Question

Answer the following questions in no more than six lines each. (3 x12 = 36) 1....

Answer the following questions in no more than six lines each. (3 x12 = 36)

1. Explain how a Smurf attack works.

2. Can the public key parameters be n = 15 and b = 6 in RSA? Why? (Of course the numbers are small and this is a toy example).

3. If you see several TCP SYN segments sent to a web server, but no ACK segments in the three-way handshake, what do you suspect may be happening? Explain.

4. Consider the network shown below.

Clients from the outside can only connect to the web server and mail server in the DMZ. Most hosts from the inside can connect only to the web server and mail server in the DMZ. Only the host 136.142.117.1 can connect to any web server on the outside. Design rules for packets entering only interface p for stateful packet filter A for the above security policy. You can use a mix of standard and extended ACLs or you can write your rules as sentences. Include details and add explanations as necessary. (24)

5. Alice and Bob are communicating using a shift cipher with the capital letters in English, but with a twist. They use two keys instead of one, and the keys alternate. So the first letter would be encrypted using the key k1 = 9, the second using the key k2 = 14, the third with k1, the fourth with k2 and so on as needed. Their hope is that this will prevent frequency analysis by Oscar.

a. Do you think this approach is secure? Why? Explain using Kerckhoff’s principle and the effort that it may take Oscar to use brute force. (15)

b. Decrypt the following ciphertext exchanged by Alice and Bob (show steps): JWAVXDYSA (25)

6. The decrypted message in Problem 3(b) corresponds to an article that was assigned for your reading. Summarize the main idea of that article in no more than 6 lines. (10)

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Answer #1

Acc. to HomeworkLib policy,i can answer only atmost 4 questions. :)

1)  A smurf attack is an exploitation of the Internet Protocol (IP) broadcast addressing to create a denial of service. The attacker uses a program called Smurf to cause the attacked part of a network to become inoperable. The exploit of smurfing, as it has come to be known, takes advantage of certain known characteristics of the Internet Protocol (IP) and the Internet Control Message Protocol (ICMP).

b) basically you need two prime numbers for generating a RSA key pair. If you are able to factorize the public key and find these prime numbers, you will then be able to find the private key. The whole security of RSA is based on the fact that it is not easy to factorize large composite numbers, that's why the length of the key highly change the robustness of the RSA algorithm.

2) Basically, you need two prime numbers for generating a RSA key pair. If you are able to factorize the public key and find these prime numbers, you will then be able to find the private key. The whole security of RSA is based on the fact that it is not easy to factorize large composite numbers, that's why the length of the key highly change the robustness of the RSA algorithm.

3) Even if that ACK was lost, there will be no resending for a very simple reason. Directly after the ACK, the host that opened the TCP protocol is likely to start sending data. That data will, as all TCP packets, have an ACK number, so the recipient would get an ACK that way. Hence, the sender of the SYN-ACK should reasonably not care that it didn't get the ACK, because it gets an "implicit" ACK in the following package.

5a)
Kerckhoff's algorithm stated that, a cryptosystem should be secure even if everything about the system,except the key,is public knowledge
Although third party doesn't know about the key,but the third party can easily judge a pattern by looking at the encrypted text.

5b)
Encrypted Text-JWAVXDYSA
J-10
W-23
A-1
V-22
X-24
D-4
Y-25
S-19
A-1

First letter is encrypted with K1=9 and second key is encrypted with K2=14 and so is third and fourth.
Assuming, we take letters no subtracting by 9 and 14.

so,

J 10-9 = 1 A
W 23-14 = 9 I
A 1-9 = -8 mod 26 = 16 P
V 22 -14 =8 H
X 24-9 = 15 O
D 4-14 =-10 mod26 = 10 J
Y 25-9 = 16 P
S 19-14 = 5 E
A 1-9 = -8 mod 26 = 16 P

6) It doesn't make sense,as there is nothing as 3b mentioned above in the question.

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