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 session key: kses = 7B (hexa);     yA = ekA(kses) = ??     yB = ekB(kses) = ??   KDC sends yA  = ?? to Alice     KDC send yB  = ?? to Bob Decrypt yA to derive kses using kA =   Decrypt yB to derive kses using kB = Message to send, m = 45 (Hexa)     Encrypyt m using session key, y = ekses(m)     Alice sends y to Bob     Decrypt y using session key to get m =     Verify that this is the message sent by Alice

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
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 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 session key: kses = 7B (hexa);

 

 

yA = ekA(kses) = ??

 

 

yB = ekB(kses) = ??

 

KDC sends y= ?? to Alice

 

 

KDC send y= ?? to Bob

Decrypt yA to derive kses using kA =

 

Decrypt yB to derive kses using kB =

Message to send, m = 45

(Hexa)

 

 

Encrypyt m using session key,

y = ekses(m)

 

 

Alice sends y to Bob

 

 

Decrypt y using session key to get m =

 

 

Verify that this is the message sent by Alice

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