A.1 Proofs for Section 2.2
Proof of Lemma 1. Let x be feasible, i.e., there exists an y such that (x, y) is a feasible solution.
The proof consists of two parts: first we show that (2.7) holds for contract k = 1, then we focus on the other contracts in the menu (k > 1).
First, realise that for an optimal y at least one IR constraint (2.5) must hold with equality. If this is not the case, we can increase all yk by adding some > 0 until at least one IR constraint is tight. This new solution is still feasible, as (2.6) only considers the difference yk− yl, which is unaffected. Moreover, the objective value of the new solution is strictly larger.
Now, suppose that y1 <
¯p1χ(x1), then for k ∈ K we have for all pk∈ [
¯pk, ¯pk] that pkχ(xk) − yk ≥
¯pkχ(xk) − yk
(2.6)
≥
¯pkχ(x1) − y1 ≥
¯p1χ(x1) − y1> 0.
Here, we use that χ is non-negative. The result implies that no IR constraint is tight, which is suboptimal as argued above. Hence, for an optimal y it must hold that y1 =
¯p1χ(x1).
Second, fix k ∈ K with k > 1 and consider the following IC constraints between contracts k and k − 1:
¯
pk−1(χ(xk) − χ(xk−1)) ≤ yk− yk−1≤
¯pk(χ(xk) − χ(xk−1)).
Since ¯pk−1=
¯pk, this implies that
yk− yk−1 =
¯pk(χ(xk) − χ(xk−1)).
Using our earlier result that y1 =
¯p1χ(x1), we obtain the following formula:
yk=
k
X
i=2¯pi(χ(xi) − χ(xi−1)) +
¯p1χ(x1), which can be rewritten into (2.7).
Proof of Lemma 2. First, we show the necessity of x1 ≤ · · · ≤ xK. Let k, k + 1 ∈ K and consider
(2.6) for Adding both IC constraints leads to (
¯pk−
Hence, all IC constraints (2.6) hold. Furthermore,
pkχ(xk) − yk ≥
Thus, all IR constraints (2.5) are satisfied and the solution is feasible.
Proof of Theorem 4. The uniform distribution (Assumption 4) implies that ωk= (¯pk−
¯pk)/(¯p −
¯p).
Therefore, we can simplify the summation in (2.8) as follows:
(¯pk−
Substituting these expressions in the result of Theorem 3 gives the desired formulation of the optimisation problem.
For the optimal solution, we first relax the constraint x1 ≤ · · · ≤ xK to obtain a separable optimisation problem. Since the objective of the relaxed problem is strictly concave and
differen-tiable (Assumption3), its optimal solution can easily be determined. Consider contract k ∈ K. We distinguish three cases.
Case I: if
d
dxk(φS+ ψ)(0) + ¯pk+
¯pk− ¯p ≤ 0,
then it is optimal for the relaxed problem to set xk= 0. Otherwise, the optimal xk for the relaxed problem satisfies xk > 0.
Case II: if
xklim→∞
d
dxk(φS+ ψ)(xk) + ¯pk+
¯pk− ¯p
≥ 0,
then it is optimal to set xk = ∞. Otherwise, a finite x is optimal. Here, we use that the above limit is zero only if it is an asymptote. This holds since φS+ ψ is strictly concave, implying that its derivative is strictly decreasing. Furthermore, this shows that Cases I and II are indeed mutually exclusive for (fixed) k.
Case III: if the above cases do not hold the optimal xkis found by setting the derivative to zero:
d dxk
φS(xk) + ψ(xk) + (¯pk+
¯pk− ¯p)xk
= 0 ⇐⇒ d
dxk(φS+ ψ)(xk) = −(¯pk+
¯pk− ¯p).
Since φS + ψ is strictly concave and differentiable, its derivative is continuous and invertible.
Furthermore, Cases I and II are excluded, so the following value is well-defined and strictly positive:
ˆ xk=
d
dxk(φS+ ψ)
−1
(¯p − ¯pk−
¯pk),
which is the optimum for the relaxed problem. Notice that the definitions of k∗ and ˆk imply that Case III corresponds to k ∈ K such that k∗ ≤ k ≤ ˆk. By strict concavity of φS+ ψ we know that its derivative is strictly decreasing. Furthermore, realise that ¯pk+
¯pk− ¯p < ¯pk+1+
¯pk+1− ¯p for all k ∈ K. Therefore, we have 0 < ˆxk∗ < · · · < ˆxˆk.
