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SJSU CS 265 - WirelessLAN

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Security in Wireless LAN 802.115 basic threats to WLANIEEE 802.11 Authentication – Open KeyShared key AuthenticationIdentity ProblemsShared key vulnerabilities (cont…)IEEE 802.11 Wired Equivalent Privacy (WEP) ProtocolWEP EncryptionWEP DecryptionSlide 10ICV WeaknessWEP Problems-with RC4WEP Problems-with IVCisco enhancements to 802.11 WEP to increase securityToday & future controlToday & future control (cont.)The EndSecurity in Wireless LAN 802.11Layla PezeshkmehrCS 265Fall 2003-SJSUDr.Mark Stamp5 basic threats to WLANSniffing - eavesdroppingInvasion – steal valid STA’s access to gain access to networkTraffic redirection – change in ARP tableDenial of service (DOS)–Flood the network–Disrupt connection between machines–Prevent a STA from connecting to WLANRogue networks and station redirection – Man- in- the- middle attacks.IEEE 802.11 Authentication – Open KeyUses null authentication, SimpleIs the default authentication 2 steps:A sends a request authentication to BB sends the result back to A If dot11 Authentication Type at B is set to "Open System"  Returns "success"  A is mutually authenticated; Otherwise A is not authenticatedAuthentication Request(Open SystemAuthentication)Authentication ResponeseSDAccess pointShared key AuthenticationProvides a better degree of authentication.Station must implements WEP(Wired Equivalent Privacy)4 steps:1. Request sends an Authentication frame to AP.2. AP replies with a random challenge text generated by the WEP engine( 128 bit).3. STA copy the challenge text, encrypt it with a shared key then send the frame to the AP.4. AP decrypt the received frame, then verifies the 32- bits CRC “ICV”, and that the challenge text matches the one it sends earlier to the station.5. Successful/negative authentication if match/mismatchAuthentication Request (Shared key Authentication)"Challenge" text string"Challenge" text stringencrypted with shared keyPositive or Negative resultbased on decryption resultWEP encryption ofchallenge textWEP decryption ofencrypted textSDAccess pointIdentity ProblemsOpen System authenticationNull authentication.Messages sent in clear.Any one can impersonate either the station or the access point.Shared key authenticationOnly station authenticates itself.No mechanism for AP to prove its identity to the station therefore malicious AP. Only the station is authenticated not the user of the station.Shared key vulnerabilities (cont…)Exchanging both challenge and response occurs over the wireless link and is vulnerable to a man-in-the-middle attack.IEEE 802.11 Wired Equivalent Privacy (WEP) ProtocolThe goal is to provide data privacy to the level of a wired network.(WEP) algorithm is used to prevent eavesdropping.An encapsulation of 802.11 data frame.64- bits key (40-bit secret key,24-bit "init" vector).Symmetric algorithm because the same key is used for cipher and decipher.Data integrity checked with CRC-32.WEP EncryptionA key shared among members of the BSS.Sender calculates CRC of the frame's data.WEP appends a new generated 24-bit initialization vector (IV) to the shared key.WEP PRNG (RC4) is used to generate a key stream.XORs key stream against (payload + CRC) to produce ciphertext.The sender also inserts the IV into frame header, and sets the WEP encrypted packet bit indicator.WEP DecryptionReceiver extracts IV from the frameappends IV to the BSS shared key, and generates the "per- packet" RC4 key sequenceciphertext is XORed against the key steam to extract plaintext.Verification: performs integrity check on plaintext Compares ICV1 result with the ICV transmitted.WEP DecryptionICV WeaknessHow is the attacker able to modify ICV to match the bit-flipped changes to the frame?WEP Problems-with RC4flip a bit in the ciphertext (C)  the corresponding bit in the plaintext will be flipped.Eavesdropper intercepts 2 ciphertext encrypted with the same key stream  possible to obtain the XOR of the 2 plaintexts. c1 = p1  b c2 = p2  b  c1  c2 = (p1  b)  (p2  b) = p1  p2WEP Problems-with IVIV is 24 bits cleartext, part of a message.A small space of initialization vectors guarantees the reuse of the same key stream. AP constantly send 1500 byte pkt at 11 Mbps will exhaust the space of IV after 1500 * 8/(11 * 10 ^ 6 ) * 2 ^ 24 = 18000s = 5hWhen the same key is used by all mobile stations more chances of IV collision.Cisco enhancements to 802.11 WEP to increase securityMutual authentication instead of one-way authenticationSecure key derivation using one way hash functionDynamic WEP keys instead of static WEP keysInitialization Vector changesToday & future control Service Set Identifier (SSID)Each AP has an SSID of the AP to identify itself. STA have to know the SSID of the AP to which it wants to connect. SSID keeps a STA from accidentally connecting to neighboring AP.This does not solve other security issues and does not keep an attacker from setting up a "rogue" AP that uses the same SSID as the valid APToday & future control (cont.)MAC filtersAP check MAC addresses of STAs before being connected to the network –AP keep a list of MAC addresses in long- term memory.–AP may send a RADIUS request with the MAC address as the userID (and a null password ) to a central RADIUS server to check the list for an address.The


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SJSU CS 265 - WirelessLAN

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