graph depends on the protocol. But basically it looks like Figure 7.10 of the book. (b) The total number of packets transmitted equals to the number of packets transmitted in the Figure 23: [Solution] Window size in Figure 7.10 of the textbook, for P5.4. slow start stage and the congestion avoidance stage. For the slow start stage, you are transferring lower than W the maximum rate possible, which cause a constant part a. In the congestion avoidance stage, you are transferring at the maximum rate possible, say 1/b, which Wmax results in a linear part bN. Wmin (c) Possible factors can be the maximum rate of Ethernet connection, efficiency of the Ethernet, RTT, routing, etc. 2T T time Figure 24: Window size W fluctuates piecewise linearly in time, from Wmin to Wmax , in periods of length T , which is assumed to be much bigger than the round-trip time. constant of 130ms (no jitter). (a) If A uses the window flow control mechanism 27 with a constant window size W , then what is the average bit rate of the connection as a function of W ? What happens as W increases to infinity? (b) Now consider the case where W fluctuates, at a much slower timescale than the round-trip time (this is not true in the TCP protocol) as in Figure 24. What is the average rate of the connection assuming Wmax is “not too big”? [Solution] (a) When window size W is small, the average bit rate x is W pkts ∗ packet size round-trip time W pkts ∗ 1KB/pkt = 130ms 8W = Mbps 130 x = This expression holds as long as 8W/130 ≤ 100, When window size is big, more specifically, when 8W/130 > 100, W packets can no longer be sent within a round-trip time. In this case, there will always be less than W unacknowledged packets after A finish transmitting a packet, and therefore A will keep transmitting packets, at the link capacity. 31 To summarize, the average bit rate is ( 4W/65Mbps if 4W/65 ≤ 100, x= 100Mbps if 4W/65 > 100. And the rate stablizes at 100Mbps as W tends to infinity. (b) Let W (t) denote the window size at time t, then the instantaneous bit rate at time t is x(t) = 4W (t)/65Mbps provided W (t) is not too big (i.e., 4W (t)/65 ≤ 100). It follows that the average rate of the connection is Z T0 1 x = lim x(t)dt T0 →∞ T0 0 Z 1 T = x(t)dt T 0 Z 1 T 4W (t) = dt T 0 65 2(Wmin + Wmax ) = Mbps. 65 5.6 TCP with AIMD (exercise), adapted from P5.3 Flows 1 and 2 share a link (the only bottleneck link) with capacity C = 20Mbps as in Figure 25. The round-trip time of flow 1 is τ1 = 0.1s while that of flow 2 is τ2 = 0.2s (assume that there is no jitter). Let x1 and x2 denote the throughput of flows 1 and 2 respectively. The hosts use AIMD to regulate their flows: • When x1 + x2 ≤ C, the throughput x1 and x2 increase linearly over time: the window of a flow increases by one packet every round-trip time. Assume that the packet size is 1.5KBytes. • When x1 + x2 > C, the hosts divide their window size by the factor α = 1.1. Assume that flow 1 and 2 decrease their window sizes at most once every 0.2ms. Throughput is estimated as the window size divided by the round-trip time. Host ﬂow 1 R R Host Host ﬂow 2 Host Figure 25: Two flows share a bottleneck link. The bottleneck link is highlighted in red. (a) Assume that at time t = 0, the window size of both flows is 1. Draw the evolution of the vector (x1 , x2 ) over time. [Hint: use matlab.] (b) What is the approximate limiting behavior for