Privacy Preserving Rechargeable Battery Policies for Smart Metering Systems Simon Li (Toronto), Ashish Khisti (Toronto), Aditya Mahajan (McGill) March 3, 2016 1/15
Motivation: Privacy Leakage through Power Profile U. Greveler, P. Glosekotter, B. Justus, and D. Loehr, “Multimedia content identification through smart meter power usage profiles,” in Int. Conf. Inform. and Knowledge Eng, 2012. 2/15
Problem Setup Battery Policy X t Y t Home Power S t +1 = S t + Y t − X t Appliances Grid q t ( Y t | X t , S t , Y t − 1 ) Figure 1: System Diagram. I User Load: X t ∈ X I Battery Update: S t +1 = S t + Y t − X t I Output Load: Y t ∈ Y 1 2 1 , Y t ≠ 1 I Policy: q t - - X t 1 , S t I Battery State: S t ∈ S Y t 1 I X , Y and S : discrete I Leakage Rate L T = 1 T I ( X T 1 ; Y T 1 ) I X t ∼ P X ( · ) (i.i.d.) I Asymptotic Leakage: L Œ 3/15
Example: Binary System X , Y , S = { 0 , 1 } , p X (0) = p X (1) = 1 / 2. I Empty State: S t = 0 I Full State: S t = 1 I X t = 1 ⇒ Y t = 1 I X t = 0 ⇒ Y t = 0 I X t = 0 ⇒ Y t ∈ { 0 , 1 } I X t = 1 ⇒ Y t ∈ { 0 , 1 } 4/15
Example: Binary System X , Y , S = { 0 , 1 } , p X (0) = p X (1) = 1 / 2. I Empty State: S t = 0 I Full State: S t = 1 I X t = 1 ⇒ Y t = 1 I X t = 0 ⇒ Y t = 0 I X t = 0 ⇒ Y t ∈ { 0 , 1 } I X t = 1 ⇒ Y t ∈ { 0 , 1 } X t = 0 , Y t = 0 X t = 0 , Y t = 1 X t = 0 , Y t = 0 S t = 0 S t = 1 X t = 1 , Y t = 1 X t = 1 , Y t = 1 X t = 1 , Y t = 0 4/15
Binary System Example Policies Policy 1: Y t = X t 1 2 3 4 5 6 7 t 0 1 1 0 1 0 1 X t 0 0 0 0 0 0 0 S t Y t 0 1 1 0 1 0 1 L T = 1 T I ( X T ; Y T ) = 1 5/15
Binary System Example Policies Policy 2: Y t = ¯ S t X t = 0 , Y t = 0 X t = 0 , Y t = 1 X t = 0 , Y t = 0 S t = 0 S t = 1 X t = 1 , Y t = 1 X t = 1 , Y t = 1 X t = 1 , Y t = 0 t 1 2 3 4 5 6 7 X t 0 1 1 0 1 0 1 I L T = 1 T I ( X T ; Y T ) ? S t 0 1 0 0 1 0 1 Y t 1 0 1 1 0 1 0 5/15
Binary System Example Policies Policy 2: Y t = ¯ S t X t = 0 , Y t = 0 X t = 0 , Y t = 1 X t = 0 , Y t = 0 S t = 0 S t = 1 X t = 1 , Y t = 1 X t = 1 , Y t = 1 X t = 1 , Y t = 0 I L T = 1 T I ( X T ; Y T ) ? t 1 2 3 4 5 6 7 I Y T 1 ⇒ S T 1 is known X t 0 1 1 0 1 0 1 0 1 0 0 1 0 1 I ( Y T 1 , S T 1 ) ⇒ X T ≠ 1 S t 1 1 0 1 1 0 1 0 Y t 1 T I ( X T ; Y T ) ≈ 1 I 5/15
Binary System Example Policies Policy 3: Randomized Policy I q ( Y t = 0 | S t = 0 , X t = 0) = q ( Y t = 1 | S t = 0 , X t = 0) = 1 / 2 I q ( Y t = 0 | S t = 1 , X t = 1) = q ( Y t = 1 | S t = 1 , X t = 1) = 1 / 2 X t = 0 , Y t = 0 X t = 0 , Y t = 1 p=1/2& p=1/2& X t = 0 , Y t = 0 S t = 0 S t = 1 X t = 1 , Y t = 1 p=1/2& p=1/2& X t = 1 , Y t = 1 X t = 1 , Y t = 0 I Equiprobable Binary Input I Leakage Rate: L Π= 0 . 5 I Optimality? 5/15
