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A Novel Privacy Preserving Authentication and Access Control Scheme for Pervasive Computing Environments Authors: Kui Ren, Wenjing Lou, Kwangjo Kim, and Robert Deng Sources: IEEE Transactions on Vehicular Technology, 55(4), pp. 1373-1384, July 2006. Reporter: Chun-Ta Li ( 李俊達 )
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2 22 Outline Pervasive computing environments (PCE) Motivations The proposed scheme Analysis Comments
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3 Pervasive computing environments Definition Integrates digital devices (such as computers, handheld devices, sensors and actuators) seamlessly with everyday physical devices (such as electrical appliances and automobiles). Three components [James Kurose and Keith Ross, 2004] Nomadic computing: wireless-technology Sensor-based smart spaces: environment-monitoring Mobile computing data management Sensor network
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4 Pervasive computing environments Service-Oriented Architecture
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5 Pervasive computing environments Sample PCE Authentication Server Router Public Internet Public Internet Gateway Access Point Printer User Fax Scanner Scientific Device
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6 Motivations Providing explicit mutual authentication between mobile user and the service Allowing mobile user to anonymously interact with the service Enabling differentiated service access control among different users Providing flexibility and scalability to both user and service sides Generating fresh session keys to secure the interaction Efficiency of communication, computation and management overheads
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7 The proposed scheme Notations
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8 The proposed scheme (cont.) System architecture Mobile User Authentication Server Service Access Point 1. Registration 2. Authorization 3. Access Request 4. Authentication Request 5. Authentication Acknowledgement 6. Access/Reject
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9 The proposed scheme (cont.) User authorization protocol Credential generation Mobile user U (a certificate CertU) Service provider S 1. Generate two nonces: r’ U and r” U 2. Sign her own ID with a nonce r” U {U, r” U } PriK U 3. Compute the anchor value C 0 h(r” U, U, {U, r” U } PriK U ) Non-repudiation property 4. Compute the credential chain C n h n (C 0 ), with length n 5. Blind C n as C U {r’ U } PubK SID * C n
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10 The proposed scheme (cont.) User authorization protocol Credential authorization Mobile user U (a certificate CertU) Service provider S U, C U, CertU, SID authorization request 6. Verify CertU with PubK S 7. Sign C U as C S {C U } PriK SID = r’ U * {C n } PriK SID CSCS authorization confirmation 8. Compute C S /r’ U (C n, {C n } PriK SID )
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11 The proposed scheme (cont.) User operational protocol Mobile user U Service provider S Access point P 1. Generate a nonce: r U 2. Send {r U, C n, {C n } PriK SID } PubK S 3. Send {r U, C n, {C n } PriK SID } PubK S secure tunnel 4. Decrypt r U, C n 5. Store C n 6. Send r U, C n secure tunnel 7. Generate a nonce: r P 8. Compute K UP =h(C n, r P, r U, 0). K’ UP =h(C n, r P, r U, 1) 9. Send r P, {r U, P} K UP access acknowledgement access request access acknowledgement
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12 The proposed scheme (cont.) User operational protocol Mobile user U Service provider S Access point P 10. Compute K UP =h(C n, r P, r U, 0), K’ UP = h(C n, r P, r U, 1). 11. Decrypt and verifies r U, C n, P 12. Encrypt X m 0 = {m 0 } K’ UP 13. Compute h K UP (X m 0 ) 14. Send r P, r U, X m 0, h K UP (X m 0 ) 15. Verify X m 0 using K UP 16. Decrypt m 0 using K’ UP … r P, r U, X m i, h K UP (X m i ) authenticated data traffic
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13 Analysis
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14 Comments Cryptanalysis of anonymity property Service provider S Step 1: Get U, C U = {r’ U } PubK SID * C n in Credential Authorization phase Step 2: Sign C U as C S {C U } PriK SID = r’ U * {C n } PriK SID Step 3: Store U, C U, C S = {C U } PriK SID = r’ U * {C n } PriK SID in their own DB Step 4: Get C n, {C n } PriK SID in User Operational phase Step 5: Compute C S / {C n } PriK SID to derive r’ U Step 6: Compute C’ U = {r’ U } PubK SID * C n to verify whether C’ U = C U holds or not. Step 7: If it holds, S confirms that mobile user U accesses the service; otherwise, S continually executes the previous Steps from 4 to 6.
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15 Comments (cont.) Efficiency improvement in user operational phase compared C j with all C j s stored in S’s DB Time complexity is O(n) if there are n users in DB solution: User i generates a T ID in access request message and sends it to service provider to store the T ID of user i Time complexity is O(1)
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16 Comments (cont.) Service abuse problem No one can derive the value of C n unless user itself and thus anyone can fabricate an invalid Cn with a valid Cert U to access the service without limits even than a valid user can deny his accesses. Cert U must keep secret for outsiders {U, C U, CertU, SID} PubK S Mobile user U (a certificate CertU) Service provider S
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