the vector? You will notice that the end-point of the vector moves on a line that go through the origin as time evolves, and you are required to give the slope of this line. 32 (c) Repeat (a) and (b) for α = 1.2, 1.3, 1.4, 1.5. Can you find out how the slope depends on α? [Solution] (a)The evolution of the vector (x1 , x2 ) over time is shown in Figure 26. The matlab code is attached throughput of flow 2 (Mbps) alpha=1.1 throughput vector line with slope 1/4 4 3 2 1 0 0 5 10 15 throughput of flow 1 (Mbps) Figure 26: [Solution] The evolution of the vector (x1 , x2 ) over time with α = 1.1. below. clear a l l ; close a l l ; clc ; %%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% i n i t i a l i z a t i o n %%%%%%%%%%%%%%%%%%%%%%%%%%%%% % t t t i n i t i a l i z e time stamps resolution = 0.1; % l e n g t h o f a time stamp . range = 60; % f i n i s h i n g time . count = floor ( t range / t r e s o l u t i o n ) ; % number o f time stamps % i n i t i a l i z e window s i z e W1 = 1 ; % window s i z e o f f l o w 1 W2 = 1 ; % window s i z e o f f l o w 2 % i n t i a l i z e round−t r i p time tau1 = 0 . 1 ; % round−t r i p time o f f l o w 1 tau2 = 0 . 2 ; % round−t r i p time o f f l o w 2 % i n i t i a l i z e throughput x1 = zeros ( t c o u n t , 1 ) ; x2 = zeros ( t c o u n t , 1 ) ; % initialize % a l l o c a t i n g space f o r throughput of flow 1 % a l l o c a t i n g space f o r thorughput of flow 2 link capacity 33 C = 20E6 ; P = 1 . 5 E3 ∗ 8 ; C = C / P; % l i n k capacity in bps % packet s i z e in b i t s % l i n k capacity in pkt /s % i n i t i a l i z e alpha alpha = 1 . 1 ; % b a c k o f f parameter used i n MD %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% time e v o l u a t i o n %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% f o r t =1: t c o u n t % compute c u r r e n t t h r o u g h p u t x1 ( t ) = W1 / tau1 ; % t h r o u g h p u t o f f l o w 1 a t ( t −1) ∗ t r e s o l u t i o n x2 ( t ) = W2 / tau2 ; % t h r o u g h p u t o f f l o w 2 a t ( t −1) ∗ t r e s o l u t i o n % u p d a t e window s i z e i f x1 ( t )+x2 ( t ) >= C % i f mod( t ,2)==0 % W1 = W1 / a l p h a ; W2 = W2 / a l p h a ; end else W1 = W1 + 1 ; % i f mod( t ,2)==0 % W2 = W2 + 1 ; end end i f throughput exceeds l i n k capacity d e c r e a s e window s i z e e v e r y 0 . 2 s e c o n d s i n c r e a s e wndow s i z e by 1 i n c r e a e window s i z e e v e r y 0 . 2 s e c o n d s end %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% %% p o s t p r o c e s s i n g %%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % change u n i t s x1 = x1 ∗P / 1E6 ; x2 = x2 ∗P / 1E6 ; % t h r o u g h p u t o f f l o w 1 i n Mbps % t h r o u g h p u t o f f l o w 2 i n Mbps % plot figure f i g = figure ( 1 ) ; c l f ; plot ( x1 , x2 , ’ l i n e w i d t h ’ , 2 ) ; hold on ; set ( gca , ’ f o n t s i z e ’ , 2 4 ) ; t i t l e ( s p r i n t f ( ’ a l p h a =%2.1 f ’ , a l p h a ) ) ; xlabel ( ’ t h r o u g h p u t o f f l o w 1 ( Mbps ) ’ ) ; ylabel ( ’ t h r o u g h p u t o f f l o w 2 ( Mbps ) ’ ) ; set ( gca , ’ x l i m ’ , [ 0 1 8 ] ) ; set ( gca , ’ y l i m ’ , [ 0 4 . 5 ] ) ; plot ( [ 0 1 8 ] , [ 0 4 . 