Most IP datagram reassembly algorithms have a timer to avoid having a lost fragment tie up reassembly buffers forever. Suppose that a datagram is fragmented into four fragments. The first three fragments arrive, but the last one is delayed. Eventually, the timer goes off and the three fragments in the receiver's memory are discarded. A little later, the last fragment
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- For NAT Network Address Translation Scenario: Jenny Bello is a small business owner selling and making customized computer peripherals. She has been finding it difficult to track her sales and inventories, however recently after an expert's advice, she adopted a Point of Sale (PoS) server. So, she can now track her sales and inventories at the store. However, she ran into another issue, the server can only be accessed within the store, because it has been assigned a private IPv4 address, it is not publicly accessible via the Internet. a) Why is not having the PoS server accessible over the Internet a problem for the business? b) What caused this problem? c) Propose a solution for the problem and explain how the solution works.R6Fragmentation of an IP datagram takes place if its size is larger than the MTU of the subnet over which the datagram will be routed. Most IP datagram reassembly algorithms have a timer to avoid having a lost fragment tie up reassembly buffers forever. Suppose a datagram is fragmented into four fragments. The first three fragments arrive, but the last one is delayed. Eventually the timer goes off and the three fragments in the receiver's memory are discarded. A little later, the last fragment stumbles in. What should be done with it?A frame containing http request is sent from Computer (A) [in whichever subnet it is now, make sure to show it in your figure], to the access point it is associated with. Draw the address fields (1, 2, and 3) of the frame travelling from Computer A to AP and the source and destination addresses of the frame travelling from AP to R1. The MAC addresses of AP1, AP2, and AP3 are M1, M2, and M3 respectively, and the Router R1’s MAC address connected to this Switch/AP is MRA. IP address values are already given, and the port addresses could also be used as given. The Router’s MAC address facing WAN side is MRW.
- In an RPC-like protocol, where numerous requests can be active at the same time and responses can be given in any sequence, the following is possible: Pretend requests are sequentially numbered, and that ACK[N] confirms the receipt of reply[N]. Should the number of ACKs be cumulative? If not, what should happen in the event that an ACK is not received?Consider a client connecting to a web server via a router as shown in Fig.Q2. Client A sends a request to the server to retrieve a 7.5 Mbytes file. Given that the segment size is 50 Kbytes, the round trip time (RTT) between the server and client is 10 ms, the initial slow-start threshold is 16 and the client's buffer always has a storage space of 1 Mbytes. Assume that TCP Reno is used, there is no loss during transmission and the headers of protocols are ignored. 400 Mbps 200 Mbps 400 Mbps Link a Link b Link c Client Web Server Fig.Q2 (a) Describe how the value of sending window changes as a function of time (in units of RTT) during the whole connection time. 2.When a file is transferred between two computers, two acknowledgement strategies are possible. In the first one, the file is chopped up into packets, which are individually acknowledged by the receiver, but the file transfer as a whole is not acknowledged. In the second one, the packets are not acknowledged individually, but the entire file is acknowledged when it arrives. Discuss these two approaches.
- In an RPC-like protocol, many requests can be active at the same time, and responses can be sent in any order: Assume that requests are sequentially numbered and that ACK[N] acknowledges reply[N]. Should ACKs be added up? If not, what should happen if an ACK is not received?Consider the following scenario in which host A is sending a file to host B over a TCP connection. Assuming that the sequence number of the first data byte sent by A is 0 and every segment always includes 1000 bytes of data, excluding the TCP header. At some point of time, bytes up to 6400 have been written into the sender’s buffer. Bytes up to 4999 have been sent out but the segment which contains bytes 2000~2999 has not arrived at host B yet. At the receiver’s side, all bytes up to 3999 have been received except for bytes 2000~2999. Bytes up to 499 have been read from the buffer by the application. Assume that the maximum size of the sender’s buffer is large enough. Consider the sliding window algorithm in TCP and answer the following questions. 1) What are the values for LastByteAcked, LastByteSent, and LastByteWritten? 2) What are the values for LastByteRead, NextByteExpected, and LastByteRcvd? 3) Assuming that the maximum size of the receiver’s buffer is 4000 byte, what would…b) Consider a scenario where fragmentation may take place in packet transfer. Discuss how such fragmentation may be reassembled at the destination. Now consider in such scenario, a datagram is fragmented into five fragments. The first four fragments arrive, but the last one is delayed. Eventually, the timer goes off and the four fragments in the receiver’s memory are discarded. A little later, the last fragment arrives. What should be done with it?
- TCP is a sliding window protocol. This means... а. the TCP retransmission timer slides up as ACKS are received. O b. there can be more than one segment "in flight" (sent but no ACK has been received) at a time. с. there can only be one segment outstanding (sent but no ACK has been received) at at time.Please solve it correctly and please provide explanation of your answers. A PC and a Web Server are communicating over a TCP connection. The PC had started the three way handshake with the initial sequence number of 3069 . The Web Server's initial sequence number is 4830 . The window size of the PC is 815 bytes and the window size of the Web Server is 463 bytes. Using the third TCP handshake ack segment the PC sends the http request of the size 396 bytes to the Web Server. Then the Web Server answers with 3 segments containing the requested data. The first segment size is 42 bytes and the second segment size is 276 bytes and the third segment size is 146 bytes respectively. The PC receives all three segments within the timer. But unfortunately the second segment was corrupted. So the PC immediately sends an acknowledgement segment. Assume that the PC uses Selective Repeat/Reject ARQ. (d) What will be the window size of PC A after receiving only the first segment from webserver? (e)…Please solve it correctly and please provide explanation of your answers. A PC and a Web Server are communicating over a TCP connection. The PC had started the three way handshake with the initial sequence number of 3069 . The Web Server's initial sequence number is 4830 . The window size of the PC is 815 bytes and the window size of the Web Server is 463 bytes. Using the third TCP handshake ack segment the PC sends the http request of the size 396 bytes to the Web Server. Then the Web Server answers with 3 segments containing the requested data. The first segment size is 42 bytes and the second segment size is 276 bytes and the third segment size is 146 bytes respectively. The PC receives all three segments within the timer. But unfortunately the second segment was corrupted. So the PC immediately sends an acknowledgement segment. Assume that the PC uses Selective Repeat/Reject ARQ. (a) What will be the sequence number of the third TCP handshake signal send from PC to webserver?…