Reliability or Inventory? An Analysis of Performance-Based Contracts for Product Support Services

Abstract

Traditional sourcing arrangements for after-sales product support have centered around physical assets. Typically, a customer would pay the supplier of maintenance services in proportion to the resources used, such as spare parts, that are needed to maintain the product. In recent years, we have witnessed the emergence of a new service contracting strategy called performance-based contracting (PBC). Under such a contractual relationship, the basis of supplier compensation is actual realized uptime of the product. In this study we build a game-theoretic model and compare the inefficiencies arising under the traditional resource-based contract (RBC) and PBC. In both cases, the customer sets the contract terms, and as a response, the supplier sets the base-stock inventory level of spares as well as invests in increasing product reliability. We find that PBC provides stronger incentives for the supplier to invest in reliability improvement, which in turn leads to savings in acquiring and holding spare product assets. Moreover, the efficiency of PBC improves if the supplier owns a larger portion of the spare assets. Our analysis advocates the view that the full benefit of a PBC strategy is achieved when suppliers are transformed into total service providers who take the ownership of physical assets.

Appendices

Appendix 1: Mathematical Preliminaries

In the proofs, we use the following conventions related to Normal approximation. To circumvent the conceptual difficulty of having negative s, we will regard 0 as the lower bound on s and, as a consequence, we define the lower bound on z as $\underline{z} \equiv (0 - 1/\tau )/\sqrt{1/\tau } = -1/\sqrt{\tau }$. This definition does not cause a problem because all quantities of our interest on the domain $(-\infty,-1/\sqrt{\tau })$ are insignificant in the range $1/N \ll \underline{\tau }<\overline{\tau } \lesssim 0.1$ we consider (see Sect. 4.3.1). Thus, $\sqrt{\tau }\phi \left (\underline{z}\right ) \simeq 0$ and $\varPhi \left (\underline{z}\right ) \simeq 0$. These approximations at the lower bound $\underline{\tau }$ require us to use the following conventions: (1) $L(\underline{z}) =\phi (\underline{z}) -\underline{ z}\overline{\varPhi }(\underline{z}) = \left [\sqrt{\tau }\phi \left (\underline{z}\right ) + \overline{\varPhi }\left (\underline{z}\right )\right ]/\sqrt{\tau }\simeq 1/\sqrt{\tau } = -\underline{z}$; (2) $\varPhi ^{-1}\left (0\right ) \simeq \underline{ z} = -1/\sqrt{\tau }$. Throughout the analysis we replace the notation $\simeq$ with an equality, with an understanding that some of them represent approximations. Under these conventions, the following relationships hold.

Lemma 3.

(i) $\partial E[B\,\vert \,\tau,s]/\partial s = -\overline{\varPhi }(z) <0$ , (ii)  $\partial E[B\,\vert \,\tau,s]/\partial \tau = -\phi (z)/2\tau ^{3/2} -\overline{\varPhi }(z)/\tau ^{2} <0$ , (iii) ∂E[I | τ,s]∕∂s = Φ(z) > 0, (iv) ∂E[I | τ,s]∕∂τ = −ϕ(z)∕(2τ ^3∕2 ) + Φ(z)∕τ ² > 0.

In the next lemma we state the property of a probability distribution exhibiting increasing failure rate (IFR) that becomes useful in the proofs.

Lemma 4.

Let X be a random variable with an IFR property whose pdf g is differentiable and vanishes at both extremes of its support $[\underline{y},\overline{y}]$ , where $\underline{y} = -\infty$ and $\overline{y} = \infty$ are permitted. Let G be the cdf of X and $\overline{G}(\cdot ) \equiv 1 - G(\cdot )$ . Then

$$\displaystyle{\omega (y) \equiv \frac{g(y)E[(X - y)^{+}]} {\overline{G}(y)^{2}} \leq 1.}$$

Proof.

IFR means

$$\displaystyle{ \frac{d} {dy}\left ( \frac{g(y)} {\overline{G}(y)}\right ) = \frac{g^{{\prime}}(y)\overline{G}(y) + [g(y)]^{2}} {\overline{G}(y)^{2}} \geq 0\,,}$$

which in turn implies $-g^{{\prime}}(y)/g(y) \leq g(y)/\overline{G}(y)$. It can easily be shown that $\omega ^{{\prime}}(y) \geq 0$. To derive the upper bound, we only need to show $m \equiv \lim _{y\rightarrow \overline{y}}\omega (y) \leq 1$. Applying l’Hopital’s rule, we obtain:

$$\displaystyle\begin{array}{rcl} m& =& \lim _{y\rightarrow \overline{y}}\frac{g(y)E[(X - y)^{+}]} {\overline{G}(y)^{2}} =\lim _{y\rightarrow \overline{y}}\frac{g^{{\prime}}(y)E[(X - y)^{+}] - g(y)\overline{G}(y)} {-2g(y)\overline{G}(y)} {}\\ & =& \frac{1} {2} -\frac{1} {2}\lim _{y\rightarrow \overline{y}}\frac{g^{{\prime}}(y)E[(X - y)^{+}]} {g(y)\overline{G}(y)} \leq \frac{1} {2} + \frac{1} {2}\lim _{y\rightarrow \overline{y}}\frac{g(y)E[(X - y)^{+}]} {\overline{G}(y)^{2}} = \frac{1} {2} + \frac{m} {2}, {}\\ \end{array}$$

where the inequality follows from the earlier result $-g^{{\prime}}(y)/g(y) \leq g(y)/\overline{G}(y)$. Arranging both sides yields m ≤ 1.