5 ] , ’ r−− ’ , ’ l i n e w i d t h ’ , 2 ) ; hold o f f ; legend ( 2 , ’ t h r o u g h p u t v e c t o r ’ , ’ l i n e with s l o p e 1/4 ’ , ’ l o c a t i o n ’ , ’ n o r t h w e s t ’ ) ; 34 print ( f i g , s p r i n t f ( ’ a l p h a %d . e p s ’ , a l p h a ∗ 1 0 ) , ’−d e p s c ’ ) ; (b) See P5.3. (c) The plots are shown in Figure 27. It can be observed that the slope does not depend on α. 3 2 1 0 0 throughput of flow 2 (Mbps) throughput vector line with slope 1/4 throughput of flow 2 (Mbps) 4 4 throughput vector line with slope 1/4 2 1 4 2 1 4 5 10 15 throughput of flow 1 (Mbps) alpha=1.5 throughput vector line with slope 1/4 3 2 1 0 0 5 10 15 throughput of flow 1 (Mbps) throughput vector line with slope 1/4 3 0 0 5 10 15 throughput of flow 1 (Mbps) alpha=1.4 3 0 0 alpha=1.3 throughput of flow 2 (Mbps) throughput of flow 2 (Mbps) alpha=1.2 5 10 15 throughput of flow 1 (Mbps) Figure 27: [Solution] The evolution of the vector (x1 , x2 ) over time with α = 1.2, 1.3, 1.4, 1.5. 5.7 TCP Vegas (exercise) 20KB A 1Mbps B Figure 28: A sends packets of 1KB to B via a 1Mbps link with 20KB buffer. Consider the case where A sends packets of 1KB to B via a 1Mbps link with 20KB buffer as in Figure 28. The propagation delay (round-trip time with the buffer being empty) is assumed to be 20ms. Assume that A uses TCP Vegas as the transmission control protocol, i.e., it keeps track of the minimum round- 35 trip time τmin and the current round-trip time τ , W + 1 W ← W −1 W and updates the window size W according to if τW −W τ < min W W if τmin − τ > otherwise α τmin β τmin where α, β are TCP parameters. The parameters are chosen α < β to introduce hysteresis, which can reduce the fluctuation in window size W . Now we try to analyze the steady-state of this scenario. We start with making the following simplifications: at steady state, one has W W (α + β)/2 − = . τmin τ τmin Assume α = 2.8 and β = 3.2. (a) If τmin observed by A is equal to the round-trip time without queueing delay, which is 20ms, then what is the window size W at steady state? How many packets are there in the buffer? (b) If τmin observed by A is equal to the round-trip time without queueing delay, plus τe = 10ms, i.e., 30ms, then what is the window size W at steady state? How many packets are there in the buffer? (c) At what value of τe will the buffer be full? Let τe∗ denote this value. What happens if τe > τe∗ ? [Solution] (a) At steady state, the traffic throughput x = W P/τ , where P is the packet size, equals link capacity C. Therefore W x C 1Mbps = = = = 125pkt/s. τ P P 1KB/pkt Then, W W (α + β)/2 − = τmin τ τmin ⇐⇒ W −3 W = τmin τ =⇒ W = 3 + τmin The round-trip time is τ= W = 3 + 20ms ∗ 125pkt/s = 5.5pkt. τ W = 44ms 125pkt/s and therefore the queuing time is tq = 44 − 20 = 24ms. Hence, the number of packets in the buffer is (average number of packets in buffer) = tq ∗ 125pkt/s = 3pkt. (b) One still have W = 3 + τmin W = 3 + 30ms ∗ 125pkt/s = 6.75pkt. τ The round-trip time is τ= W = 54ms 125pkt/s and therefore the queuing time is tq = 54 − 20 = 34ms. Hence, the number of packets in the buffer is (average number of packets in buffer) = tq ∗ 125pkt/s = 4.25pkt. (c) For the buffer to be full, one needs 20pkt = tq ∗ 125pkt/s =⇒ tq = 160ms. 36 Then, the round-trip time is τ = 160 + 20 = 180ms. The window size would be W = τ ∗ 125pkt/s = 22.5pkt. Substitute in W = 3 + τmin W/τ to obtain τmin = W −3 τ = 156ms. W Then τe∗ = τmin − 20 = 136ms. When τe > τe∗ , the buffer will not be able to hold all packets, and has to drop some packets. Then, TCP Vegas will work as TCP Reno (since packet drop occurs). 37