Once the server sends the success packet, it will begin sending the entire satellite dump over the socket. In order to help understand the transmission, we have stolen an excerpt of a secret document that may help. 2.1 Each packet has the following structure: 0 4 8 12 | SEQ | CHK | LEN |...LEN bytes of raw Linear PCM | CHK (Checksum) The 32-bit checksum is calculated by XOR'ing each 32-bit chunk of the 4 byte big-endian sequence number (SEQ) and data. If LEN is not a multiple of 4, the byte 0xAB shall be used in place of the missing bytes only for the purpose of calculating the checksum.

Once the server sends the success packet, it will begin sending the entire satellite dump over the socket. In order to help understand the transmission, we have stolen an excerpt of a secret document that may help. 2.1 Each packet has the following structure: 0 4 8 12 | SEQ | CHK | LEN |...LEN bytes of raw Linear PCM | CHK (Checksum) The 32-bit checksum is calculated by XOR'ing each 32-bit chunk of the 4 byte big-endian sequence number (SEQ) and data. If LEN is not a multiple of 4, the byte 0xAB shall be used in place of the missing bytes only for the purpose of calculating the checksum.

Database System Concepts

7th Edition

ISBN:9780078022159

Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan

Chapter1: Introduction

Section: Chapter Questions

Problem 1PE

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mov edx, esp mov ecx, esp nx: mov edi, [ebp+8] add edi, 6 ; Welcome message ; position to where we put name s1: mov al, byte ptr [ecx]; get start of name from local mov byte ptr [edi], al; store it in welcome message inc edi inc ecx cmp al,0 jnz s1 add sp, 32 ret 8 sco endp .data db 32 dup(0) .code ; start of name, for saving now ; start of name, for reading later end start ; keep going until zero delimeter ; release local storage ; release call parameter xs: MsgBox 0, "We never get here", "Buffalo", MB_OK invoke ExitProcess,0
6. Consider the following simplified version of the CFB mode. The plaintext is broken into 32-bit pieces: P = [P₁, P2, ,], where each P, has 32 bits, rather than the 8 bits used in CFB. Encryption proceeds as follows. An initial 64-bit X₁ is chosen. Then for j = 1,2,..., the following is performed: CjPjL32 (Ek (Xj)) Xj+1 = R32(X₁)||C₁ where L32(X) denotes the 32 leftmost bits of X, R32 (X) denotes the rightmost 32 bits of X, and XY denotes the string obtained by writing X followed by Y. Find the decryption algorithm.
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?…
ASN.1 is an ISO standard used in many internet‑related protocols, especially those in the area of network management. In ASN.1, BER (Basic Encoding Rule) uses a so‑called TLV (Type, Length, Value) approach to encoding data for transmission. According to this BER encoding rule, what would be the encoding of {weight, 178} {lastname, “Philips”}?
When Snort generates an alert message, it will usually start with a line like the one shown below: [**] [116:56:1] (snort_decoder): T/TCP Detected [**] These values/numbers are used to track the rules that Snort uses for a particular alert. Describe briefly what the above three sets of numbers shown in the square bracket mean. i.e. [116:56:1] Answer: a. The first number set (i.e.116) is: b. The second number (56) is: c. The third number is (i.e.1) is:
42. Though there are practical reasons why we don't complicate our lives this way in the real world, from a theoretical perspective we could borrow non-consecutive bits for subnetting. So, when borrowing the first host bit, the second host bit and the last two host bits (counting from the left of the host bit field) from 172.1.0.0 can you determine the following: d. number of remaining host bits and then the number of useable hosts on a subnet?e. First useable host on the 2nd useable subnet?f. broadcast address for 2nd useable subnet?
You have an originating plain text message (in hexadecimal): 41 72 61 62 Use your Student ID number to obtain a 10-bit key as follows: Key (in decimal) = (ID mod 1024) +1.You have to manually cipher the first block of the following plain text message (first 8 bits). The encrypting process will be done using Simplified DES (S-DES). you must explain all steps in details [mandatory: write down all the calculations].
using c++ write a program program:- Suppose that you are receiving a packet number of a data travelling over a network in array. First of all input number of packets a data will be in. Received packets must be in an order and there will be difference of 4 among packet numbers. If there is any missing packet than display message packet missing. If no packets are missing but are not in order arrange the packets in an order.
Write a code in C that prints the following: Prints default TCP send and receive buffer sizes.Prints default UDP send and receive buffer sizes.Prints default SCTP send and receive buffer sizes.
> موعد التسلیم السبت، 1:59 1 م HW2 computer network protocols äböi 5 إضافة تعليق صف Q1) Compare OSI and TCP/IP models ? Q2) Assume a block code has k=3 and n=6. Determine the code word of a data word (101). Discuses how the code word is derived from this dada word ? تم تعيینه عملك نصف تعليقا خاصا
SInternet Check sum Consider the figure below (simplex communication scenario). The TCP sender sends an initial window of 4 segments. Suppose the initial value of the sender ན༽)། sequence number, SEQ1, is 100 and the first 4 segments each contain 400 bytes of payload. As you can see ACK 3 is lost. What is the SEQ number (SEQ2, SEQ3, SEQ4, SEQ5) from the Sender and ACK Number (ACK1, ACK2, ACK3, ACK4, ACK5) from the receiver for each segment? Sender Seq1 Seq2 Seq3 Seg4 ACK1 ACK2 ACK3 ACK4 Seq5 ACK5 Receiver