EECS 122: Introduction to Communication Networks Homework 3 Solutions

Transcription

1 EECS 22: Introduction to Communication Networks Homework 3 Solutions Solution. a) We find out that one-layer subnetting does not work: indeed, 3 deartments need 5000 host addresses, so we need 3 bits ( , so 2 bits are not enough) for the host art of the subnetwork. Since we have 6 bits available (it s a class B address), this leaves only bits for the subnetwork ID art, which can accommodate only deartments. So we have to use two-layer subnetting, that is, divide some subnetworks into subsubnetworks. At the first layer, we use 8 subnetworks of hosts each. Three of these subnetworks are allocated to the three deartments that need 5000 hosts each. For examle, deartment gets the subnetwork with ID 00, deartment 2 gets the subnetwork with ID 00 and deartment 3 gets the subnetwork with ID 0. Deartment can then number its hosts 00xxxxxxxxxxxxx, where the x s are either or 0. We are left with subnetworks of 892 hosts each. We need host addresses for each of the 7 small deartments, so we can divide one of the 892-host subnetworks into subsubnetworks of hosts each. For examle, we can divide the subnetwork with ID 00, and let 0000 be the subsubnetwork ID of deartment 4, 0000 be the subsubnetwork ID of deartment 5, 000 be the subsubnetwork ID of deartment 6, and so on. Deartment 0 can then name its hosts 00xxxxxxxxxx, where the x s are either or 0. We are still left with 4 subnetwork IDs (0 ) for future use. (In real life it s slightly more comlicated: address fields that are all 0s or all s are reserved, so subnetwork 00 can have only 6 subsubnetworks, 0000 through 000. For Deartment 0 we could divide subnetwork 0 the same way, and give it subsubnetwork 000.) b) If we are using class C network addresses, then for the deartments with 5000 hosts we need class C addresses er deartment. For the deartments with 600 hosts we need class C addresses er deartment. The total number of class C network addresses needed is Solution 2. In this case acket format is s d S D data, where s d are Ethernet addresses (of course, the Ethernet header also includes other fields not relevant to this roblem) and S D are IP addresses (of course, the IP header also includes other fields not rel-

2 evant to this roblem). Note that since B and C are switches, all nodes are in fact on one local network, so the link layer addresses remain the same throughout. Solution 3. a) Assuming full-dulex interfaces, the maximum total transmission rate is achieved when all the nodes are sending and receiving data at the line rate, 00 Mbs. This is achieved when no more than 00 Mbs of data is destined to a single node (for examle, when distinct airs of nodes are communicating to each other). The resulting maximum total transmission rate over the network is thus 6 00 Mbs 6 Gbs. For half-dulex interfaces, data can only be flowing in one direction at a time on each link, so the answer is half as much (800 Mbs). b) Consider the case where 5 nodes want to transmit data to the last one. In that case the total transmission rate is limited, in the best case, to the line rate of the last node, which is 00 Mbs. Solution 4. a) The maximum transmission rate from the server is 00 Mbs, but the maximum transmission rate to all the comuters is 6 0 Mbs 60 Mbs. Hence the achievable rate is simly min 00 Mbs 60 Mbs 00 Mbs. b) In this case, the maximum transmission rate from the server is 0 Mbs, while the rate to the comuters remains the same. Hence the achievable rate is min 0 Mbs 60 Mbs 0 Mbs. We can see that in both cases the achievable rate is limited by the link to the server for this configuration. c) Once the switch starts forwarding the acket to the server, it must always have bits to send, otherwise there is going to be a ga in the transmission of the acket, and the acket will be lost. The simlest (and in fact, best) solution is for the switch to comletely receive the whole acket before it begins forwarding it to the server. A more comlicated, less robust, and thus worse solution would be the following. The switch figures out the length of the acket (from the header), say L bits. The switch waits until it has received at least 9 0 L bits (nine tenths of the acket) before it starts forwarding the acket to the server. Since the incoming rate is 0 the outgoing transmission rate to the server, the buffer of the switch will not be drained until the whole acket is received. Solution 5. ARP tables translate the network address of a node to its Ethernet address. For comuter t such a air would be [(c.k:u)], which is the entry that aears in ARP for subnet c. Similarly, the entries in ARP2 for subnet b are [(b.i:z),(b.s:w)], corresonding to comuters n and r resectively. 2

