Consider three networking scenarios N1, N2, and N3. In scenario N1, 10 users deploy SKC for secure communications. In scenario N2, 20 users use PKC. In scenario N3, 50 users use SKC with a central key distribution center.

Consider three networking scenarios N1, N2, and N3. In scenario N1, 10 users deploy SKC for secure communications. In scenario N2, 20 users use PKC. In scenario N3, 50 users use SKC with a central key distribution center.

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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Consider the following simple authentication dialogue: C → AS: IDc II Pc II IDv AS → C: Ticket C → V: IDc II Ticket; Ticket = E(Kv, [IDc II ADc II IDv]) Where: C = client AS = authentication server V = server IDc = identifier of user on C IDv = identifier of V Pc = password of user on C ADc = network address of C Kv = secret encryption key shared by AS and V ***Explain the process of this protocol.
Alice wants to set up a business company that securely deals with her clients. Bob is a potential client that Alice wants to have a good connection. As a security consultant, your job is to devise a plan to establish a secure information exchange. The assumption is made that Alice and Bob already have their public and public keys; (PrivA, PubA) and (PrivB, PubB) respectively, and both of them have exchanged their public key beforehand. Your plan includes comprehensive protocols to secure all messages, according to Alice’s requirements below. Discuss following 3 in relation with Asymmetric Encryption . i) Integrity: Alice and Bob want to make sure that every message sent cannot be forged by anyone. ii) Authenticity: Alice and Bob can verify that messages actually came from iii) Confidentiality: Alice exchanges business-related documents with Bob in a way, so that Alice and Bob may open, read or modify the documents.
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Computer Science Before the introduction of IEEE 802.11i, the security scheme for IEEE 802.11 was Wired Equivalent Privacy (WEP). Assume that the initialization vector (IV) is not used in WEP. Instead, RC4 is applied using K as input and producing s as output. Assume RC4(K, n) is applied for each packet and that all packets have the same length. If an attacker intercepts a single packet with ciphertext C1 and also has a known plaintext/ciphertext pair (P2, C2). Show how the attacker can find the plaintext P1. You can make use of an example by assuming some random data is being encrypted.
Shared session key establishment using a Key Distribution Center (KDC). Using the following table, illustrate how Alice can initiate a secure session with Bob with the help of KDC. Here, KEKs are the long term key establishment keys used to transport the session keys across the network securely. Assume the encryption process to be as follows: Block (LB || RB) is 8 bits; Encryption Key (LK||RK) is 8 bits; Ciphertext = LC|| RC where LC=LB ⊕ RK; and RC=RB ⊕ LK. For example, if plaintext=A7 (Hexa) and Key = 6D; then LC=A ⊕ D = 1010 ⊕ 1101 = 0111 = 7 (Hexa); and RC = 7 ⊕ 6 = 0111 ⊕ 0110 = 0001 = 1 (Hexa); so Ciphertext = 71 (Hexa). To decrypt, it does the reverse operation: Given ciphertext of C=LC||RC, it finds plaintext B=LB||RB, by finding LB=LC ⊕ RK and RB = RC ⊕ LK. Fill in the values at "??" below. Alice KDC Bob KEK: kA= D6 (hexa) KEK: kA = D6 (hexa); kB = A8(hexa); KEK: kB= A8 (hexa); Alice sends a message to KDC requesting a session key between Alice and Bob…
Shared session key establishment using a Key Distribution Center (KDC). Using the following table, illustrate how Alice can initiate a secure session with Bob with the help of KDC. Here, KEKs are the long term key establishment keys used to transport the session keys across the network securely. Assume the encryption process to be as follows: Block (LB || RB) is 8 bits; Encryption Key (LK||RK) is 8 bits; Ciphertext = LC|| RC where LC=LB ⊕ RK; and RC=RB ⊕ LK; For example, if plaintext=A7 (Hexa) and Key = 6D; then LC=A ⊕ D = 1010 ⊕ 1101 = 0111 = 7 (Hexa); and RC = 7 ⊕ 6 = 0111 ⊕ 0110 = 0001 = 1 (Hexa); so Ciphertext = 71 (Hexa). To decrypt, it does the reverse operation: Given ciphertext of C=LC||RC, it finds plaintext B=LB||RB, by finding LB=LC ⊕ RK and RB = RC ⊕ LK. Alice KDC Bob KEK: kA = A6 (hexa) KEK: kA = A6 (hexa); kB = D8(hexa); KEK: kB = D8 (hexa); Alice sends a message to KDC requesting a session key between Alice and Bob Generate a random…
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