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Microsoft Challenge Handshake Authentication Protocol CS265 Spring 2005 ChungShun Wei.

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Presentation on theme: "Microsoft Challenge Handshake Authentication Protocol CS265 Spring 2005 ChungShun Wei."— Presentation transcript:

1 Microsoft Challenge Handshake Authentication Protocol CS265 Spring 2005 ChungShun Wei

2 Private Network Restrict from outside access Highly secure if no bad guy has access to the physical LAN But you are also blocked if not locally Even Internet will not help

3 Virtual Private Network (VPN) Through VPN server Remote user can connect to intranet through public internet

4 VPN Authentication Password Authentication Protocol (PAP) – Username & password in clear text – Use it only when VPN server only support PAP Challenge Handshake Authentication Protocol (CHAP) – Encrypt password

5 Microsoft Challenge Handshake Authentication Protocol (MS-CHAP) Base on CHAP MS-CHAP version 1 and version 2 MS-CHAP v2 is an improvement over MS-CHAP v1

6 Request Login Challenge Request login challenge 16-byte random challenge

7 Generate 8-byte Challenge Client

8 Generate 24-byte MS-CHAP Reply Client

9 Retrieve Password From DB 24-byte reply & Peer Authenticator Challenge & client’s username Client username Password

10 Authenticate VPN Server Match

11 Authenticator Response 20-byte Authenticator Response -VPN Server will use 16-byte Peer Authenticator Challenge and Client’s hashed password to create 20-byte Authenticator Response -Client computes its own Authenticator Response to compare with Server’s. If match, server is authenticated

12 Find Out 8-byte Challenge Although 8-byte challenge did not send through in clear text Attack can easily compute 8- byte challenge by listening 16-byte random challenge from server, Peer Authenticator Challenge, and client’s username

13 Analysis MS-CHAP Reply sanjose askjKeL35h2k49kj (16 byte) NT hash askjKeL35h2k49kj00000 (21 byte) Pad with 0 to 21 byte askjKeL35h2k49 kj00000 Iwe652nWn8mxhUw0xjO82nzx Encrypt challenge n8mxhUw0Iwe652nW xjO82nzx

14 Attack on MS-CHAP Reply Attackers do not need 2 192 effort But 2 56 + 2 56 + 2 16 ≈ 2 57 Iwe652nWn8mxhUw0xjO82nzx askjKeL35h2k49kj (16 byte) NT hash askjKeL35h2k49kj00000 (21 byte) askjKeL35h2k49 kj00000 Encrypt challenge Iwe652nWn8mxhUw0xjO82nzx Iwe652nW n8mxhUw0 xjO82nzx Iwe652nW xjO82nzx n8mxhUw0 xx00000 2 16 tries Encrypt challenge xxxxxxx 2 56 tries xxxxxxx 2 56 tries sanjose


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