– Leakage of Output Power (xj): At t = 0, i receives λj(0) and yj(0).
As i knows the local power demand Dj, i can simply get the value of xj(0).
Then, i can find out the output power estimate xj(t) at every round as i
can getP
k∈Nj+qjkyk(t) from each k in Nj+at round t and yj(t + 1) from
j at round t + 1: xj(t + 1) = X k∈N+ j qjkyk(t) − yj(t + 1) + xj(t)
If the protocol ends at round T , xj(T ) is the final power output estimate
for node j. This can be easily estimated by node i.
– Leakage of Cost Function Parameters (aj, bj): i knows values of
λj(t) and xj(t), from two rounds t and t + 1:
xj(t) = βjλj(t) + αj
xj(t + 1) = βjλj(t + 1) + αj
Solving two linearly independent equations, node i can find the values of αj and βj which will give values of aj and bj.
– Leakage of Generator Constraint Parameters (xj and xj): At
initial round t = 0, i observes xj(0), so i might get one of the values of xj
or xj.
8.4
Privacy-preserving ED (PPED) Protocol
To prevent the privacy leakage, we present our PPED protocol which uses a pri- vacy layer in each round using a secure sum protocol. We can use the consensus- based algorithm proposed by Yang et al. [8] described in section 8.2.3 as the basis. The secure sum protocol used in PPED is similar to the n − private protocol for summation presented by Benaloh in [41] but uses (n − 1) partitions instead of n.
8.4.1
System Model
We consider a complete synchronous network G = (V, E) with n nodes and a secure and reliable point to point communication channel between every node (no eavesdropping). As G is a complete graph, every element in P and Q matrix will be 1n.
A practical assumption would be that all numerical values we want to compute are fixed point values for ED algorithms. We can multiply any fixed point elements with a suitable constant and convert them into integers. In Step 2 (next section), we convert λi(t) and yi(t) to integer and in Step 7 we convert
80 CHAPTER 8. PRIVACY-PRESERVING ED PROTOCOL I
Figure 8.1: PPED protocol for n = 4
8.4.2
PPED protocol
The diagram of our PPED protocol with 4 nodes is illustrated in figure 8.1. The protocol is outlined as follows:
– Step 1: Initialization of every node at t = 0:
xi(0) = xi, if xi< Di Di, if xi≤ Di≤ xi xi, if Di< xi ∀i ∈ V λi(0) = xi(0) − αi βi yi(0) = Di− xi(0)
– Step 2: In step 2 to step 6 we map λi(t) and yi(t) values to integer values
and map it back in step 7. We consider ZM as the additive group of inte-
gers from 0 to M − 1 (M is a large integer such that M >Pn
i=1λi(t) and
M >Pn
i=1yi(t)). Every node i, chooses (n − 1) numbers independently
with uniform distribution from ZM, such that their sum (in modulo M ) is
equal to λi(t). The same is done for breaking yi(t) into n − 1 parts where
8.4. PRIVACY-PRESERVING ED (PPED) PROTOCOL 81
segment y(j)i (t) for the node j. All calculations are done in modulo M . λi(t) = X ∀j∈V ,j6=i λ(j)i (t) yi(t) = X ∀j∈V ,j6=i y(j)i (t)
– Step 3: Every node i sends segments λ(j)i (t) and yi(j)(t) to the respective j in the network. We can use some token to distinguish between λi(t) and
yi(t) segment values.
– Step 4: Each node i will receive a total of n−1 data segments for λ(t) and y(t) respectively from the other n − 1 nodes in the graph. All the nodes add all their received segments for λ(t) and y(t) separately. At node i, the total received sums λiS4(t) and yiS4(t) is calculated as follows:
λiS4(t) = X ∀j∈V ,j6=i λ(i)j (t) yiS4(t) = X ∀j∈V ,j6=i yj(i)(t) – Step 5: Every node i sends λi
S4(t) and yS4i (t) to the remaining n − 1
nodes in the network.
– Step 6: Every node i adds all received λjS4(t)’s from the other n − 1 nodes and its own λi
S4(t). Hence, every node gets sum value without knowing
the individual inputs:
X ∀i∈V λi(t) = X ∀j∈V ,j6=i λjS4(t) + λiS4(t)
Do the same for y(t).
– Step 7: Node i finds:
λi(t + 1) = 1 n X ∀i∈V λi(t) + yi(t)
– Step 8: For all nodes i, the power output is calculated as:
xi(t + 1) = βiλi(t + 1) + αi
– Step 9: For all nodes i, the power difference is compared as:
yi(t + 1) = 1 n X ∀i∈V yi(t) − (xi(t + 1) − xi(t))
82 CHAPTER 8. PRIVACY-PRESERVING ED PROTOCOL I
– Step 10: Check: if
∀i ∈ V , yi(t + 1) ≈ 0
then ED solution found break
else t = t+1, Repeat ∀i ∈ V Step 2 to Step 10.
Remark. (Correctness of PPED) If Yang et al.’s ED protocol [8] is correct, the PPED protocol is also correct.
Proof. Correctness of Yang et al.’s ED protocol is shown in [8, Thm 2 and 3]. Hence the correctness of PPED protocol follows from, for any round for t = {0, 1, . . . , T }, P ∀i∈Vλi(t) =P∀j∈V ,j6=iλ j S4(t) + λ i S4(t) and P ∀i∈V yi(t) = P ∀j∈V ,j6=iy j S4(t) + y i S4(t).