Appendix 2: Proofs of the Main Results

Proof (Proof of Lemma 1).

The supplier’s expected profit under RBC is $u(\tau,s) = E[T\,\vert \,\tau,s]-\psi (\tau,s) = w-K(\tau )+(r-\kappa )/\tau +\left (p - c -\delta h_{b}\right )s-\delta \left (h_{g} - h_{b}\right )E[I\,\vert \,\tau,s]$. Differentiating this with respect to s, (see Lemma 3) we have $\partial u/\partial s = p - c -\delta h_{b} -\delta \left (h_{g} - h_{b}\right )\varPhi (z)$ and $\partial ^{2}u/\partial s^{2} = -\delta \left (h_{g} - h_{b}\right )\sqrt{\tau }\phi (z) <0$, i.e., u(τ, s) is concave in s for a fixed τ. It can be shown that a finite feasible solution of $(\widetilde{\mathcal{C}})$ exists only if c +δ h _b  < p < c +δ h _g (proof is omitted). With δ > 0 and c +δ h _b  < p < c +δ h _g , the profit-maximizing s is found in the interior. The first-order condition $\frac{\partial u} {\partial s} = 0$ yields z ^∗ = Φ ⁻¹[1 − (c +δ h _g − p)∕(δ(h _g − h _b ))], which is independent of τ. Substituting $s^{{\ast}} = 1/\tau + z^{{\ast}}/\sqrt{\tau }$ and noting $E[I\,\vert \,\tau,s] = (\phi (z) + z\varPhi (z))/\sqrt{\tau }$, u(τ, s) becomes

$$\displaystyle{\widetilde{u}(\tau ) \equiv u(\tau,s^{{\ast}}) = w - K(\tau ) + \frac{r + p -\kappa -c -\delta h_{b}} {\tau } -\frac{\delta \left (h_{g} - h_{b}\right )\phi (z^{{\ast}})} {\sqrt{\tau }}.}$$

Note that $\lim _{\tau \rightarrow \overline{\tau }}\widetilde{u}(\tau ) = -\infty$. Differentiating $\widetilde{u}(\tau )$ yields

$$\displaystyle\begin{array}{rcl} \widetilde{u}^{{\prime}}(\tau )& =& -K^{{\prime}}(\tau ) -\frac{r + p -\kappa -c -\delta h_{b}} {\tau ^{2}} + \frac{\delta \left (h_{g} - h_{b}\right )\phi (z^{{\ast}})} {2\tau ^{3/2}} \quad \text{and} {}\\ \widetilde{u}^{{\prime\prime}}(\tau )& =& -K^{{\prime\prime}}(\tau ) + \frac{2(r + p -\kappa -c -\delta h_{b})} {\tau ^{3}} -\frac{3\delta \left (h_{g} - h_{b}\right )\phi (z^{{\ast}})} {4\tau ^{5/2}}. {}\\ \end{array}$$

Let $\widehat{\tau }$ be the solution of the first-order condition $\widetilde{u}^{{\prime}}(\tau ) = 0$. Multiplying this condition by $2/\widehat{\tau }$ and adding it to $\widetilde{u}^{{\prime\prime}}(\tau )$, we get

$$\displaystyle\begin{array}{rcl} \widetilde{u}^{{\prime\prime}}(\widehat{\tau })& =& -K^{{\prime\prime}}(\widehat{\tau }) -\frac{2K^{{\prime}}(\widehat{\tau })} {\widehat{\tau }} + \frac{\delta \left (h_{g} - h_{b}\right )\phi (z^{{\ast}})} {4\widehat{\tau }^{5/2}} {}\\ & <& -\frac{2K^{{\prime}}(\widehat{\tau })} {\widehat{\tau }} + \left (-2 + \frac{\widehat{\tau }} {4\sqrt{2\pi }}\right )\frac{h_{g} - h_{b}} {\widehat{\tau }^{3}} <-\frac{2K^{{\prime}}(\widehat{\tau })} {\widehat{\tau }} -\left (1.99\right )\frac{h_{g} - h_{b}} {\widehat{\tau }^{3}} <0. {}\\ \end{array}$$

