Which three statements about NSAP addresses are true? (Choose three.)
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- Suppose the DES Mangler function mapped every 32-bit value to zero, regardless of the value of its input, what function would DES compute? Solve this question by answering the following 3 sub-questions. What is the result of each round? What is the result after the 16 rounds? Remind DES = Initial permutation + 16 rounds + swap halves + final permutation? Assume the original message is a 64-bit message, i.e., 1 2 3 4 … 64. Specify the bit positions by then end of these 4 operation. Use 8x8 matrix for this. Must provide detailed explanation of the calculation steps.Computer Science Can you tell whether the datagram is carrying UDP, TCP, or ICMP data? How? What is the SPI for this SA from your Ubuntu to DVWA-Ubuntu? What is the SPI for this SA from DVWA-Ubuntu to Ubuntu? How are the sequence numbers changing in each of the SAs? In your own words, describe in what cases is manual keying feasible and in what cases it is not.What is an IP address exactly? An abbreviation for "memory address" is a mnemonic address. How many distinct domains are feasible when using a 32-bit representation? How many computers may be included in a domain?
- Suppose Alice and Bob are going to communicate using AES in CBC mode. Unfortunately Alice's message length (in bytes) is not a multiple of 16. Suppose the last block of her message is just a single zero byte. How can she pad out the last block so that she can use CBC mode? Since this needs to be a reversible operation, how does Bob recognize the padding and remove it?Which pairs of addresses have the same network address if the netmask used for the network is 255.255.248.0? ( Choose two) a. 192.168.10.66 and 192.168.10.71 b. 192.168.10.130 and 192.168.10.133 c. 192.168.10.1 and 192.168.10.6 d. 192.168.10.27 and 192.168.10.32 e. 192.168.10.57 and 192.168.10.61The very last block of a message encrypted with AES-256 in CBC mode is 15 bytes long and ends with the bytes: 0102030405060708090a0b0c0d0e0f When padded in accordance with PKCS #7 with 16-byte blocks, what is the complete value of the last block? *Please explain how you have solved this, thank you!!
- What exactly is the purpose of this endeavour if you need a private network address? Is it at all possible for a datagram to make its way from a private network to the public Internet after being transmitted from that network? Explain.Compute CBC-MAC for a message of 16 bits, “ABCD” (in Hexa). Assume a block size of 8 bits with an IV=C9 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be D8 (in Hexa).a. Compute CBC-MAC for a message of 16 bits, “8642” (in Hexa). Assume a block size of 8 bits with an IV=F1 (in hexa). For simplicity, assume the encryption to be a simple XOR of the key with the plaintext. Let the encryption key be B4 (in Hexa). (Hint: Divide the message into blocks of 8 bits each; XOR each block with the previous cipher output; then encrypt this with the key. For the first block, XOR it with IV. Details in pages 325-326 Ch.12 of the textbook) b. Suppose Alice computes the Secret prefix MAC (page 322: secret prefix MAC(x) = h(key || x)) for the message ”AM” (in ASCII) with key “G” (in ASCII) that both Alice and Bob know. The hash function that is used is h(x1x2x3)= g(g(x1 XOR x2) XOR x3 ) where each xi is a character represented as 8 bits, and g(x) is a 8-bit string that is equal to the complement of bits in x. For example, g(10110011) = 01001100. The MAC is 8 bits. (8-bit ASCII representation of the characters is given below.) What is the Secret prefix MAC…
- To reduce waste when subnetting, borrow exactly half the host bits. True or False?What is the range of addresses for jump and jump and link in MIPS (M 1024K)?1. Addresses between 0 and 64M 12. Addresses between 0 and 256M 13. Addresses up to about 32M before the branch to about 32M aft er4. Addresses up to about 128M before the branch to about 128M aft er5. Anywhere within a block of 64M addresses where the PC supplies the upper 6 bits6. Anywhere within a block of 256M addresses where the PC supplies the upper 4 bitsUDP and TCP use 1’s complement for their checksums to detect errors. Suppose you have the following 8-bit bytes: 11011001, 01010010, 11001010, 10100100 and 01011001. A. What is the 1’s complement of the sum of these 8-bit bytes? Show all the details of your work. B. Why do UDP and TCP take the 1’s complement of the sum as their checksum? That is, what do UDP and TCP not simply use the sum of these bytes as checksum? C. With the 1’s complement scheme, how does the receiver detect errors? D. With this checksum scheme, is it possible that any l-bit error will go undetected? How about a two-bit error? Explain your answer.