Interplay of security and clock synchronization"

Transcription

1 July 13, 2010, P. R. Kumar " This work is licensed under a Creative Commons Attribution-Noncommercial-No Derivative Works 3.0 Unported License." See Interplay of security and clock synchronization" Yih-Chun Hu and P.R. Kumar Dept. of Electrical and Computer Engineering, and" Coordinated Science Lab" University of Illinois, Urbana-Champaign" Sep 4-5, 2008" 1/20

2 Clock synchronization over networks" Knowledge of time is important in Networks" Communication network protocols" Sensor network applications" Networked control" However no two clocks agree Several issues" How to synchronize clocks in wireless networks?" Can clock synchronization be helpful vis-à-vis security?" And what about security of clock synchronization itself?" 2/20

3 It is impossible to synchronize two clocks" Theorem (Graham & K ʻ04)" It is impossible to determine (d 12, d 21, a 2, b 2 ) through any packet exchanges" Reference Clock 1" d 12 d 21 Clock 2" r 1 = a 2 (s 1 + d 12 ) + b 2 τ 2 r 1 s 2 r 3 s 4 Skew a 2 τ 2 Offset b 2 τ 1 τ 1 d12 d 21 d 12 d 21 r 4 s 1 r 2 s 3 s 2 = a 2 (r 2 d 21 ) + b 2 r 1 s 2 r 3 = s 4... s r a 2 a s d 12 a r d 21 b "Rank 3: Cannot estimate 4 parameters" 3/20

4 So what is determinable?" r â 1 s * s a r a 2 â 2 ˆd12 a * * = 2 d 12 + â ˆb 1 a r 3 = s d 12 2 ˆd21 a * * 2 s 4 2 d 21 a 1 r d 21 ˆb b a * 2 := r (k ) (l ) 1,2 r 1,2 (k s ) (l ) 1 s 1 d * 12 := r (k ) 1,2 a * (k ) 2 s 1 d * * 21 := a * (l 2r ) (l ) 2,1 s 2 * a 2 a 2 ˆd 12 0 and ˆd 21 0 ˆb 2 [ a * 2 d * 21,a * 2 d * 12 ] (k r ) (k 1,2 = a 2 s ) 1 + a 2 d 12 + b 2 = a * (k 2 s ) 1 + a * * 2 d 12 4/20

5 Interplay between Clock Synchronization and Security (Hu & K ʼ08 )" 5/20

6 Man in the middle attack" What must a Man in the Middle do to remain undetected?" What resources does a Man in the Middle need to remain undetected?" How to challenge the Man in the Middle?" R" Can we synchronize clocks in spite of the Man in the Middle? " M" M provides a logical channel between S and R" M cannot decrypt any messages between S to R" M cannot alter any messages between S and R" S" M cannot create any fake messages between S and R" M can occasionally discard messages between S and R" 6/20

7 What can Man-in-the-Middle do?" 7/20

8 Affine forwarding policy" Without Man in the Middle" Time received is affine in" a SR! S Coefficient is estimate of skew" a SR (! S + d SR ) + b SR! S + d SR R" With Man in the Middle" Mʼs forwarding policy" Packet received at τ Forwarded at F(τ) Receipt time a SR M"! S + d SM has to be affine in"! S So F(τ) has to be affine in τ " ( F (! S + d SM ) + d ) MR + b SR! S! S a SR F (! S + d SM ) S" ( F (! S + d SM ) + d MR ) + b SR R" S" 8/20

9 Expansionary affine forwarding policy" Consider affine forwarding policy " F (! ) = " F! + # F Causality" Forwarding packet can only take place after receiving packet"! S R"! S + d SM M"! F (" S + d SM ) + # F S"! F (" S + d SM )+ # F $" S + d SM for all " S So "! F " 1 9/20

10 M can only add a delay to packets" Estimate of skew = Coefficient of " So skew estimate made by R with reference to S is" a SR! F Backward path skew estimate made by S with reference to R is" a RS! B But product of skew estimates has to be 1"! S! S ( ) + b SR a SR! F (" S + d SM ) + # F + d MR! S + d SM! F (" S + d SM ) + # F R" M" S" a SR! F a RS! B =! F! B = 1 a SR! F R" But! F " 1 and"! B " 1! F! B M" So"! F =! B = 1 Forwarding time is pure delay: F!" ( ) =! + " F S" a RS! B