In the first inequality of this result, we used the assumption $2\left (h_{g} - h_{b}\right ) <\underline{\tau } ^{3}K^{{\prime\prime}}(\underline{\tau })$ from Sect. 4.3.2, δ ≤ 1, and the upper bound on the standard normal pdf, i.e., $\phi (z) \leq \sqrt{1/(2\pi \text{Var} [O(\tau )])} = \sqrt{\tau /(2\pi )}$. In the second inequality, we used the upper bound $\widehat{\tau }<\overline{\tau } \lesssim 0.1$ as specified in Sect. 4.3.1. Suppose that the solution is in the interior, i.e., $\underline{\tau }<\widehat{\tau }<\overline{\tau }$. Then $\widetilde{u}^{{\prime\prime}}(\widehat{\tau }) <0$ implies that any interior critical point, if it exists, should be a maximizer. Let us consider two cases: $\widetilde{u}^{{\prime}}(\underline{\tau })> 0$ and $\widetilde{u}^{{\prime}}(\underline{\tau }) \leq 0$. If $\widetilde{u}^{{\prime}}(\underline{\tau })> 0$ then $\widehat{\tau }>\underline{\tau }$ since $\widetilde{u}(\tau )$ initially increases and approaches $-\infty$ as $\tau \rightarrow \overline{\tau }$. Therefore $\widehat{\tau }$ is a unique maximizer since more than one maximizer requires at least one interior minimizer (as $\widetilde{u}(\tau )$ is continuous), which contradicts our earlier observation that any interior critical point should be a maximizer. Now suppose $\widetilde{u}^{{\prime}}(\underline{\tau }) \leq 0$. Again, if an interior critical point exists, then it should be a maximizer. But this means a minimizer should exist to the left of the maximizer, since $\widetilde{u}(\tau )$ initially decreases. This leads to a contradiction, and therefore, no interior critical point exist in this case; $\widetilde{u}(\tau )$ decreases monotonically if $\widetilde{u}^{{\prime}}(\underline{\tau }) \leq 0$. Then $\widetilde{u}(\tau )$ is maximized at $\tau =\underline{\tau }$. Summarizing, the supplier uniquely chooses $\tau ^{{\ast}} =\underline{\tau }$ if $\widetilde{u}^{{\prime}}(\underline{\tau }) \leq 0$ and $\tau ^{{\ast}} =\widehat{\tau }>\underline{\tau }$ if $\widetilde{u}^{{\prime}}(\underline{\tau })> 0$, where $\widehat{\tau }$ is obtained from the first-order condition (4.1). To establish a connection between this result and the three cases stated in the lemma, observe that, for $p_{\min } \equiv c +\delta h_{b} <p <c +\delta h_{g} \equiv p_{\max }$ and a fixed τ, we have: (i) $\overline{m}(\tau ) \equiv \lim _{p\rightarrow p_{\min }}\widetilde{u}^{{\prime}}(\tau ) = -K^{{\prime}}(\tau ) - (r-\kappa )/\tau ^{2}$; (ii) $\underline{m}(\tau ) \equiv \lim _{p\rightarrow p_{\max }}\widetilde{u}^{{\prime}}(\tau ) = -K^{{\prime}}(\tau ) - (r -\kappa +\delta (h_{g} - h_{b}))/\tau ^{2} <\overline{m}(\tau )$; (iii) $\partial \widetilde{u}^{{\prime}}(\tau )/\partial p = -1/\tau ^{2} - z^{{\ast}}/(2\tau ^{3/2}) = -1/(2\tau ^{2}) - s^{{\ast}}/(2\tau ) <0$. At $\tau =\underline{\tau }$, $\overline{m}(\underline{\tau }) = (b - r)/\underline{\tau }^{2}$ and $\underline{m}(\underline{\tau }) = (a - r)/\underline{\tau }^{2}$, where a and b are defined in the lemma. If 0 ≤ r ≤ a, we have $\overline{m}(\underline{\tau })>\underline{ m}(\underline{\tau }) \geq 0$. Since $\widetilde{u}^{{\prime}}(\underline{\tau })$ decreases in p from $\overline{m}(\underline{\tau })> 0$ to $\underline{m}(\underline{\tau }) \geq 0$, $\widetilde{u}^{{\prime}}(\underline{\tau })> 0$ for all $p \in (p_{\min },p_{\max })$, implying that $\widetilde{u}(\tau )$ is increasing initially at $\tau =\underline{\tau }$. Therefore, the optimal τ is found in the interior, i.e., $\tau ^{{\ast}} =\widehat{\tau }>\underline{\tau }$. Next, assume a < r < b. Then $\overline{m}(\underline{\tau })> 0$ and $\underline{m}(\underline{\tau }) <0$, implying that there exists $\overline{p}\left (r\right ) \in (p_{\min },p_{\max })$ such that $\widetilde{u}^{{\prime}}(\underline{\tau })> 0$ for $p \in (p_{\min },\overline{p}\left (r\right ))$ and $\widetilde{u}^{{\prime}}(\underline{\tau }) \leq 0$ for $p \in [\overline{p}\left (r\right ),p_{\max })$. As we found above, $\tau ^{{\ast}} =\widehat{\tau }>\underline{\tau }$ in the former case and $\tau ^{{\ast}} =\underline{\tau }$ in the latter case. $\overline{p}\left (r\right )$ is determined from the equation $\widetilde{u}^{{\prime}}(\underline{\tau }) = 0$. Finally, assume r ≥ b. Then we have $0 \geq \overline{m}(\underline{\tau })>\underline{ m}(\underline{\tau })$, which implies that $\widetilde{u}^{{\prime}}(\underline{\tau })$ remains in the negative region as p increases from $p_{\min }$ to $p_{\max }$. So $\widetilde{u}^{{\prime}}(\underline{\tau }) \leq 0$ for all $p \in (p_{\min },p_{\max })$. Then by the finding above, $\tau ^{{\ast}} =\underline{\tau }$. The comparative statics results in the lemma are shown via implicit differentiation of the first-order condition $\widetilde{u}^{{\prime}}(\widehat{\tau }) = 0$.