Combining all cases leads to a solution satisfying 0 ≤ x1 ≤ · · · ≤ xK, which is feasible for the non-relaxed problem. Hence, this is the optimal solution to our original problem.
A.2 Proofs for Section 2.4
Proof of Lemma 6. Consider an optimal partition ∆ that does not satisfy the properties stated in the lemma. First, recall the definition of k∗ = mink ∈ K : δk+ δk−1 > α−1α . Suppose k∗(∆) > 2, which requires K > 2. Construct a new partition ˆ∆ with ˆδ1 = δk∗−1, ˆδk= δk∗ for 1 < k < k∗ and δˆk = δk otherwise. By construction, the partition ˆ∆ leads to the same objective value as ∆ and is therefore an optimal partition. Furthermore, we have
δˆ1+ ˆδ0 = δk∗−1+ δ0≤ δk∗−1+ δk∗−2 ≤ α−1α
δˆk+ ˆδk−1 ≥ δk∗+ δk∗−1 > α−1α , for k = 2, . . . , K.
Thus, k∗( ˆ∆) = 2. Therefore, by applying this transformation, we can assume without loss of generality that the optimal partition ∆ satisfies k∗(∆) ∈ {1, 2}.
Second, we modify the partition ∆ into a strictly better one, which is a contradiction. The details require two cases to be analysed.
Case I: there exists an index i ∈ {1, . . . , K − 1} such that δi−1 = δi < δi+1. Notice that i + 1 ≥ 2 ≥ k∗(∆) ∈ {1, 2}. Therefore, δi+1+ δi> α−1α and there exists an 0 < < 1 such that
(1 − )δi+1+ (1 + )δi> α−1α .
Construct a new partition ˆ∆ by setting ˆδi = (1−)δi+1+δiand ˆδk= δkotherwise. By construction, we have ˆδi−1 < ˆδi < ˆδi+1, i ≥ k∗( ˆ∆), and k∗( ˆ∆) ≤ k∗(∆). The normalised objective value corresponding to ˆ∆ differs from that of ∆ as follows: the terms
i+1
X
k=max{i,k∗(∆)}
(δk− δk−1) α1 + δk+ δk−1− 1n+1n
= (δi+1− δi) α1 + δi+1+ δi− 1n+1n
are replaced by
(ˆδi− ˆδi−1)
1
α + ˆδi+ ˆδi−1− 1n+1n
+ (ˆδi+1− ˆδi)
1
α + ˆδi+1+ ˆδi− 1n+1n
= (1 − )(δi+1− δi) α1 + (1 − )δi+1+ (1 + )δi− 1n+1n + (δi+1− δi) α1 + (2 − )δi+1+ δi− 1n+1n
> (δi+1− δi) α1 + δi+1+ δi− 1n+1n .
The inequality follows from strict convexity of the function (·)n+1n on R≥0. This implies that ˆ∆ is strictly better than ∆, which contradicts the optimality of ∆.
Case II: δK−1 = δK = 1. Define i = min{k ∈ {1, . . . , K} : δk = δK} to be the first partition point that coincides with δK. Notice that δ0 < δi and δi+ δi−1= 1 + δi−1≥ 1 > α−1α , so i ≥ k∗(∆).
Therefore, there exists an 0 < < 1 such that (1 − )δi+ (1 + )δi−1> α−1α .
Construct a new partition ˆ∆ by setting ˆδi = (1 − )δi + δi−1 and ˆδk = δk otherwise. By construction, we have ˆδi−1 < ˆδi < ˆδi+1 and k∗( ˆ∆) = k∗(∆). The normalised objective value corresponding to ˆ∆ differs from that of ∆ as follows: the terms
i+1
X
k=max{i,k∗(∆)}
(δk− δk−1) α1 + δk+ δk−1− 1n+1n
= (δi− δi−1) α1 + δi+ δi−1− 1n+1n
are replaced by
(ˆδi− ˆδi−1)
1
α + ˆδi+ ˆδi−1− 1n+1n
+ (ˆδi+1− ˆδi)
1
α + ˆδi+1+ ˆδi− 1n+1n
= (1 − )(δi− δi−1) α1 + (1 − )δi+ (1 + )δi−1− 1n+1n + (δi− δi−1) α1 + (2 − )δi+ δi−1− 1n+1n
> (δi− δi−1) α1 + δi+ δi−1− 1n+1n .