11 Detecting a Half-duplex Man-in-the- Middle" 11/20

12 Detecting a Half-Duplex Man-in-the-Middle" Send a long packet of duration greater than d RS +d SR R" d SM " d RS +d SR +d SM" d SM +β F " M" 0" d RS +d SR" S" Simultaneous send to R while receiving from S" M will need to simultaneously receive and send for a positive duration: (d RS +d SR +d SM ) (d SM +β F ) = d RS +d SR β F > 0" 12/20

13 Detecting a Full-Duplex Man-in-the-Middle" 13/20

14 The Simultaneous Receive, Send, Receive Challenge (SRSR Challenge)" Let C S = Time taken by S to switch from Transmit to Receive mode" Let C R = Time taken by R to switch from Transmit to Receive mode" R can verify that M is forwarding if C S - d RS + C R < C S - d RS + d RM + d MR " d RS" C S -d RS" d SR " R" Send me a pkt from -100 to 0." I will send you a pkt at C S.. " Send me a pkt from -100 to 0." I will send you a pkt at C S.. " C S -d RS +d RM" d SM" M" S" -1000" -100" 0" C S " Simultaneous send to R and receive from R as well as S" If C R < d RM + d MR and C S d RS + d RM < d SM then M gets caught" 14/20

15 The Switch Time condition for detecting the Man-in-the-Middle" IF C R < d RM + d MR and C S d RS + d RM < d SM" " " " OR" X" X" +" If both violated, then C S + C R d RS + d SR " "C S < d SM + d MS and C R d SR + d SM < d RM" then M gets caught in one of the two directions" Suppose C R < d RM +d MR AND C S <d SM +d MS " Then C R +C S < d RM +d MR +d SM +d MS < d SR + d RS" Hence EITHER C S d RS + d RM < d SM OR C R d SR + d SM < d RM holds " Thus M gets caught in one of the two directions" 15/20

16 When is Man-in-the-Middle impossible to detect?" 16/20

17 Impossibility condition for detecting Man-in-the-Middle" Theorem" If C R > d MR +d RM OR C S > d MS +d SM Then Man-in-the-Middle can evade detection" Proof" Suppose C R > d MR +d RM wlog" Consider the following RS-Priority Detection Prevention Strategy for M" Choose forwarding delay 0 β F < C R d MR d RM for both R to S and S to R" Conflict resolution strategy: Give priority to RS packets (from R to S)" When packets from both R and S are incoming, listen only to R and not S" When M needs to transmit to both R and S, transmit only to S and not R! 17/20

18 RS Priority Detection Avoidance Strategy" All packets from R are received and sent to S" We only need to consider packets from S to R: Recall β F < C R d MR d RM " Receiver conflict" R cannot check reception" Transmitter conflict" R cannot check reception" β F +d MR +d RM <C R" β F +d MR +d RM <C R" -d RM" β F +d MR" R" -β F -d RM" d MR" R" X" 0" β F " X" M" S" -β F " 0" X" M" S" What M doesnʼt hear doesnʼt hurt" What M doesnʼt transmit doesnʼt hurt" 18/20

19 Man in the Middle Attack: Clocks, Detectability and Consequences" Theorem" A Half Duplex Man-in-the-Middle can always be detected" A Full Duplex Man-in-the-Middle will be detected by the SRSR Challenge if both turnaround times are short: C R < d RM +d MR AND C S <d SM +d MS " The Full Duplex Man-in-the-Middle can avoid detection by:" An RS-Priority Strategy with Low Forwarding Delay (RSLFD) if C R > d RM +d MR," By an SRLFD Policy if C S > d SM +d MS " The Double Full Duplex Man-in-the-Middle can avoid detection" Two simultaneous transmitters, two simultaneous receivers with shielding between transmitters and receivers, and two tunnels" Even when he can avoid detection" A Man-in-the-Middle can only add a pure delay in each direction" The delay should be small enough if Man-in-the-Middle is Full Duplex" So time-based applications are still temporally consistent! 19/20

20 Thank you" 20/20