200.210.56.76/28 find the subnet mask, subnet id broadcast address and also find total number of subnet and total number of host in s
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Let IP address 200.210.56.76/28 find the subnet mask, subnet id broadcast address and also find total number of subnet and total number of host in subnet. And explain Remote monitoring with neat diagram.
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- Consider the network setup in the following figure where the router is NAT-enabled. Suppose each host has two ongoing TCP connections, all to port 80 at host 128.119.40.86. Provide the six corresponding entries in the NAT translation table. Assume that the router assigns port number starting from 3000 and each host assigns port number starting in 5000For the HSRP configuration, we will use the sample network topology below. We are going to configure HSRP on both routers (R1 and R2) for subnet 192.168.1.0/24, so that PC1 is not affected by failure of any of the routers1. What is the difference between packet fragmentation (i.e., at network layer) and frame frag- mentation (i.e., at link layer) in terms of purpose? 2. Suppose that host A is connected to a router R1, R1 is connected to another router, R2, and R2 is connected to host B. Suppose that a TCP message that contains 800 bytes of data and 20 bytes of TCP header is passed to the IP function at host A for delivery to B. Show the Total length, DF, MF, and Fragment offset fields of the IP header in each packet transmitted over the three links. (Assume that link A-R1 can support a maximum frame size of 1024 bytes including a 14-byte frame header, link R1-R2 can support a maximum frame size of 512 bytes, including an 8-byte frame header, and link R2-B can support a maximum frame size of 432 bytes including a 12-byte frame header.) (*hint: the Fragment offset field is denominated by 8-bytes, not bytes) 3. What is the purpose of the path MTU discovery process (see textbook Figure 5-42) and why does…
- The maximum transmission unit on an Ethernet link is 4500 bytes. This means that the IP packets sent over Ethernet cannot be larger than 4500 bytes including the IP header. Suppose the application layer sends a 6500-byte message. The transport layer uses TCP with no options. The network layer is using IP version 4. Obviously, the IP layer will have to fragment the data. Provide the length of new datagrams (after fragmentation). Provide the Flag and offset of each of the new datagrams.Consider the network shown below. The IP and MAC addresses are shown for hosts A, B, C and D, as well as for the router's interfaces. Consider an IP datagram being sent from node B to node D. 77-34-F1 EF-14-72 128,119,97,18 A (1) 68-01-BC-58-AF-24 128,119,50,107 B 49-FA-B0-3C-E2-7C 128,119,50,60 72-9E-40-31-9C-42 128.119.240.15 CC-A5-81-08-AF-33 128,119.97.194 40-00-A0-74-06-1F 128,119,240,52 (5) C D5-AO-EE-9A-73-06 128.119.240.116 A. What is the source/destination network- or link-layer address at the location (3)? B. What is the source/destination network- or link-layer address at the location (4)? C. What is the source/destination network- or link-layer address at the location (5)? Consider an IP datagram being sent from node A to node C. D. What is the source/destination network- or link-layer address at the location (2)? E. What is the source/destination network- or link-layer address at the location (4)? F. What is the source/destination network- or link-layer address at the…If I have a Network-ID 127.91.220.100 of subnet , What is the subnet mask? can you explain it
- Consider the network setup in Figure 1. Suppose that the ISP assigns the router a public address of 44.33.111.245 and the network address of the home network is 192.168.1/24.Figure 1a. Assign addresses to all the interfaces of the router. (Consider router to be the respectable candidate for the first address in the home network, and let’s assign the addresses to the hosts in the order after that.)b. Suppose each host has two ongoing TCP connections, all to port 80 at host 123.34.56.78. Provide the six corresponding entries in the NAT translation table (Image 2). Use only dynamic ports for NAT purpose. (Hint: Choose Port Numbers in the range (and in the order) 50000 to 50005 for WAN side and 60000 to 60005 for LAN side addresses e.g. 50000 for connection1, 50001 for connection2 of Host1, and so on.TCP sessions are full-duplex, which means that data can be sent in either direction during the lifetime of the session. Consider a session in which the connection is established, the client sends 100 data segments, all of them are ACKed, and all of the ACKS are received by the sender, then the session is ended by both sides closing the connect. How many segments in total have the SYN bit of the header set to 1? а. 209 O b. 2 6 с. d. 4 О е. 1A router interconnects a subnet, where all the interfaces of the subnets must have the prefix 164.132.62/24. suppose that the Subnet is required to support up to 60 interfaces, provide a network address that satisfies the above.
- Consider the subnet address 184.84.0.0 divided into 9 subnets. If we want to maximize the number of hosts per subnet, what subnet mask should we use? What would the numbers of the first 2 and last 2 subnets be, along with their address range? Explain your answer.Consider an IP packet with a length of 4,500bytes that includes a 20-byteIPv4 header ans 40-byte TCP header. The packet is forwarded to an IPv4 router that supports a Maximum Transmission Unit (MTU) of 600bytes. Assume that the length of the IP header in all the outgoing fragments of this packet is 20bytes. Assume that the fragmentation offset value stored in the first fragment is 0. The fragmentation offset value stored in the third fragment isConsider the figure below with hosts A to F, and interfaces assigned IP and MAC addresses. Suppose all of the ARP tables are up to date and Host A sends a datagram to Host F. Give the source and destination MAC and IP addresses in the frame encapsulating this IP datagram as the frame is transmitted (i) from A to the left router, (ii) from the left router to the right router, and (iii) from the right router to F. E 192.168.1.001 00-00-00-00-00-00 192.168.2.001 44-44-44-44-44-44 192.168.3.001 77-77-77-77-77-77 Router 1 192.168.1.002 192.168.2.002 192.168.2.003 22-22-22-22-22-22 33-33-33-33-33-33 55-55-55-55-55 192.168.2.004 192.168.1.003 D 66-66-66-66-66 11-11-11-11-11-11 from A to Router 1 Router 1 to Router 2 LAN B Source MAC address Destination MAC address Source IP Destination IP LAN Router 2 LAN 192.168.3.002 88-88-88-88-88-88 F from Router 2 to F 192.168.3.003 99-99-99-99-99-99