3 Solution 6. Suose there are K levels in the domain hierarchy, root domain being level, subroot being level 2, and so on until level K (the leaves of the tree, that is, hosts). Let every name domain at level i, i K, have L i subdomains (domains at level i ). Then each name server for a domain at level i, i K, needs L i entries, one for the name of each subdomain. The total number of names needed is K L i L L 2 L K i This roduct must be equal to the total number of names, 0 9. Therefore, if K 2, then L 0 9. If K 3, L L 2 0 9, and assuming L L 2, we get L L If K 0, we only need 0 entries er server. We see that the greater K is (the deeer the tree, the more levels in the hierarchy) the smaller the number of entries each name server needs. On the other hand, the deeer the tree, the larger the look-u time of a name becomes: for a tree of deth K, the worst-case look-u time is on the order of K (see roblem 7). Solution 7. Let be the robability of a name to be found in the local server. Let K be the number of levels, L i be the number of entries at level i, and L K i L i (see roblem 6). Then the exected looku time is: time for local looku time for non-local looku The time taken for a local looku is αlog L K, since local servers are in level K of the hierarchy (level K is hosts). A non-local looku consults K servers: the local server, lus K servers from the root to the K st server, which is resonsible for the non-local name. So the time taken for a non-local looku is αlog L K K K αlog L i i If we use caching, increases, so the exected looku time decreases. The rice to ay is memory for the cache and time for cache oerations. Solution 8. Let N A, N B, and N C be the number of networks that use class A, B, or C network addresses, resectively. A network with k hosts uses: a class A address if 2 6 a class B address if 2 8 k 2 24 ; k 2 6 ; a class C address if k

4 We know that the number of networks with k hosts is: N Therefore, we have: 2 24 N A N k N B k 2 8 N 2 8 N C N k N N N k 2 6 k 2 8 k N N N Each class A network uses u 2 24 addresses (see roblem 9). Each class B network uses u 2 6 addresses. Each class C network uses u 2 8 addresses. Therefore, the total number of addresses used is: U N A 2 24 N B 2 6 N C 2 8 Now, a network with k hosts assigns only k addresses, even if it has used u more than k addresses. Therefore, the number of addresses really assigned to hosts in class A, B, and C networks is: 2 24 V k N k N k k N That is, U V is the number of addresses wasted. Knowing V and U, we can comute the efficiency of this addressing scheme. The efficiency is defined as: number of addresses really assigned number of addresses reserved (used u) Notice that N will cancel out. Deending on the value of, we find that the efficiency varies between about 2% and 2%. Solution 9. a) If the comany uses three class B networks, then it will use u IP addresses. since each class B network address comrises 2 6 IP addresses. Note that to use u means to reserve so that others cannot use it, that is, to consume. So even if the comany will actually assign to its 500 hosts only 500 of the IP addresses it has been given, it has still used u all of them. V U N 4

5 b) If the comany uses one class B network address with subnetting, it uses u only 2 6 IP addresses. c) If the comany adots CIDR addressing, it could get class C network addresses, which means it uses u IP addresses. The comany could use a network ID of 2 bits to identify its suernetwork. The remaining bits are used as follows: 2 bits identify the LAN (we actually have only 3 LANs, so one LAN ID is left available for future use) and the remaining 9 bits identify the host in the LAN ( ). Solution 0. a) The known facts are the following: All hosts/routers belong to the same network (0xxxx). Host b and router interface r are both on subnet. Host c is on subnet 2. Host a is on neither subnet nor 2. From the first two facts, we know that subnet must have a 3-bit mask (like interface r), so its ID is 0. Therefore host b must have the same mask (000) and an address of 0xxx, where the x s are either or 0 but xxx (because that is host a s address). For host c, things are a bit more comlicated. Suose the mask of subnet 2 is 3 bits long (000). The ID of subnet 2 cannot be 00, because then hosts on subnet 2 would think that host a was on their local network, because its address begins with 00. The ID of subnet 2 cannot be 0 either, because then hosts on subnet 2 would think host b was on their local network, because its address begins with 0. Since the ID of subnet 2 must begin with 0, we have eliminated all ossibilities for a 3-bit mask, so the mask must be at least 4 bits long (00). The ID cannot be 000, because again host a would aear to be on subnet 2, so the ID must be 00. Therefore a ossible mask for host c is 00, in which case its address can be anything of the form 00xx, like 000. b) 0xxxx c) Since subnet has a 3-bit mask, there are 3 bits left for distinguishing hosts, so there can be at most 8 nodes on subnet. 5