Proof (Proof of Proposition 2).

The following statement can be proved (proof is omitted). Under RBC, one of the following three outcomes emerges in equilibrium, along with the condition $L(z^{{\ast}})/\sqrt{\tau ^{{\ast}}} =\beta$: (i) $\tau ^{{\ast}}>\underline{\tau }$  that solves (4.1) with r = 0, (ii) $\tau ^{{\ast}}>\underline{\tau }$  that solves $\left (\tau ^{{\ast}}\right )^{2}K^{{\prime}}(\tau ^{{\ast}}) =\kappa -h_{g} + h_{b}$ and (4.1), or (iii) $\tau ^{{\ast}} =\underline{\tau }$. That the backorder constraint binds in equilibrium follows directly from this statement. Binding constraint reduces the original two-dimensional problem to a single-dimensional one, as the equation $L(z)/\sqrt{\tau } =\beta$ establishes a one-to-one correspondence between τ and s. Writing this condition as $z = L^{-1}\left (\beta \sqrt{\tau }\right ) =\zeta \left (\tau \right )$ and substituting it in the customer’s expected cost expression yields the reduced cost function $\widetilde{C}(\tau ) =\underline{ u} + h_{g}N + K(\tau ) + (\kappa +c + h_{b})/\tau + (c + h_{g})\zeta \left (\tau \right )/\sqrt{\tau }$. As expected, this function is convex and minimized at the first-best solution τ ^FB that we derived in Proposition 1. (This can be verified by noting that $\zeta ^{{\prime}}\left (\tau \right ) = (-f\left (\zeta \left (\tau \right )\right ) +\zeta \left (\tau \right ))/(2\tau )$.) It remains to choose the optimal solution among the three candidates for the equilibrium identified in the above statement. Consider each case (i)–(iii):

• (i) In Lemma 1 we derived the optimality condition z ^∗ = Φ ⁻¹[1 − (c + δ h _g − p)∕(δ(h _g − h _b ))]. Inverting this and letting $z^{{\ast}} =\zeta \left (\tau \right )$, i.e., the necessary condition for the binding backorder constraint, yields $p = c +\delta h_{b} +\delta \left (h_{g} - h_{b}\right )\varPhi \left (\zeta \left (\tau \right )\right )$. Substituting this in (4.1) and setting r = 0, we get

  $$\displaystyle{ \tau ^{2}K^{{\prime}}(\tau ) =\kappa -\delta \left (h_{ g} - h_{b}\right )\left (\varPhi \left (\zeta \left (\tau \right )\right ) - (\sqrt{\tau }/2)\phi (\zeta \left (\tau \right ))\right ). }$$

  (4.5)

  Note that $\varPhi \left (\zeta \left (\tau \right )\right ) - (\sqrt{\tau }/2)\phi (\zeta \left (\tau \right )) = 1 -\overline{\varPhi }\left (\zeta \left (\tau \right )\right ) - (\sqrt{\tau }/2)\phi (\zeta \left (\tau \right )) <1$. Hence, the solution of (4.5), which we call τ ₁, satisfies $\tau _{1}^{2}K^{{\prime}}(\tau _{1})>\kappa -\delta \left (h_{g} - h_{b}\right )$.

• (ii) Optimal τ for this case, called τ ₂, is given by the stated condition $\tau ^{2}K^{{\prime}}(\tau ) =\kappa -\left (h_{g} - h_{b}\right )$. It is clear that $\tau _{2}^{2}K^{{\prime}}(\tau _{2}) =\kappa -\left (h_{g} - h_{b}\right ) <\kappa -\delta \left (h_{g} - h_{b}\right ) <\tau _{ 1}^{2}K^{{\prime}}(\tau _{1})$, where the last inequality is from (i) above.

