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- The maximum packet size (Maximum Transmission Unit or MTU) of an IP packet including IPv4 header on 100 Mbps Ethernet is usually set at 1500 bytes. A typical IPv4 header consists of 20 bytes, and a UDP header consists of 8 bytes. If we split up a file of 20,000,000 bytes so we can send it as a series of UDP payloads, how many IP packets do we have to send in order to transfer the entire file? Enter an integer number without formatting (no commas). Answer:The maximum packet size (Maximum Transmission Unit or MTU) of an IP packet including IPv4 header on 100 Mbps Ethernet is usually set at 1500 bytes. A typical IPv4 header consists of 20 bytes, and a UDP header consists of 8 bytes. If we split up a file of 20,000,000 bytes so we can send it as a series of UDP payloads, how many IP packets do we have to send in order to transfer the entire file? Enter an integer number without formatting (no commas). Answer: The MTU for IP packets on 100 Mbps Ethernet is typically set to 1500 bytes. A typical IPv4 header consists of 20 bytes, and a UDP header consists of 8 bytes. If we split up a file of 25,000,000 bytes so we can send it as a series of UDP payloads, how many bytes do we have to send at the network layer in order to transfer the entire file? Enter an integer number (no commas). Answer: You are sending a 27,000,000 byte file using UDP over IP over an Ethernet with MTU 1500 bytes. The Ethernet header is 14 bytes and the frame checksum is 4…A packet switch receives a packet and determines the outbound link to which the packet should be forwarded. When the packet arrives, one other packet is halfway done being transmitted on this outbound link and four other packets are waiting to be transmitted. Packets are transmitted in order of arrival. Suppose all packets are 1,500 bytes and the link rate is 2 Mbps. What is the queuing delay for the packet? More generally, what is the queuing delay when all packets have length L, the transmission rate is R, x bits of the currently-being-transmitted packet have been transmitted, and n packets are already in the queue?
- An IP address can be written as a 32-bit number. For a class B network, the two most significant bits are set to 10. The 16 most signi cant bits are used as a network ID, and the 16 least signi cant bits are used as a host ID. However, the host ID cannot be all 0's or all 1's. How many hosts (i.e., host IDs) can there be on a class B network?please answer with proper explanation and step by step solution. Question: Consider a fictional six-layer protocol hierarchy where layer 1 is the lowest layer, and layer 6 is the highest layer. Therefore, an application sends a message M by passing it to layer 6. All the even-numbered layers attach a trailer Ti (i = 2, 4, 6) to their payload, and all the odd-numbered layers attach a header Hi (i = 1, 3, 5) to their payload. Draw the headers, trailers, and original message M in the order they are sent out.An IP address can be written as a 32-bit number. For a class B network, the two most significant bits are set to 10. The 16 most significant bits are used as a network ID, and the 16 least significant bits are used as a host ID. However, the host ID cannot be all 0's or all 1's. How many hosts (i.e., host IDs) can there be on a class B network?
- An Internet user would like to transfer a message from his computer (host A) to another computer (host B). The size of the message passed to the IP layer is 3800 bytes. The datagram(s) carrying the message will have to cross two routers (R1 and R2) and 3 networks (NET1, NET2, and NET3) as described in the above figure. Each of these networks has a specific MTU (for example the MTU of network 1 is 1500 bytes. MTU here stands for Maximum Transmission Unit, the largest data that a frame can carry). Fill the table with the IP datagram(s) needed right after passing through NET3 (no need to show your fragmentation for N1 or N2). Assume a 2-byte options in the header. Show all your details (for partial points in case your numbers are incorrect). NET1 MTU=1500 R1 NET2 MTU=640 R2 NET3 MTU 1500 V BIU Fragment # Format Offset Flag (3 bits) Data Length Total Length ...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.1. 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…
- Host A sends the information of size 3000 bytes to Host B using TCP protocol. As long as Host B receives each packet in size 1000 bytes, Host B must acknowledge Host A. We assume the header size in each packet is 40 bytes, the starting sequence number used by Host A is 1000, and the window size of the sender side is 3000 bytes. What is the value of SND.NXT when the second packet is sent out, but only acknowledged for the first 300 bytes of the first packet? Question 7 options: 1000 2000 3000 1300Communication between two programs running ondifferent terminals is possible because these programsuse the services of the application layer. At the same time,Application layer protocols usepresentation and session layer services. TheseThe last two use the services of the protocols of thetransport layer, which depends on the correctoperation of network layer protocols. Thetransmission of data over a link depends on thelink layer protocols. Finally, the data issent at the bit level according to the physical layer standardcorresponding to the communication link between a pairof nodes. All these concepts are very difficult to deal with, sothat your mission in this case is to explain how it is that twoprograms executed on two different terminalsThey can communicate using the OSI protocol stack.Use the following diagram as the basis for your explanation.Note that the diagram corresponds to the TCP/IP model, not to theOR IF.A system has 7 layers protocol hierarchy. Applications generate message of length "500" Bytes. At each of the layers, a "20" byte header is added. The percentage of the network bandwidth wasted on headers is