Aveneu Park, Starling, Australia

New amount 1126387Token =3C F1 70 C8

New York Institute of TechnologySchool of Engineering and ComputingSciencesINCS-741 Cryptography – Assignment3Submitted by:Abhilash Basheerbad Student Id:1126387a. Calculate the messages sent in steps 1 to 4 if A is sending $ amountequals to your NYIT ID to C.AnsStep 1: NYIT ID: 1126387Amount: 1126387Step 2:KeyBank:AES128 Key: 670DBA40AB1F3752EF0DC1D0F8FB4958token = AES128(KeyBank , amount )After encrypting the amount using the amount 1126387Token =3C F1 70 C8 7D E9 35 38 A4 E4 4C 0B 72 CA 6B 17check = RSA (PuC , token)check =8c ec a3 4b 2c dd 1a 11 d6 b7 70 cb 1b d0 96 0b 6f 6f 93 80 f8 a0 ef23 0e eb ff 18 d1 c7 17 db ba 68 76 be 4d 2a 71 2c f9 c1 af 5e f9 d3 8b db 6501 6c 99 1f 4e 18 a4 4a b8 db 6c 61 01 0d 3d b9 9c 83 87 f9 0a 53 87 36 65f1 76 55 5e d7 bd 1b 47 13 b4 6f 74 9e ab ec 97 2a aa d4 57 6a 5b eb e1 3654 c2 f1 14 c0 97 56 12 48 c8 37 98 ef 46 0f 37 bb 31 bf 75 db e4 ba bf 6050 f0 94 99CR = RSA( PuA , check )CR = 87 70 c7 81 79 6a f7 ef 72 d5 97 aa be b2 e4 e4 e3 7f ce 96 a9 38 42ef 12 28 0e 64 7c 77 f3 37 f3 44 7f 92 5d 4f 1b da b9 7c 0f e4 4d 05 08 be 06f6 f8 09 ee 6f 7d 36 e1 c5 be ba 03 d0 65 19 61 7a f7 00 39 82 62 5a c3 70d2 94 a6 7d 34 b5 dd 61 7e 59 55 f6 38 57 93 4f d8 2a e5 06 c9 e3 7d 5f 7e0e 87 6c 89 f7 7e eb 1b ab 7c 91 34 9d b1 d1 e4 b7 15 1f 1c 4a 39 b7 8c dad8 0d a7 d9Step 3:A will decrypt the CF received from BankCheck = RSA(PrA , CR)Check = 8c ec a3 4b 2c dd 1a 11 d6 b7 70 cb 1b d0 96 0b 6f 6f 93 80 f8 a0ef 23 0e eb ff 18 d1 c7 17 db ba 68 76 be 4d 2a 71 2c f9 c1 af 5e f9 d3 8b db65 01 6c 99 1f 4e 18 a4 4a b8 db 6c 61 01 0d 3d b9 9c 83 87 f9 0a 53 87 3665 f1 76 55 5e d7 bd 1b 47 13 b4 6f 74 9e ab ec 97 2a aa d4 57 6a 5b eb e136 54 c2 f1 14 c0 97 56 12 48 c8 37 98 ef 46 0f 37 bb 31 bf 75 db e4 ba bf60 50 f0 94 99Step 4From C to Bank: RSA( PrC , token )Token = 3c f1 70 c8 7d e9 35 38 a4 e4 4c 0b 72 ca 6b 17From C to Bank : 1e 17 94 57 b2 f2 fc db f6 1c d0 4b 4a ed 68 b0 c9 2d 2d0d 15 e3 6f 2f d3 f0 9a 86 86 37 7c 6c a0 3f 27 ee 03 5e 8d eb 3c 21 d7 577a 58 07 6e 54 47 ae 24 b9 76 83 54 8a 2a e9 75 8e 0d fe cc 55 e6 24 cf 8467 99 97 55 93 bb 25 1f 47 16 22 f5 87 43 d9 29 45 b2 ac cc 0e c3 cb 6e d8dd 77 a6 a9 81 ad cc 95 b6 de 97 b1 02 45 d4 d9 1c 58 e9 ad 47 5e 20 a9 a0a3 1b cc f3 af 89 b2 6e cca. Explain two different vulnerabilities and attack scenarios that areapplicable to the above design.Ans Different vulnerabilities and attack scenarios are as follows:? In above mentioned technique, keys are more vulnerable to bruteforce attacks. On the basis of the algorithm, the unintended person cancrack out the private keys? Man in the middle attack is also one of the vulnerabilities that theabove technique may face. In this situation, the malicious user canintercept the message in between and can pass the message along withhis or her public key and claim to be the original sender.b. Propose an enhancement to the above design to make it more secure.Ans Above process can be more efficient and robust by using the public keycryptography with the secret key In this mechanism, the sender canincorporate the extra message that can be generated using the secret keyencryption and the user the verify the extra message using the same secretkey technique.Hash(M) = Encrypt (Message, Secret Key)Total Message = Message + Hash(M)The receiver will verify the hash by computing the hash on the message if itcomes out be the same then it is the correct message otherwise the receiverwill neglect it.References:http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.112.7813&rep=rep1&type=pdfhttps://itstillworks.com/disadvantages-publickey-encryption-1980.htmlhttp://www.dcs.ed.ac.uk/home/adamd/essays/crypto.html

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