• (iii) In this case we have $\tau ^{{\ast}} =\underline{\tau }$. From the assumption $\underline{\tau }^{2}K^{{\prime}}(\underline{\tau }) <\kappa -\left (h_{g} - h_{b}\right )$ found in Sect. 4.3.2, we have $\underline{\tau }^{2}K^{{\prime}}(\underline{\tau }) <\kappa -\left (h_{g} - h_{b}\right ) =\tau _{ 2}^{2}K^{{\prime}}(\tau _{2})$.

Combining the inequalities we derived above, we have $\underline{\tau }^{2}K^{{\prime}}(\underline{\tau }) <\tau _{ 2}^{2}K^{{\prime}}(\tau _{2}) <\tau _{ 1}^{2}K^{{\prime}}(\tau _{1})$. Since $\tau ^{2}K^{{\prime}}(\tau )$ is increasing, this implies $\underline{\tau }<\tau _{2} <\tau _{1}$. Next, we show τ ₁ < τ ^FB. Using the definition of Γ(τ), the optimality condition (4.5) can be rewritten as

$$\displaystyle{ \varGamma (\tau _{1}) -\frac{\left (1-\delta \right )h_{b} + p} {\tau _{1}^{2}} -\frac{\left (1-\delta \right )h_{g} + p} {2\tau _{1}^{3/2}} f(\zeta \left (\tau _{1}\right )) = K^{{\prime}}(\tau _{ 1}). }$$

(4.6)

The derivative of the customer’s cost is $\widetilde{C}^{{\prime}}(\tau ) = K^{{\prime}}(\tau ) -\varGamma (\tau )$. Evaluating this at τ ₁ using (4.6), it is immediate that $\widetilde{C}^{{\prime}}(\tau _{1}) <0$. Since $\widetilde{C}(\tau )$ is convex and minimized at τ ^FB, $\widetilde{C}^{{\prime}}(\tau _{1}) <0$ implies τ ₁ < τ ^FB. Therefore, we have $\underline{\tau }<\tau _{2} <\tau _{1} <\tau ^{FB}$ and among the three candidate equilibrium outcomes (i)–(iii) above, (i) produces the lowest customer cost at τ ₁ since $\widetilde{C}(\tau )$ is decreasing for τ < τ ^FB. The optimality condition (4.2) is obtained by rearranging (4.5).

Proof (Proof of Lemma 2).

The supplier’s expected profit under PBC is $u(\tau,s) = E[T\,\vert \,\tau,s]-\psi \left (\tau,s\right ) = w-K(\tau )-\kappa /\tau -\left (c +\delta h_{b}\right )s-\delta \left (h_{g} - h_{b}\right )E[I\,\vert \,\tau,s]-vE[B\,\vert \,\tau,s]$. Differentiating this with respect to s (see Lemma 3), we get $\partial u/\partial s = v - c -\delta h_{b} -\left (v +\delta \left (h_{g} - h_{b}\right )\right )\varPhi (z)$ and $\partial ^{2}u/\partial s^{2} = -\left (v +\delta \left (h_{g} - h_{b}\right )\right )\sqrt{\tau }\phi (z) <0$, i.e., for a fixed τ, the supplier’s expected profit is concave in s. It can be shown that a finite feasible solution of $(\widetilde{\mathcal{C}})$ exists only if v > c +δ h _b (proof is omitted). With v > c +δ h _b , the profit-maximizing s is found in the interior. The first-order condition ∂ u∕∂ s = 0 yields z ^∗ = Φ ⁻¹[1 − (c +δ h _g )∕(v +δ(h _g − h _b ))], which is independent of τ. With $s^{{\ast}} = 1/\tau + z^{{\ast}}/\sqrt{\tau }$, the supplier’s expected profit becomes $\widetilde{u}(\tau ) \equiv u(\tau,s^{{\ast}}) = w - K(\tau ) - (\kappa +c +\delta h_{b})/\tau -\left (v +\delta \left (h_{g} - h_{b}\right )\right )\phi (z^{{\ast}})/\sqrt{\tau }$. Note that $\lim _{\tau \rightarrow \overline{\tau }}\widetilde{u}(\tau ) = -\infty$. Differentiating $\widetilde{u}(\tau )$, we get

$$\displaystyle\begin{array}{rcl} \widetilde{u}^{{\prime}}(\tau )& =& -K^{{\prime}}(\tau ) + \frac{\kappa +c +\delta h_{b}} {\tau ^{2}} + \frac{\left (v +\delta \left (h_{g} - h_{b}\right )\right )\phi (z^{{\ast}})} {2\tau ^{3/2}} \quad \text{and} {}\\ \widetilde{u}^{{\prime\prime}}(\tau )& =& -K^{{\prime\prime}}(\tau ) -\frac{2(\kappa +c +\delta h_{b})} {\tau ^{3}} -\frac{3\left (v +\delta \left (h_{g} - h_{b}\right )\right )\phi (z^{{\ast}})} {4\tau ^{5/2}} <0, {}\\ \end{array}$$

