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Communication Networks

Communication Networks. Recitation 11 Security. RSA: The problem. A wants to send B a message, but A and B cannot meet so cannot decide on a common key. RSA: The solution. B sends A a public key e B . The public key can be used to encrypt, but not to decrypt

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Communication Networks

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  1. Communication Networks Recitation 11 Security Comnet 2010

  2. RSA: The problem • A wants to send B a message, but • A and B cannot meet so cannot decide on a common key Comnet 2010

  3. RSA: The solution • B sends A a public key eB. The public key can be used to encrypt, but not to decrypt • A encrypts the message m with the public key and sends eB(m) to B • B uses his own private key dB to decrypt: dB(eB(m))=m Comnet 2010

  4. RSA obtaining keys • Choose two large prime numbers p, q. • Compute n = pq  give to others. • Calculate z=(p-1)(q-1). Destroy p,q. • Choose 1<e<n that has no common factors with z give to others. • Compute d such that (de-1) is evenly divisible by z. Find an integer K which causes d = (Kz+ 1)/e to be integer, and use d  keep secret. Destroy z. Public key (n,e). Private key (n,d). Comnet 2010

  5. RSA example • p = 61, q = 53 • n = 3233 • z=3120 • 7, 11, 17 will do. We choose e=17. • d=2753: (3120K+1)/17 integer  K=15. Public key (3233,17).Private key (3233,2753). Comnet 2010

  6. RSA encryption/decryption • Encryption : c = memod n • Decryption : m = cdmod n Comnet 2010

  7. RSA encryption/decryption example • encrypt(m) = m17 mod 3233 • Encrpyt(123) = 12317 mod 3233 = 855 • decrypt(c) = c2753 mod 3233 • Decrpyt(855) = 8552753 mod 3233 = 123 Comnet 2010

  8. RSA with a pocket calculator • 2753 = 101011000001 base 2 • 2753 = 1 + 26 + 27 + 29 + 211 = 1 + 64 + 128 + 512 + 2048 8551 = 855 (mod 3233) 8552 = 367 (mod 3233) 8554 = 3672 (mod 3233) = 2136 (mod 3233) 8558 = 21362 (mod 3233) = 733 (mod 3233) 85516 = 7332 (mod 3233) = 611 (mod 3233) 85532 = 6112 (mod 3233) = 1526 (mod 3233) 85564 = 15262 (mod 3233) = 916 (mod 3233) 855128 = 9162 (mod 3233) = 1709 (mod 3233) 855256 = 17092 (mod 3233) = 1282 (mod 3233) 855512 = 12822 (mod 3233) = 1160 (mod 3233) 8551024 = 11602 (mod 3233) = 672 (mod 3233) 8552048 = 6722 (mod 3233) = 2197 (mod 3233) Comnet 2010

  9. 8552753 (mod 3233) = 855^(1 + 64 + 128 + 512 + 2048) (mod 3233) = 8551 * 85564 * 855128 * 855512 * 8552048 (mod 3233) = 855 * 916 * 1709 * 1160 * 2197 (mod 3233) = 794 * 1709 * 1160 * 2197 (mod 3233) = 2319 * 1160 * 2197 (mod 3233) = 184 * 2197 (mod 3233) = 123 (mod 3233) = 123 Comnet 2010

  10. RSA: Signatures • How can B know the message was from A? • A produces a hash H(m) • A encrypts with his private key dA(H(m)) and sends with m. • B produces H(m), decrypts dA(H(m)) with A’s public key eA :eA(dA(H(m)))=H(m)and compares them. Comnet 2010

  11. RSA Signature example • A wants to send “This is a very important message”. • p=5, q=7  n = 35, z = 24 • e = 5; d = 29 • Public key: (35, 5) Private key: (35, 29) • H(“This is a very important message”)=26 • 2629 mod 35 = 31 • A sends “This is a very important message”, 31 • B gets public key 5, 315 mod 35 = 26. • Compares to H(“This is a very important message”)=26 Comnet 2010

  12. RSA: Authorization • How can B know this is really the A he knows? • Certification Authority has public key eCA and private key dCA • A proves to CA that he is A using some identity proof, and gets dCA(eA) • B can now use eCA(dCA(eA))=eA Comnet 2010

  13. RSA Authorization example • A wants to send “This is a very important message”. • ... • CA has Public key (3337, 79), Private key (3337, 1019) • B gets A’s authorized public key 51019 mod 3337 = 199 • B uses 19979 mod 3337 = 5 • … Comnet 2010

  14. Firewall • Isolates organization’s internal net from larger Internet, allowing some packets to pass, blocking others • Firewall is usually implemented as a router • Router filters packets, based on: • source IP address • destination IP address • TCP/UDP source and destination port numbers • ICMP message type • TCP SYN and ACK bits • “Smart filtering” Comnet 2010

  15. Example firewall rules • “Allow outgoing traffic only on ports HTTP, HTTPS, FTP and TELNET” • Used in a work place to make sure people aren’t using dangerous/illegal sharing • Too wide • “Do not allow incomingng traffic on port TCP/4661 (edonkey)” • Too narrow Comnet 2010

  16. More rules • “Allow incoming traffic only on port HTTP/HTTPS” • Is it blocking enough? • What other applications? • “Do not allow traffic from bezeqint.net” • Sad but true Comnet 2010

  17. Firewalls prevented SYN DDOS attack • An external host sends a syn packet. • Firewall responds with a syn+ack to the external host (at this point, the internal server doesn’t even know that there is something going on like this). • If the external hosts sends an ack packet, then the firewall creates a new session by syn to ack to the internal server. • Then it connects them together so that the communication works. Can this be circumvented? What else can the firewall do? Comnet 2010

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