Hence, ˆ∆ is strictly better than ∆, which contradicts the optimality of ∆. This concludes the proof.
A.3 Proofs for Section 2.5
Proof of Lemma 8. By Lemma 6 we know that k∗ ∈ {1, 2} for the optimal partition. First, we analyse the objective function ΓK when we consider k∗ as a parameter independent of the chosen partition (which it is not). To simplify notation, we use the normalised νΓK, which does not affect the optimality of a partition. Suppose k∗ = 1, then the normalised objective function is
νΓK|k∗=1 =
(δ1− δ0)(α1 + δ1+ δ0− 1)2+ (δ2− δ1)(α1 + δ2+ δ1− 1)2
+ (δ3− δ2)(α1 + δ3+ δ2− 1)2+ (δ4− δ3)(α1 + δ4+ δ3− 1)2+ · · · + (δK−1− δK−2)(α1 + δK−1+ δK−2− 1)2
+ (δK− δK−1)(α1 + δK+ δK−1− 1)2 .
Since terms cancel out, this is a quadratic function for each δk, k = 1, . . . , K − 1. Note that δ0 = 0 and δK = 1 are fixed for any partition. Setting the gradient to zero gives (δk+1 − δk−1)(δk+1+ δk−1− 2δk) = 0 for all k ∈ {1, . . . , K − 1}. By Lemma6 we know that δk+1> δk−1 must hold for the optimal partition, so the only possibility is δk = 12(δk+1+ δk−1) for all k ∈ {1, . . . , K − 1}. The solution to this linear system of equalities is
δk= k
K ≡ δequik , which is the equidistant partition ∆equi.
Likewise, suppose k∗ = 2, then we have νΓK|k∗=2 =
(δ2− δ1)(α1 + δ2+ δ1− 1)2
+ (δ3− δ2)(α1 + δ3+ δ2− 1)2+ (δ4− δ3)(α1 + δ4+ δ3− 1)2+ · · · + (δK−1− δK−2)(α1 + δK−1+ δK−2− 1)2
+ (δp − δK−1)(α1 + δK+ δK−1− 1)2 .
This is cubic in δ1 and quadratic in the other δk (k ∈ {2, . . . , K − 1}). Setting the gradient to zero gives
(1 −α1 + δ2− 3δ1)(α1 − 1 + δ2+ δ1) = 0
and (δk+1− δk−1)(δk+1+ δk−1− 2δk) = 0 for k ∈ {2, . . . , K − 1}. The roots of the derivative of the cubic function are
δ1 = 13(1 −α1 + δ2) and δ1 = 1 −α1 − δ2.
By closer investigation of the shape of this cubic function, the larger value of these two corresponds to the maximum. Solving the system of linear equations for both cases results in:
δ1 = 13(1 −α1 + δ2) =⇒ δk= K + k − 1
2K − 1 − K − k 2K − 1
1 α,
δ1 = 1 −α1 − δ2 =⇒ δk= K + k − 3
2K − 3 − K − k 2K − 3
1 α.
Since δ1 is larger in the first case, this is the correct solution. Hence,
δk = K + k − 1
2K − 1 − K − k 2K − 1
1
α = 1 − K − k 2K − 1
1 α + 1
≡ δkcubic.
We refer to this partition as ∆cubic.
Finally, we check which of these partitions is feasible. First, we check correctness with k∗. We have (∆equi ⇒ k∗ = 1) if and only if δ1equi > α−1α , which is K1 > α−1α or equivalently α < K−1K . Likewise, (∆cubic ⇒ k∗ = 2) if and only if δ1cubic≤ α−1α and δcubic2 + δ1cubic> α−1α . These conditions require the following:
δ1cubic≤ α−1α ⇐⇒ 1 −2K−1K−1 1α+ 1 ≤ α−1α ⇐⇒ α ≥ K−1K , δcubic2 + δ1cubic> α−1α ⇐⇒ 2 −2K−32K−1 1α+ 1 > α−1α ⇐⇒ α > −1.
