(Solved) : Use Ban Logic Analyze Three Key Distribution Protocols Given Slides Q42705677 . . .

Use BAN logic to analyze the three key distribution protocolsgiven in the slides.

KEY DISTRIBUTION SCENARIO Key Distribution Center (KDC) a.k.a. supplicant Initiator A Responder B (1) IDA || IDB || N Names and nonce J Key distribution steps (2) E(KA [K, II IDAII ID|| ND) || E(K), [K, || IDA) Session key, etc. For each of A&B (encrypted) (3) E(K), [K, ILIDAD) Forward key with name — – – – – – — – — – – – — (4) E(K.N2) Challenge nonce k Authentication steps Response (5) E(Ks, f(N2)) DECENTRALIZED PROTOCOL No protection at all Name and challenge nonce (in clear) -(1) IDAN Initiator Responder Shows B saw nonce (fresh), B sent msg Session key, nonces, names encrypted with master key —(2) E( K K | IDA I IDg || f(N) || N2 ) – B (3) E(Ks, f(N)) Second challenge nonce encrypted with session key Shows supplicant was able to read contents (nonces and key) Is willing to use key PKCS-BASED PROTOCOL Assumes A and B have each other’s public keys – (1) E(PU): [N || IDAD — (2) E(PUQ [N, I N2]) — Initiator Responder в —(3) E(PU), N2) (4) E(PU), E(PRQ,K)) What is wrong with this protocol ? Show transcribed image text KEY DISTRIBUTION SCENARIO Key Distribution Center (KDC) a.k.a. supplicant Initiator A Responder B (1) IDA || IDB || N Names and nonce J Key distribution steps (2) E(KA [K, II IDAII ID|| ND) || E(K), [K, || IDA) Session key, etc. For each of A&B (encrypted) (3) E(K), [K, ILIDAD) Forward key with name — – – – – – — – — – – – — (4) E(K.N2) Challenge nonce k Authentication steps Response (5) E(Ks, f(N2))
DECENTRALIZED PROTOCOL No protection at all Name and challenge nonce (in clear) -(1) IDAN Initiator Responder Shows B saw nonce (fresh), B sent msg Session key, nonces, names encrypted with master key —(2) E( K K | IDA I IDg || f(N) || N2 ) – B (3) E(Ks, f(N)) Second challenge nonce encrypted with session key Shows supplicant was able to read contents (nonces and key) Is willing to use key
PKCS-BASED PROTOCOL Assumes A and B have each other’s public keys – (1) E(PU): [N || IDAD — (2) E(PUQ [N, I N2]) — Initiator Responder в —(3) E(PU), N2) (4) E(PU), E(PRQ,K)) What is wrong with this protocol ?

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