Prior Work I Kalogridis et al (2010): Rechargeable Battery for Privacy. Metrics such as clustering and regression. I Varodayan and Khisti (2011): Mutual Information as a Privacy Metric, Binary Smart Meters Model, Randomized Policies, Numerical Simulation Technique I Giaconi, Gunduz, and Poor (2015): Energy Harvesting/ Alternative Sources and Smart Meter Privacy I Yao and Venkitasubramaniam (2013): Markov Decision Process I Li-Khisti-Mahajan (2015, 2016): MDP, Identified optimal policy as a solution to a fixed point equation. This Work: Information Theoretic Proof of Optimality 6/15
Problem Setup Admissible charging policies: q = ( q 1 , q 2 , · · · ) ∈ Q A where q t ( y t | x t , s t , y t ≠ 1 ) Battery constraints: ÿ q t ( y | x t , s t , y t ≠ 1 ) = 1 y œ Y ¶ ( s t , x t ) where: Y ¶ ( s t , x t ) = { y ∈ Y : s t − x t + y ∈ S} . 7/15
Problem Setup Admissible charging policies: q = ( q 1 , q 2 , · · · ) ∈ Q A where q t ( y t | x t , s t , y t ≠ 1 ) Battery constraints: ÿ q t ( y | x t , s t , y t ≠ 1 ) = 1 y œ Y ¶ ( s t , x t ) where: Y ¶ ( s t , x t ) = { y ∈ Y : s t − x t + y ∈ S} . Joint probability distribution: P q ( S T = s T , X T = x T , Y T = y T ) T 5 Ÿ = P S 1 ( s 1 ) P X 1 ( x 1 ) q 1 ( y 1 | x 1 , s 1 ) 1 s t { s t ≠ 1 − x t ≠ 1 + y t ≠ 1 } t =2 6 × P X ( x t ) q t ( y t | x t , s t , y t ≠ 1 ) . T I q ( S 1 , X T ; Y T ) 1 Finite horizon Leakage rate: 7/15
Main Result Problem Find q ∈ Q A to minimize the infinite horizon leakage rate 1 T I q ( S 1 , X T ; Y T ) . L Œ ( q ) := lim T æŒ 8/15
Main Result Problem Find q ∈ Q A to minimize the infinite horizon leakage rate 1 T I q ( S 1 , X T ; Y T ) . L Œ ( q ) := lim T æŒ Theorem The minimum leakage rate is given by: L ı Œ = min P S ( · ) I ( S − X ; X ) (1) where X ∼ P X ( · ) is independent of S. The optimal policy is a time invariant, memoryless policy: q ı ( y | x , s ) = P X ( y ) P ı S ( y + s − x ) P ı S ≠ X ( s − x ) where P ı S ( · ) achieves the minimum above. 8/15
Remarks Properties of Optimal Policy: q ı ( y | x , s ) = P X ( y ) P ı S ( y + s − x ) P ı S ≠ X ( s − x ) I Stationary and Memoryless: q t ( y t | x t , s t ) = q ( y t | x t , s t ) S ( · ) and S t ⊥ Y t ≠ 1 I Invariance: S 1 ∼ P ı S ( · ) then S t ∼ P ı I P Y ( · ) must have the same support as P X ( · ). Thus it su ffi ces to use Y = X . 9/15
Remarks Properties of Optimal Policy: q ı ( y | x , s ) = P X ( y ) P ı S ( y + s − x ) P ı S ≠ X ( s − x ) I Stationary and Memoryless: q t ( y t | x t , s t ) = q ( y t | x t , s t ) S ( · ) and S t ⊥ Y t ≠ 1 I Invariance: S 1 ∼ P ı S ( · ) then S t ∼ P ı I P Y ( · ) must have the same support as P X ( · ). Thus it su ffi ces to use Y = X . Example: I Binary System X = Y = S = { 0 , 1 } I Equiprobable Input: P X ( X = 0) = P X ( X = 1) = 1 / 2 I P ı S 1 ( S 1 = 0) = P ı S 1 ( S 1 = 1) = 1 / 2 I q ı ( Y 1 = 0 | X 1 = S 1 ) = q ı ( Y 1 = 1 | X 1 = S 1 ) = 1 / 2 9/15
Stationary Posterior Policies Definition (Invariance property) Given an initial battery state distribution P S 1 , a stationary memoryless policy q satisfies the invariance property if P q ( S 2 = s 2 | Y 1 = y 1 ) = P S 1 ( S 1 = s 2 ) , ∀ s 2 ∈ S , y 1 ∈ ˆ Y where ˆ Y := { y : P Y 1 ( y 1 ) > 0 } . We call such a policy as a Stationary Posterior Policy . 10/15