which shows that $\widetilde{u}(\tau )$ is concave. Evaluating $\widetilde{u}^{{\prime}}(\tau )$ at $\tau =\underline{\tau }$ and using the assumption $\underline{\tau }^{2}K^{{\prime}}(\underline{\tau }) <\kappa -\left (h_{g} - h_{b}\right )$ from Sect. 4.3.2 and the condition v > c +δ h _b in Lemma 2, we have

$$\displaystyle\begin{array}{rcl} \widetilde{u}^{{\prime}}(\underline{\tau })& =& -K^{{\prime}}(\underline{\tau }) + \frac{\kappa +c +\delta h_{b}} {\underline{\tau }^{2}} + \frac{\left (v +\delta \left (h_{g} - h_{b}\right )\right )\phi (z^{{\ast}})} {2\underline{\tau }^{3/2}} {}\\ &>& \frac{c + h_{g} -\left (1-\delta \right )h_{b}} {\underline{\tau }^{2}} + \frac{\left (c +\delta h_{g}\right )\phi (z^{{\ast}})} {2\underline{\tau }^{3/2}}> 0. {}\\ \end{array}$$

Since $\widetilde{u}(\tau )$ is a concave function that initially increases at $\tau =\underline{\tau }$ and approaches $-\infty$ as $\tau \rightarrow \overline{\tau }$, we conclude that it is maximized at $\widehat{\tau }>\underline{\tau }$ which is uniquely determined from the first-order condition $\widetilde{u}^{{\prime}}(\tau ) = 0$, as written in (4.3). To obtain sensitivity results, first note that

$$\displaystyle\begin{array}{rcl} \frac{\partial z^{{\ast}}} {\partial v} = \frac{c +\delta h_{g}} {\left (v +\delta \left (h_{g} - h_{b}\right )\right )^{2}} \frac{1} {\phi \left (z^{{\ast}}\right )}> 0.& & {}\\ \end{array}$$

Using this result and via implicit differentiation of the first-order condition $\widetilde{u}^{{\prime}}(\widehat{\tau }) = 0$, we can show that

$$\displaystyle\begin{array}{rcl} \frac{\partial \widehat{\tau }} {\partial v} = \frac{L(z^{{\ast}})} {2\widehat{\tau }^{3/2}} \left (-\widetilde{u}^{{\prime\prime}}(\widehat{\tau })\right )^{-1}> 0.& & {}\\ \end{array}$$

To show ∂ s ^∗∕∂ v > 0, we first derive two intermediate results. Observe

$$\displaystyle\begin{array}{rcl} -\widetilde{u}^{{\prime\prime}}(\widehat{\tau })& =& K^{{\prime\prime}}(\widehat{\tau }) + \frac{2(\kappa +c +\delta h_{b})} {\widehat{\tau }^{3}} + \frac{3\left (v +\delta \left (h_{g} - h_{b}\right )\right )\phi (z^{{\ast}})} {4\widehat{\tau }^{5/2}} {}\\ &>& K^{{\prime\prime}}(\widehat{\tau }) + \frac{2(\kappa +c +\delta h_{b})} {\widehat{\tau }^{3}} + \frac{3\left (c +\delta h_{g}\right )} {4\widehat{\tau }^{5/2}} z^{{\ast}} {}\\ &>& \frac{8\kappa + 5\left (c +\delta h_{g}\right )} {4\widehat{\tau }^{3}} + \frac{3\left (c +\delta h_{g}\right )} {4\widehat{\tau }^{2}} s^{{\ast}}> \frac{c +\delta h_{g}} {4\widehat{\tau }^{2}} \left (\frac{1} {\widehat{\tau }} + s^{{\ast}}\right ), {}\\ \end{array}$$

where we used

$$\displaystyle{L(z^{{\ast}}) =\phi (z^{{\ast}}) - z^{{\ast}}\overline{\varPhi }\left (z^{{\ast}}\right ) =\phi (z^{{\ast}}) - \frac{c +\delta h_{g}} {v +\delta \left (h_{g} - h_{b}\right )}z^{{\ast}}> 0}$$

in the first inequality and the assumption $2\left (h_{g} - h_{b}\right ) <\underline{\tau } ^{3}K^{{\prime\prime}}(\underline{\tau })$ from Sect. 4.3.2 in the second. Also from Lemma 4, we obtain

$$\displaystyle{-L\left (z^{{\ast}}\right ) + \left ( \frac{c +\delta h_{g}} {v +\delta \left (h_{g} - h_{b}\right )}\right )^{2} \frac{1} {\phi \left (z^{{\ast}}\right )} \geq 0.}$$

Using these two results, we can verify ∂ s ^∗∕∂ v > 0.

