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- In a TCP data packet, if the sequence number has the value 001111, ACK number has the value 010110 and the urgent pointer has the value 001100 assuming that URG flag is set. What is the last urgent byte in the data section of the packet? All numbers are in binary.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?Please solve it correctly and please provide explanation of your answers. Please answer parts g, h and i. 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. g) The second segment from webserver was corrupted. So the PC…
- 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?…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.
- 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 ACKIN] 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?|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?Complete the following protocol message so that it represents a potentiometer reading of 987: (write values in hex, with the leading 0x) Ох21 Ох33 Ох0з Оxdb
- In IPv4, the address format is a 32-bit numeric number that is frequently written as four octets separated by periods from the range [0, 255]. At the time of writing, Internet Protocol version 6 (IPv6), with 128-bit addresses, was gradually replacing IPv4. How many IP addresses may be assigned in IPv4 and IPv6 in theory?CIS269 Packet Translation Lab Break down the following IP datagram into its individual fields, listing the value of each. Hint: It’s an IP datagram containing a TCP segment containing another protocol… 45 00 00 31 94 00 40 00 80 06 1B 9F 80 A3 C9 29 D8 45 29 15 04 09 00 15 00 00 C3 A1 DF 65 A8 45 50 18 20 AA 8E 8C 00 00 43 57 44 20 66 75 6E 0D 0AThe filed “window size” in TCP header is in length of 16 bit. It allows 64KB as the maximal size for a TCP segment by default. Under some circumstances, the sending rate is pretty high and it results in a very short sending time (<2ms). But the end to end transmission delay is 50ms. Therefore the idle rate of a channel will be more than ⅔. How does TCP solve this problem?