Stationary Posterior Policies Definition (Invariance property) Given an initial battery state distribution P S 1 , a stationary memoryless policy q satisfies the invariance property if P q ( S 2 = s 2 | Y 1 = y 1 ) = P S 1 ( S 1 = s 2 ) , ∀ s 2 ∈ S , y 1 ∈ ˆ Y where ˆ Y := { y : P Y 1 ( y 1 ) > 0 } . We call such a policy as a Stationary Posterior Policy . Lemma For a stationary posterior policy: L Œ ( q ) = I q ( S 1 , X 1 ; Y 1 ) , where ( S 1 , X 1 , Y 1 ) ∼ P S 1 ( s 1 ) P X ( x 1 ) q ( y 1 | x 1 , s 1 ) . 10/15
Stationary Posterior Policies Lemma An initial battery distribution P S 1 and a stationary memoryless policy q = ( q , q , . . . ) satisfies the invariance property i ff for each ( s 2 , y 1 ) ∈ S × X , we have ÿ P S 1 ( s 2 ) P X ( y 1 ) = q ( y 1 | ˜ x 1 , ˜ s 1 ) P X (˜ x 1 ) P S 1 (˜ s 1 ) . (˜ x 1 , ˜ s 1 ) œ D ( s 2 ≠ y 1 ) where D ( w ) := { ( x , s ) ∈ X × S : s − x = w } . 11/15
Optimal Stationary Posterior Policy Lemma (Optimal stationary posterior policy) Given a fixed P S 1 the optimal policy satisfying the invariance property is q ú ( y | x , s ) = P X ( y ) P S 1 ( y + s − x ) P S 1 ≠ X 1 ( s − x ) achieving a leakage rate of L Œ ( q ú ) = I ( S 1 − X 1 ; X 1 ) where ( S 1 , X 1 ) ∼ P S 1 ( s 1 ) Q ( x 1 ) . 12/15
Converse T I ( S 1 , X T ; Y T ) = ÿ I ( S 1 , X T ; Y t | Y t ≠ 1 ) t =1 T I ( S 1 , X t ; Y t | Y t ≠ 1 ) ÿ = (Causality) t =1 T ÿ I ( S t , X t ; Y t | Y t ≠ 1 ) = ( S t +1 = S t − X t + Y t ) t =1 T ÿ I ( S t , X t ; Y t | Y t ≠ 1 ) ( I ( ; ) ≥ 0) ≥ t =1 T I ( S t − X t ; Y t | Y t ≠ 1 ) ÿ ≥ (Data Processing Inequality) t =1 13/15
Converse Lemma If X t is i.i.d. and S t +1 = S t − X t + Y t then: I ( S t − X t ; Y t | Y t ≠ 1 ) = H ( S t − X t | Y t ≠ 1 ) − H ( S t +1 − X t +1 | Y t , X t +1 ) I ( S t − X t ; Y t | Y t ≠ 1 ) = H ( S t − X t ; Y t | Y t ≠ 1 ) − H ( S t − X t | Y t ) = H ( S t − X t | Y t ≠ 1 ) − H ( S t − X t + Y t | Y t ) = H ( S t − X t | Y t ≠ 1 ) − H ( S t +1 | Y t ) = H ( S t − X t | Y t ≠ 1 ) − H ( S t +1 | Y t , X t +1 ) = H ( S t − X t | Y t ≠ 1 ) − H ( S t +1 − X t +1 | Y t , X t +1 ) 13/15
Converse T I ( S 1 , X T ; Y T ) ≥ ÿ I ( S t − X t ; Y t | Y t ≠ 1 ) t =1 T I ( S t − X t ; X t | Y t ≠ 1 ) − H ( S T − X T | Y T ) ÿ ≥ H ( S 1 − X 1 ) + t =2 1 T I q ( S 1 , X T ; Y T ) L ú Œ = min q œ Q A lim T æŒ C T D 1 ÿ I q ( S t − X t ; X t | Y t ≠ 1 ) ≥ min q œ Q A lim T T æŒ t =2 ≥ min ◊ œ P S I ( S − X ; X ) 13/15
Numerical Example: Binomial distributed power demand Leakage Rates for Binomial Distributed Power Demand 1 10 J eq for m x = 5 J * for m x = 5 J eq for m x = 10 J * for m x = 10 0 10 J eq for m x = 20 Leakage Rate J * for m x = 20 − 1 10 − 2 10 0 5 10 15 20 25 30 35 40 45 50 Battery Size 14/15
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