CSEC Coursework

Question 1: Correct answer is (b) {(KPbB) KPrS mod KPbS , G2 , KBS , KPrB , {(KPbB) KPrS mod KPbS}K1, NA , NB)

The principal K sends its identity to KPbS. NA prepare the nonce and this nonce sends to the P. then encryption takes place and P then encrypts NA and NB with the help of key K1 and after that, P sends this encrypted value x to NA. NB then encrypts P, KPbB, x with KPbB and NB then send this encrypted value to Y and it also sends this encrypted key to the authentication server S. The most important consideration is that S knows both the secret keys, it starts the decryption of y with KPrBS, and recovers K, then it starts the decryption of x with KPS and by the help of this process it recovers P, Q, n. Now, the server S that we discussed earlier, server S then start encryptions with P, Q, n with k1, and then it forwards the encrypted value to KPbS. KPbS then starts computing and calculates KPbS, Na, Nb and the verification process start with g1, g2, g3. These three values g1, g2, and g3 then verifies the scenario that if this value is equal to the value received from K. After the computation, if both values are equal, then authentication succeeds, else it fails.

Question 2: Correct answer is (h) Both A and B authenticate each other by knowing each other’s identities

A want to establish connection with B. In this context A send details to the B and B acknowledge the details and verify the details if there is no issue with the details then B send their detail to the A. When A and B both receive details then they exchange certificate and acknowledgment done.

Question 3: Correct answer is (g) At time point 3, S can produce evidence that that A has access to key K 

S provides the evidence that the B received the message. It is generated by the delivery agent

(usually TTP) and it will be sent to A that B is alive.

Question 4: Correct answer is (d) The secrecy of P3. K2 is compromised, but P2. K2 and P1. K2 would remain secret since K1 is refreshed after each run, therefore P3. K1 is different from P2.K1 and is different from P1.K1 

P1 and P2 are two separated functions and they would not be affected k2. If P3 compromises at any stage, then there is no affect because they can be refreshed.

Question 5: Correct answer is (a) 3. B ® S: (A, {(B, {K2}KpbA)}KpbS)

A sends its identity to S, it also sends the packet which is encrypted with the key, Kpb, and the key shared with S. The current timestamp contains this packet that we discussed above, A is the preferred partner in the communication channel. After that the key KpbA is generated randomly, and this is used by the communication process between A and B. S tries to obtain the KAB and this can be achieved when S decrypts. If S decrypts, then it obtains the KpbS and then it forward this key to the B and this encrypted key forward in the encrypted packet which also contains the current timestamp and the identity of A. Now, B start decrypting the message with the help of key and then it shares with the S, then S retrieve the identity of the other party and the key K.

Question 6: Correct answer is (a) 1. A ® B: (C, {NA}KAB)

C wants to enter in the communication of A and B to give it harm. So, C is the harmful packet.

Question 7: Correct answer is (a) (l read from z), (l write to g), (g read from l), (g read from e), (e write to c), (c write to b), (a read from b)

As it is mentioned above that access attain because of proper sequence. Z goes into sequence so that why information leak.

Question 8: Correct answer is (d) Unified Host and Network Dataset

Unified host resist duplication and network data avoid the system and harmful person to enter in the system.

Remember, at the center of any academic work, lies clarity and evidence. Should you need further assistance, do look up to our Computer Science Assignment Help

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