3. Suppose a client uses UDP to send data to a server and suppose that the length of the UDP datagram (including the UDP r) is 2048 bytes. ate the efficiency of this transmission at the UDP level. s, the ratio of data bytes (i.e. bytes that are not in the UDP header) to total bytes. ss your answer as a percentage accurate to the nearest tenth of a percent (e.g. in the form 56.7%)| erl
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- TCP / IP uses the client-server communication model, in which a user or computer (a client) receives a service (such as sending a web page) from another computer (a server) on the network. In short, the TCP / IP protocol suite is classified as stateless, which means that each client request is considered new because it has nothing to do with previous requests. Being stateless, network routes are released so that they can be used continuously. Answer the following questions briefly. How does resource reservation work in the TCP / IP model? Which TCP / IP layer is responsible to reserve resources across a network using the TCP / IP model? What protocol assigns an IP address to the client connected to the Internet? Checksum is used by various protocols on the Internet, but not at the one of TCP / IP Explain.9.Correct expression tor UDP user datagram length is length of UDP = length of IP- length of IP header's length of UDP= length of UDP-length of UDP header's length of UDP = length of IP+length of IP header's length of UDP = length of UDP+length of UDP header's.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)…
- A portion of the header of a UDP segment is captured as 8-bit hexadecimal numbers as shown below. 00 35 D2 71 00 86 For this UDP header, the checksum in hexadecimal is Ox[a] and the length of the UDP segment in decimal is [b]. Note: do not include 'Ox' in your first answer.Suppose that the average object size is 850,000 bits and that the average request rate from the institution’s browsers to the origin servers is 16 requests per second. Also suppose that the amount of time it takes from when the router on the Internet side of the access link forwards an HTTP request until it receives the response is three seconds on average (see Section 2.2.5). Model the total average response time as the sum of the average access delay (that is, the delay from Internet router to institution router) and the average Internet delay. For the average access delay, use Δ/(1 – Δ ), where Δ is the average time required to send an object over the access link and is the arrival rate of objects to the access link. a. Find the total average response time.b. Now suppose a cache installed in the institutional LAN, Suppose the miss rate is 0.4. Find the total response 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. (a) What will be the sequence number of the third TCP handshake signal send from PC to webserver?…
- In a bit-oriented link-layer protocol, the start and end of a frame are each marked by a flag, which is a sequence of exactly 6 consecutive 1 bits. Bit stuffing is used during the rest of the frame: after sending 5 consecutive 1 bits, a 0 is added. The bits below contain an example frame, with the leftmost bit received first. The content of the frame is a sequence of bytes, sent most significant bit first. Put the values of the first four bytes of the frame content, in order, in the four answer boxes below. You may enter the values in either decimal (e.g. 76, 123) or hexadecimal (e.g. 2a, f7). 00101011111010001111110011100111110000111101111101010110100010010011111101101100100 57−240−123−229Suppose we have a web page which contains 10 objects including: a html main file, 8 jpg image, and a .mp3 background music. For simplicity, assume each of the 10 objects has a size of 4 Kbits and that each object can be completely transferred in one TCP packet. Furthermore, assume the round-trip time between the server and client is RTT= 250 ms, and the download rate is 1 Mbps. What will be the time it takes to complete transferring the web page running in the non-persistent mode? Total delay = msRegistration Number: 193046 Consider a network with 3 hosts and a NAT router. Suppose that the network address of the home network is 192.168.NNN/24, where NNN are the last 3 digits of your Reg#. For example, if your Reg# is BSE193046, NNN = 046 and therefore the network address wil be 192168.46/24. (If NNN>255, consider NNN=254) (a) Assign any public IP address to NAT router’s outgoing interface. (b) Assign IP addresses to all 3 hosts and local interface of the NAT router in the home network. (
- Q1.Suppose a client, say C, has established a TCP connection with a server, say S. After establishing the connection, the sender sends two TCP messages with data back-to-back ( the first TCP message with sequence number 32 and the second TCP message with sequence number 100). a. What is the sequence number chosen by the client? What could be the reason for choosing this value of sequence number?b. What is the data size of first and second TCP data messages?c. Suppose the server receives the first and second TCP messages with data with the difference of 100 milliseconds delay. What could be the ACK number for both the TCP data messages from the server side as per the TCP RFC.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…Consider Figure 2.12, for which there is an institutional network connected to the Internet. Suppose that the average object size is 1,000,000 bits and that the average request rate from the institution’s browsers to the origin servers is 16 requests per second. Also suppose that the amount of time it takes from when the router on the Internet side of the access link forwards an HTTP request until it receives the response is three seconds on average (see Section 2.2.5). Model the total average response time as the sum of the average access delay (that is, the delay from Internet router to institution router) and the average Internet delay. For the average access delay, use ∆/(1 - ∆b), where ∆ is the average time required to send an object over the access link and b is the arrival rate of objects to the access link. Find the total average response time. I get ∆=1000000/15000000= 1/15, then ∆/(1 - ∆b)= (1/15)/(1-(1/15)*16)=-1, so the finally answer is : 3+(-1)=2s or 3+0=3s?