amp;nbsp;⟶ **shortage** incurs *opportunity* cost $C_u$ | $(1-\color{red}{q})\color{red}{q}amp;nbsp;⟶ **match** gains value $+V$ | | $\color{skyblue}{D}=0$ (no customer) | $\color{red}{q}(1-\color{red}{q})amp;nbsp;⟶ **match** (no cost) | $(1-\color{red}{q})^{2}amp;nbsp;⟶ **excess** incurs *overage* cost $C_o$ | $L(q)=C_uq^{2}+C_o(1-q)^{2}-Vq(1-q).$ #### A.1.2 Optimal Quality $ q^{*}=\frac{V+2C_o}{2\,(C_u+C_o+V)}, $ ### Appendix A.1.2 Asymmetric Commitments |Symbol|Interpretation| |---|---| |$q \in [0,1]$|product quality (decision)| |$\beta_r, \beta_c$|resource partner and customer responsiveness parameters| |$P_c(q) = \beta_c \cdot q$|customer commitment probability| |$P_r(q) = 1 - \beta_r \cdot q$|resource partner commitment probability| |$C_u, C_o$|under‑ / over‑mismatch unit costs| |$V$|match bonus when both commit| #### A.1.2.1 Expected Loss The 2×2 outcome matrix becomes: | | $\color{skyblue}{Q}=0$ (partner absent) | $\color{skyblue}{Q}=1$ (partner present) | |---|---|---| | $\color{skyblue}{D}=1$ (customer shows) | $(\beta_c q)^2$ ⟶ **shortage** incurs _opportunity_ cost $C_u$ | $(\beta_c q)(1-\beta_r q)$ ⟶ **match** gains value $+V$ | | $\color{skyblue}{D}=0$ (no customer) | $(1-\beta_c q)(\beta_r q)$ ⟶ **no match** (no cost) | $(1-\beta_c q)(1-\beta_r q)$ ⟶ **excess** incurs _overage_ cost $C_o$ | $L(q,\beta_r,\beta_c) = C_u(\beta_c q)^2 + C_o(1-\beta_c q)(1-\beta_r q) - V(\beta_c q)(1-\beta_r q)$ #### A.1.2.2 Expanded Form $L(q,\beta_r,\beta_c) = C_u\beta_c^2 q^2 + C_o(1-\beta_c q-\beta_r q+\beta_r\beta_c q^2) - V\beta_c q(1-\beta_r q)$ $= q^2(C_u\beta_c^2 + C_o\beta_r\beta_c + V\beta_r\beta_c) + q(-C_o\beta_c - C_o\beta_r - V\beta_c) + C_o$ #### A.1.2.3 Optimal Quality (Primal Problem) Given $\beta_r, \beta_c$, minimize with respect to $q$: $\frac{\partial L}{\partial q} = 2q(C_u\beta_c^2 + C_o\beta_r\beta_c + V\beta_r\beta_c) + (-C_o\beta_c - C_o\beta_r - V\beta_c) = 0$ $q^* = \frac{C_o(\beta_c + \beta_r) + V\beta_c}{2(C_u\beta_c^2 + C_o\beta_r\beta_c + V\beta_r\beta_c)}$ #### A.1.2.4 Optimal Responsiveness (Dual Problem) Given $q$, the first-order conditions for $\beta_r, \beta_c$ are: $\frac{\partial L}{\partial \beta_r} = C_o q(1-\beta_c q) + V\beta_c q^2 = 0$ $\frac{\partial L}{\partial \beta_c} = 2C_u\beta_c q^2 - C_o q(1-\beta_r q) - Vq(1-\beta_r q) = 0$ #### A.1.2.5 Verification When $\beta_r = \beta_c = 1$, this reduces to the symmetric case: $L(q,1,1) = C_u q^2 + C_o(1-q)^2 - Vq(1-q)$ --- ## Appendix A.2 Quality-Non-Linear Commitments ### Appendix A.2.1 Non-Linear Symmetric Commitments | Symbol | Meaning | | ----------------------------- | ------------------------------- | | $q \in \mathbb{R}$ | quality level (decision) | | $P_c(q) = \frac{1}{1+e^{-q}}$ | customer commitment probability | | $P_r(q) = \frac{1}{1+e^{q}}$ | partner commitment probability | | $C_u, C_o$ | under-/over-mismatch unit costs | | $V$ | match bonus when both commit | #### A.2.1.1 Expected Loss $L(q) = C_u P_c(q)[1-P_r(q)] + C_o P_r(q)[1-P_c(q)] - V P_c(q)P_r(q)$ #### A.2.1.2 Optimal Quality The first-order condition $L'(q) = 0$ yields: $C_u P_c(1-P_c)(1-P_r) - C_u P_c P_r(1-P_r) + C_o P_r(1-P_r)(1-P_c) - C_o P_r P_c(1-P_c) - V P_c(1-P_c)P_r - V P_c P_r(1-P_r) = 0$ Factoring and simplifying with $P_c = \frac{e^{-q}}{1+e^{-q}}$ and $P_r = \frac{1}{1+e^{-q}}$: $P_c(V + 2C_o) = P_r(V + 2C_u)$ Substituting the sigmoid forms and solving: $q^* = \ln\left(\frac{2C_o + V}{2C_u + V}\right)$ This closed-form solution exists due to the symmetric responsiveness parameters ($\beta_c = 1, \beta_r = -1$). ---- ### Appendix A.2.2 General Non-Linear with Asymmtric $β$ Parameters | Symbol | Meaning | |---------------------------------------------|---------------------------------------------------| | $q\in\mathbb R$ | quality level (decision) | | $P_c(q)=\dfrac1{1+e^{-β_c q}}$ | customer commitment probability | | $P_r(q)=\dfrac1{1+e^{+β_r q}}$ | partner commitment probability | | $β_c,β_r>0$ | steepness of each sigmoid | | $C_u,C_o,V>0$ | under-commit cost; over-commit cost; match bonus | #### A.2.2.1 Expected Loss We define $ L(q) = C_u\,P_c(q)\bigl[1 - P_r(q)\bigr] + C_o\,P_r(q)\bigl[1 - P_c(q)\bigr] - V\,P_c(q)\,P_r(q). $ #### A.2.2.2 First-Order Condition 1. Compute the derivatives $ P_c'(q) = β_c\,P_c(1-P_c), \quad P_r'(q) = -β_r\,P_r(1-P_r). $ 2. Differentiate $L(q)$ term-by-term: $ \frac{d}{dq}\bigl[C_uP_c(1-P_r)\bigr] = C_u\bigl[P_c'(1-P_r) - P_c\,P_r'\bigr], $ $ \frac{d}{dq}\bigl[C_oP_r(1-P_c)\bigr] = C_o\bigl[P_r'(1-P_c) - P_r\,P_c'\bigr], $ $ \frac{d}{dq}\bigl[-V\,P_cP_r\bigr] = -V\bigl[P_c'P_r + P_c\,P_r'\bigr]. $ 3. Assemble $L'(q)$: $ L'(q) = P_c'\Bigl[C_u(1-P_r) - (C_o+V)P_r\Bigr] + P_r'\Bigl[C_o(1-P_c) - (C_u+V)P_c\Bigr]. $ 4. Substitute $P_c',P_r'$ and set $L'(q)=0$: $ β_cP_c(1-P_c)\bigl[C_u(1-P_r) - (C_o+V)P_r\bigr] -β_rP_r(1-P_r)\bigl[(C_u+V)P_c - C_o(1-P_c)\bigr] =0. $ 5. Rearranged into the implicit equation $ \boxed{ \frac{β_c\,P_c(1-P_c)}{β_r\,P_r(1-P_r)} = \frac{\,C_o(1-P_c)\;-\;(C_u+V)P_c\,} {\,C_u(1-P_r)\;-\;(C_o+V)P_r\,} }. $ #### A.2.2.3 Special Limits — Closed-Form Solutions We now derive the four tractable cases step by step. --- ##### 1. Symmetric Responsiveness ($β_c=β_r=β$) - Let $x = e^{βq}$. Then $ P_c = \frac{x}{1+x},\quad P_r = \frac{1}{1+x}. $ - Implicit FOC becomes $ (1-P_c)(C_u+V) \;=\; P_c\,(C_o+V). $ - Substitute $1-P_c = 1/(1+x)$ and $P_c=x/(1+x)$: $ \frac{1}{1+x}(C_u+V) = \frac{x}{1+x}(C_o+V) \;\Longrightarrow\; x = \frac{C_u+V}{C_o+V}. $ - Hence $ \boxed{ q^* = \frac{1}{β}\, \ln\!\Bigl(\frac{C_u+V}{C_o+V}\Bigr). } $ --- ##### 2. Customer-Dominant Responsiveness ($β_c\gg β_r$) from $\boxed{\; \beta_c P_c(1-P_c)\,\bigl[\,C_u(1-P_r)-(C_o+V)P_r\bigr] = \beta_r P_r(1-P_r)\,\bigl[(C_u+V)P_c-C_o(1-P_c)\bigr] \; } \tag{FOC}$ 1. Divide the FOC by $\beta_r$. As $\beta_c/\beta_r\to\infty$, the left-hand factor $\beta_cP_c(1-P_c)$ explodes **unless** its square-bracket term is driven to 0. Hence to leading order $C_u(1-P_r)-(C_o+V)P_r = 0.$ 2. Solve for $P_r$: $P_r^\star = \frac{C_u}{C_u+C_o+V},\qquad \frac{1-P_r^\star}{P_r^\star}=\frac{C_o+V}{C_u}.$ 3. Invert the (decreasing) logistic for $P_r$: $P_r(q)=\frac{1}{1+e^{\beta_r q}} \;\Longrightarrow\; q^\star =\frac{1}{\beta_r}\, \ln\left(\frac{C_o+V}{C_u}\right).$ $\boxed{\; q^{*}_{\text{cust-dom}} =\frac{1}{\beta_r}\, \ln\left(\frac{C_o+V}{C_u}\right)\;}$ --- ##### 3. Partner-Dominant Responsiveness ($β_r\gg β_c$) - By symmetry, swap roles: $ C_o(1-P_c) - (C_u+V)P_c = 0 \;\Longrightarrow\; P_c = \frac{C_o}{C_u+C_o+V}. $ - With $P_c = 1/(1+e^{-β_c q})$, we get $ e^{-β_c q} = \frac{1-P_c}{P_c} = \frac{C_u+V}{C_o} \;\Longrightarrow\; q^* = \frac{1}{β_c}\, \ln\!\Bigl(\frac{C_o}{C_u+V}\Bigr). $ - Hence $ \boxed{ q^* = \frac{1}{β_c}\, \ln\!\Bigl(\frac{C_o}{C_u+V}\Bigr). } $ --- ##### 4. High-Match-Value Limit ($V\gg C_u,C_o$) - Approximate $L(q)\approx -V\,P_cP_r$, so FOC becomes $ β_c P_c(1-P_c) = β_r P_r(1-P_r). $ - Solving this logistic marginal equality yields $ \boxed{ q^* = \frac{1}{β_c + β_r}\, \ln\!\Bigl(\frac{β_r}{β_c}\Bigr). } $