Proof (Proof of Proposition 3).

We first show that the backorder constraint binds at the optimum. First observe that, for $\tau ^{{\ast}} =\widehat{\tau }>\underline{\tau }$ and s ^∗ > 0 found in Lemma 2,

$$\displaystyle{\frac{\partial E[B\,\vert \,\widehat{\tau },s^{{\ast}}]} {\partial v} = \frac{\partial } {\partial v}\left (\frac{L(z^{{\ast}})} {\sqrt{\widehat{\tau }}} \right ) = -\frac{L(z^{{\ast}})} {2\widehat{\tau }^{3/2}} \frac{\partial \widehat{\tau }} {\partial v} -\frac{\overline{\varPhi }(z^{{\ast}})} {\sqrt{\widehat{\tau }}} \frac{\partial z^{{\ast}}} {\partial v} <0,}$$

since $\partial \widehat{\tau }/\partial v> 0$ and ∂ z ^∗∕∂ v > 0, as we showed in the proof of the lemma. Combined with $\lim _{v\rightarrow c+\delta h_{b}}E[B\,\vert \,\widehat{\tau },s^{{\ast}}]>\beta$ and $\lim _{v\rightarrow \infty }E[B\,\vert \,\widehat{\tau },s^{{\ast}}] = 0$, which are straightforward to show, $\partial E[B\,\vert \,\widehat{\tau },s^{{\ast}}]/\partial v <0$ implies that the feasible region for the backorder constraint $E[B\,\vert \,\widehat{\tau },s^{{\ast}}] \leq \beta$ can be expressed as $v \geq v_{\min }$, where $v_{\min }> c +\delta h_{b}$ solves $E[B\,\vert \,\widehat{\tau },s^{{\ast}}] = L(z^{{\ast}})/\sqrt{\widehat{\tau }} =\beta$. Second, differentiating the customer’s expected cost $C(\widehat{\tau },s^{{\ast}}) =\underline{ u} + h_{g}N + K(\widehat{\tau }) + (\kappa -h_{g} + h_{b})/\widehat{\tau } + (c + h_{g})s^{{\ast}}$ and substituting the supplier’s optimal response $\widehat{\tau }$ given by the first-order condition (4.3) yields

$$\displaystyle\begin{array}{rcl} \frac{\partial C} {\partial v} & =& \left (K^{{\prime}}(\widehat{\tau }) -\frac{\kappa -h_{g} + h_{b}} {\widehat{\tau }^{2}} \right ) \frac{\partial \widehat{\tau }} {\partial v} + (c + h_{g})\frac{\partial s^{{\ast}}} {\partial v} {}\\ & =& \left (\frac{c + h_{g} -\left (1-\delta \right )h_{b}} {\widehat{\tau }^{2}} + \frac{v +\delta \left (h_{g} - h_{b}\right )} {2\widehat{\tau }^{3/2}} \phi (z^{{\ast}})\right ) \frac{\partial \widehat{\tau }} {\partial v} + (c + h_{g})\frac{\partial s^{{\ast}}} {\partial v}> 0, {}\\ \end{array}$$

since $\partial \widehat{\tau }/\partial v> 0$ and ∂ s ^∗∕∂ v > 0. This monotonicity implies that the $C(\widehat{\tau },s^{{\ast}})$ is minimized at the smallest feasible value of v, i.e., it is optimal to set $v^{P} = v_{\min }$, at which the backorder constraint binds ($L(z^{{\ast}})/\sqrt{\widehat{\tau }} =\beta$). The equilibrium values v ^P and τ ^P, determined from (4.4), are obtained by combining the optimality conditions (4.3), $L(z^{{\ast}})/\sqrt{\widehat{\tau }} =\beta$, and z ^∗ = Φ ⁻¹[1 − (c +δ h _g )∕(v +δ(h _g − h _b ))].

Proof (Proof of Proposition 4).

In all three cases (FB, RBC, and PBC) the backorder constraint binds in equilibrium, i.e., $L(z)/\sqrt{\tau } =\beta$. Applying this, the customer’s expected cost $C(\tau,s) =\underline{ u} + h_{g}N + K(\tau ) + (\kappa -h_{g} + h_{b})/\tau + (c + h_{g})s$ becomes $\widetilde{C}(\tau ) =\underline{ u} + h_{g}N + K(\tau ) + (\kappa +c + h_{b})/\tau + (c + h_{g})\zeta (\tau )/\sqrt{\tau }$, which is convex and minimized at τ ^FB (see the proof of Proposition 2 for more details). Its derivative is $\widetilde{C}^{{\prime}}(\tau ) = K^{{\prime}}(\tau ) -\varGamma (\tau )$.