Moreover, we need to check δ1cubic> δ0 = 0:
δ1cubic> 0 ⇐⇒ 1 −2K−1K−1 α1 + 1 > 0 ⇐⇒ α > K−1K . This condition is trivially satisfied for the range α ≥ K−1K corresponding to ∆cubic.
To conclude, for α < K/(K − 1) the optimal partition is ∆equi, whereas for α ≥ K/(K − 1) the optimal partition is ∆cubic.
A.4 Proofs for Section 2.6
Proof of Lemma 10. First, we consider k∗ = 1, for which
νΓ2 = δ1 1
α + δ1− 132
+ (1 − δ1) α1 + δ1
32 .
Setting the derivative with respect to δ1 to zero and solving the equation for δ1 results in two solutions:
δ1+= 301
q
36α12 − 15 + 15 − 6α1
and δ−1 = 301
− q
36α12 − 15 + 15 − 6α1
. The partition point δ1+is a local maximum, whereas δ1−does not maximise the objective. Further-more, δ+1 is valid for α ∈ (0,25√
15], i.e., it is feasible and corresponds to k∗= 1.
Second, consider k∗ = 2, where the normalised optimal objective is
νΓ2 = (1 − δ1) α1 + δ1
32 .
Again, setting its derivative to δ1 to zero, leads to a single feasible solution δ1∗= 1 − 25(α1 + 1).
The partition point δ∗1 is valid for α ∈ [32, ∞).
In contrast to the DMU-1 problem, the valid intervals overlap: α ∈ (0, 1.5491] for k∗ = 1 and α ∈ [1.5, ∞) for k∗ = 2. It turns out that the optimal δ1 switches from δ+1 to δ1∗ (a discontinuous jump) as α increases. The partitions δ+1 and δ∗1 are both optimal for αtrans ≈ 1.5371 (the exact expression for αtrans is too verbose). This is illustrated in Figure7, where the maximum on the left corresponds to δ1∗ and that on the right to δ1+.
Finally, the lower bound on δ1opt is attained at the switch to δ1∗ (at αtrans) and the upper bound is reached for α → ∞.
Figure 7: DMU-2: pooling performance Γ2/Γ∞ at αtrans in terms of the partition δ1. The shown points are δ∗1 (left), δ1− (middle), and δ+1 (right).
A.5 Numerical solver
We describe the used methodology to numerically optimise the partition for the DMU-2 problem of Section 2.6. For each α, we have to maximise ΓK or equivalently ΓK/Γ∞, so we can use the
formulas of the normalised objective values νΓK and νΓ∞. However, the formula for ΓK contains index k∗, which depends on the partition. From Lemma6we know that k∗∈ {1, 2} for the optimal partition. Therefore, we simply optimise twice: for k∗ = 1 in the formula with the restriction δ1 > α−1α , and for k∗ = 2 with the restrictions δ1≤ α−1α and δ1+ δ2> α−1α . The optimal partition is the best of the resulting partitions. Note that k∗ = 1 is always optimal for 0 < α ≤ 1, since k∗ = 2 is infeasible for this range.
For DMU-1 and DMU-2, we have inspected the shape of ΓK as a function of the used partition for K = 2 and K = 3. From our observations, ΓK with k∗ fixed to 1 or 2 is a smooth function with a clear maximum on the respective domain that can be found using a gradient-based search. Thus, we apply a gradient-based search to maximise ΓK with k∗ fixed to either 1 or 2. To be precise, we use Maple’s built-in solver ‘NLPSolve’ for non-linear programs with only bounds on the variables, which uses the Modified-Newton method, and verify the feasibility of the obtained partition. That is, we check if 0 = δ1 < δ2 < · · · < δK−1 < δK = 1 and if the partition indeed results in the used value for k∗.
We have also used Maple’s built-in solver ‘NLPSolve’ for constrained non-linear programs, which uses Sequential Quadratic Programming. With this solver we can enforce the required constraints on δk directly. Note that we still separately solve for k∗ fixed to either 1 or 2. Both solvers give the same results, also when specifying different starting solutions.
The used methodology only guarantees to find a local maximum. However, the numerical solver always finds the same local optimum, i.e., it is stable. Furthermore, the results are consistent with our available theoretical results, such as for DMU-1 with any K and for DMU-2 with K = 2.
Therefore, all results indicate that the numerical solver is able to find the global maximum.