4. Given the IP address 192.168.7.67/28, find the following: 128 64 32 16 8 4 2 1 (a) Number of subnets = (b) bits need to borrow from the host = (c) Usable hosts IP addresses in each subnet
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- 52- A network for a local office requires 4 LAN subnets as follows: · Subnet A - North Office with 31 hosts; · Subnet B - East Office with 15 hosts; and · Subnet C - South Office with 7 hosts. · Subnet D – West Office with 4 hosts. Given the IP Address 192.168.15.0 /24 and above described network requirements: 1. Design an optimal address scheme using Variable Length Subnet Mask (VLSM) method; What will be the subnet mask of Subnet A? Select one: A. 255.255.255.192 B. 255.255.128.255 C. 255.255.255.128 D. 255.255.192.192 E. 255.255.192.054- A network for a local office requires 4 LAN subnets as follows: · Subnet A - North Office with 31 hosts; · Subnet B - East Office with 15 hosts; and · Subnet C - South Office with 7 hosts. · Subnet D – West Office with 4 hosts. Given the IP Address 192.168.15.0 /24 and above described network requirements: 1. Design an optimal address scheme using Variable Length Subnet Mask (VLSM) method; and What will be the subnet mask of Subnet B? Select one: A. 255.255.128.255 B. 255.255.192.0 C. 255.255.192.192 D. 255.255.255.192 E. 255.255.255.224Suppose 198.53.202.0 is a network address and we want 4 subnets. Find the following: - Number of bits required for subnetting - Subnet mask - Range of address in each subnet - Broadcast address for each subnet
- 3.1 An end device was given the IP address of 198.168.4.35/28. Consider this address and indicate: 3.1.1 The network ID/address to which the host belongs. 3.1.2 The network broadcast address to which the host belongs. 3.1.3 The total number of hosts available in the network. 3.1.4 The total number of networks. 3.2 Consider two neighbours, Alice and Bob. Each have wireless IPv4 routers with integrated NAT. Each neighbour connects their laptop to their own wireless router, and each uses appropriate utilities to examine the IP address of each laptop. They realise the laptops have the same IP address. 3.2.1 How is that possible? 3.2.2 Justify one reason that wide-spread deployment of IPv6 would remove the need for the NAT devices.1.4.12.2 End-to-end Delay. Consider again the network shown above. The links again have transmission rates of R1 = R2 = 100 Mbps (i.e., 100 x 106 bits per second), and each packet is 1 Mbit (106 bits) in size. Assume that the propagation delay is 1 msec per link. What is the end-to-end delay of a packet from when it first begins transmission at the sender, until it is received in full by the server at the end of the rightmost link. Assume store-and forward packet transmission. You can assume the queueing delay is zero. Answer choices: A. 2 x 106 msec b. 2.01 msec C. 1.1 msec D. 2.02 msec 1.4.12.3 Maximum Throughput. Consider again the network shown above. The links again have transmission rates of R1 = R2 = 100 Mbps Assume that the link R2 is fairly shared (as we've seen is done via TCP) between the two sessions. What is the maximum end-to-end throughput achieve by each session, assuming both sessions are sending at the maximum rate possible? Answer choices:…Consider the following diagram: A,B,C and D are subnets, R1, R2, and R3 are routers. Default signifies theentire rest of the internet. All 4 subnets each contain the same number of hosts.Consider the following routing table at R3: CIDR Mask Link 0.0.0.0/0 L4 128.4.6.0/24 L3 128.4.8.0/24 L5 1. If 128.4.6.0 is in A, what is the routing table at R1? 2. The ISP that owns all these networks is short on IP addresses, so theyuse a NAT router for R2. How many public IP addresses will C andD use together assuming they use as many private 10.X IP addresses asthey can? How many total public IP addresses will be used by the entirenetwork? Why doesn’t the ISP care about how many private IP addressesare used? 3. Fix R3’s routing table now that we are using NAT: 4. A node in C is given the IP address 10.0.0.1 and only sends and receivespackets on port 1234. A packet destined for this node reaches R2 witha source of (IP=128.4.6.6, Port=3000) and destination of (IP=128.4.8.0,Port=10000).…
- 1-In a network based on the bus topology, the bus is a non-shareable resource for which the machines must compete in order to transmit messages. How is deadlock controlled in this context? 2-Using 32-bit Internet addresses was originally thought to provide ample room for expansion, but that conjecture is not proving to be accurate. IPV6 uses 128-bit addressing. Will that prove to be adequate? Justify your answer’ (for example, you might compare the number of possible addresses to the population of the world).q no 5: In the above network diagram, IP and MAC addresses are given for network nodes. Suppose Host1 (Shahid) sends a datagram to Host2 (Khalid). What IP and MAC addresses (source and destination) will be written in IP and Data link layer headers respectively, for the following intermediate points from Host1 to Host2? Host1 to Router1 (R1) Router1 to Router2 (R2) Router2 to Host2Solve the Subnetting using FLSM & VLSM " Solve the following : a . 192.168.33.0/24 Requirement 6, 15 , 22 b . 10.0.0.0/8 subnet 14 , 22,30 c . 172.16.0.0/16 hosts 25,96 , 2040 d . 10.0.0.0/8 hosts 512 , 32,000
- A large number of consecutive IP addresses are available starting at 198.16.0.0. Sup- pose that four organizations, A, B, C, and D, request 4000, 2000, 4000, and 8000 ad- dresses, respectively, and in that order. For each of these, give the first IP address as- signed, the last IP address assigned, and the mask in the w.x.y.z/s notation. A router has just received the following new IP addresses: 57.6.96.0/21, 57.6.104.0/21, 57.6.112.0/21, and 57.6.120.0/21. If all of them use the same outgoing line, can they be aggregated? If so, to what? If not, why not? The set of IP addresses from 29.18.0.0 to 19.18.128.255 has been aggregated to 29.18.0.0/17. However, there is a gap of 1024 unassigned addresses from 29.18.60.0 to 29.18.63.255 that are now suddenly assigned to a host using a different outgoing line. Is it now necessary to split up the aggregate address into its constituent blocks, add the new block to the table, and then see if any reaggregation is possible? If not, what can be…Given a Class C subnet Mask 255.255.255.192. what will be the subnet address and host number for a machine with IP address 197.1.2.67. Select one: a. second subnet and host 2 b. Third subnet and host number 3 c. second subnet and host 4 d. First subnet and host number 3 e. None f. Second subnet and host number 3Computer Networks Assume the given class C IP Address, 192.168.0.X, where; X is 49 )Apply subnetting of 3 bits and write about: First and last valid IP of the subnet in which this IP 192.168.0.X belongs to. Sub-net mask of this IP after subnetting.