(i) Substituting the optimality conditions for RBC and PBC found in (4.6) and (4.4) in $\widetilde{C}^{{\prime}}(\tau )$, along with the feasibility condition c +δ h _b  < p from Lemma 1, it is easy to verify $\widetilde{C}^{{\prime}}(\tau ^{R}) <\widetilde{ C}^{{\prime}}(\tau ^{P}) \leq \widetilde{ C}^{{\prime}}(\tau ^{FB}) = 0$. This, combined with convexity of $\widetilde{C}(\tau )$, implies τ ^R < τ ^P ≤ τ ^FB. The next result follows immediately from the binding backorder constraint, as τ ^R < τ ^P ≤ τ ^FB and $\beta = L(z^{R})/\sqrt{\tau ^{R}} = L(z^{P})/\sqrt{\tau ^{P}} = L(z^{FB})/\sqrt{\tau ^{FB}}$ imply z ^R > z ^P ≥ z ^FB, and from $s^{{\ast}} = 1/\tau ^{{\ast}} + z^{{\ast}}/\sqrt{\tau ^{{\ast}}}$, we have s ^R > s ^P ≥ s ^FB. The next result C ^R > C ^P ≥ C ^FB is implied by convexity of $\widetilde{C}(\tau )$, $\widetilde{C}^{{\prime}}(\tau ^{FB}) = 0$, and τ ^R < τ ^P ≤ τ ^FB.

(ii) Implicit differentiation of the first-order condition (4.2) yields

$$\displaystyle{\frac{\partial \tau ^{R}} {\partial \delta } = \frac{(h_{g} - h_{b})\varPhi (\zeta (\tau ^{R}))} {2(\tau ^{R})^{2}\widetilde{C}^{{\prime\prime}}(\tau ^{R})} {\biggl ( - 2 + \frac{\sqrt{\tau ^{R}}\phi (\zeta (\tau ^{R}))} {\varPhi (\zeta (\tau ^{R}))} \biggr )}.}$$

In the lower limit $z \rightarrow \underline{ z} = -\frac{1} {\sqrt{\tau }}$ (see Sect. 4.5),

$$\displaystyle{\lim _{z\rightarrow \underline{z}}\frac{\phi (z)} {\varPhi (z)} =\lim _{z\rightarrow \underline{z}}\frac{-z\phi (z)} {\phi (z)} = \frac{1} {\sqrt{\tau }},}$$

where we used l’Hopital’s rule. Since ϕ(z)∕Φ(z) is decreasing, $\phi (z)/\varPhi (z) <1/\sqrt{\tau }$ for any $z \equiv \sqrt{\tau }s - 1/\sqrt{\tau }$ that corresponds to given τ and s. Hence,

$$\displaystyle{-2 + \frac{\sqrt{\tau ^{R}}\phi (\zeta (\tau ^{R}))} {\varPhi (\zeta (\tau ^{R}))} <-2 + \frac{\sqrt{\tau ^{R}}} {\sqrt{\tau ^{R}}} = -1.}$$

Together with $\widetilde{C}^{{\prime\prime}}(\tau ^{R})> 0$, this implies ∂ τ ^R∕∂ δ < 0. Then ∂ s ^R∕∂ δ > 0 immediately follows from the binding backorder constraint. ∂ C ^R∕∂ δ > 0 can be shown using the envelope theorem. Similarly, implicit differentiation of (4.4) yields

$$\displaystyle{\frac{\partial \tau ^{P}} {\partial \delta } = \frac{1} {\widetilde{C}^{{\prime\prime}}(\tau ^{P})}\left (\frac{h_{b}} {\tau ^{2}} + \frac{h_{g}} {2\tau ^{3/2}}f\left (\zeta \left (\tau ^{P}\right )\right )\right )> 0,\quad \frac{\partial s^{P}} {\partial \delta } <0,\quad \text{and}\quad \frac{\partial C^{P}} {\partial \delta } <0.}$$

(iii) In the limit $\delta \rightarrow 1$, (4.4) becomes $\varGamma (\tau ^{P}) = K^{{\prime}}(\tau ^{P})$, the same optimality condition as the first-best. Hence, $\tau ^{P} \rightarrow \tau ^{FB}$ in this limit. Then $s^{P} \rightarrow s^{FB}$ and $C^{P} \rightarrow C^{FB}$ follow. From (i) and (ii), we have τ ^R < τ ^FB and ∂ τ ^R∕∂ δ < 0, implying that the gap between τ ^R and τ ^FB is narrowed with smaller δ. In the limit $\delta \rightarrow 0$, (4.2) becomes

$$\displaystyle{\varGamma (\tau ^{R}) -\frac{c + h_{b}} {(\tau ^{R})^{2}} - \frac{c + h_{g}} {2(\tau ^{R})^{3/2}}f(\zeta (\tau ^{R})) = K^{{\prime}}(\tau ^{R}),}$$

which is not equal to the first-best condition. Therefore, τ ^R never approaches its first-best counterpart and neither do s ^R and C ^R.