Analyzing information-enabled stockout management under vendor-managed inventory

Abstract

We develop a mechanism under vendor-managed inventory (VMI) by which a manufacturer provides an incentive contract to a retailer to convert lost sales stockouts into backorders. An incentive contract is required since the retailer’s efforts are not directly observable. We first show that when there are no limits on order quantities or inventory levels imposed on the manufacturer, the manufacturer will push inventory onto the retailer. The manufacturer minimizes the possibility for lost sales stockouts by maintaining high inventory levels at the retailer rather than by paying incentives to the retailer. However, modern information systems (IS), such as radio frequency identification (RFID), allow the retailer to monitor inventory at its premises and to enforce limits on order quantities. With strict limits on order quantities, the manufacturer will provide incentives to the retailer to convert lost sales stockouts to backorders. We analyze the conditions under which these incentive payments are likely to be highest.

Appendix

1.1 Proof of Lemma 1

Apply a uniform distribution U(0, 1) to the demand and the linear contract transfer, and restate the manufacturer and retailer profit functions:

$$ E_x \pi _M =-\frac{l}{2}(1-\theta)q^2+\left[ {w+l(1-\theta)} \right]q-\frac{l}{2}(1-\theta)-\beta \theta $$

(A-1)

$$ E_x \pi _R =-\frac{1}{2}pq^2+(p-w)q-e^2+\beta \theta $$

(A-2)

Solve for the optimal effort:

$$ \frac{\partial \pi _R }{\partial e}=-2e+\beta $$

(A-3)

$$ \frac{\partial ^2\pi _R }{\partial e^2}=-2 $$

(A-4)

The negative second order condition (i.e., −2) indicates a maximum. Set first order condition (i.e., A-3) equal to 0, we obtain the optimal effort:

$$ e^\ast =\frac{1}{2}\beta $$

(A-5)

Insert (A-5) into (A-2), set it to 0, and solve for w, we have:

$$ wq=pq-\frac{1}{2}pq^2+\frac{1}{4}\beta ^2 $$

(A-6)

Insert (A-5) and (A-6) into (A-1), the manufacturer’s profit function becomes:

$$ \mathop {\max }\limits_{\beta,q} E_{\theta,x} \pi _M =-\frac{l}{2}(1-\frac{1}{2}\beta )q^2+pq-\frac{1}{2}pq^2+\frac{1}{4}\beta ^2+l\left( {1-\frac{1}{2}\beta } \right)q-\frac{l}{2}\left({1-\frac{1}{2}\beta } \right)-\frac{1}{2}\beta ^2 $$

(A-7)

We can obtain the first order conditions:

$$ \frac{\partial \pi _M }{\partial \beta }=\frac{l}{4}(1-q)^2-\frac{1}{2}\beta $$

(A-8)

$$ \frac{\partial \pi _M }{\partial q}=l\left({1-\frac{1}{2}\beta } \right)(1-q)+p(1-q) $$

(A-9)

Setting the first order conditions equal to 0, we obtain the optimal solutions: β ^*₁ = 0, q ^*₁ = 1, and $w_1^\ast =\frac{p}{2}$.

Checking the Hessian matrix for the second order conditions, we conclude the it is negative definite at the point of optimal solution, indicating the optimal solutions are the maximum. □

1.2 Proof of Lemma 2

IR1 is no longer binding but IR2 is binding, q ^*₂ = q ₀ and $w_2^\ast =w_1^\ast =\frac{p}{2}$, because the manufacturer always has incentives to increase q and w. By inserting IR2 and (A-5) into (A-1), the manufacturer’s profit function becomes:

$$ \mathop {\max }\limits_{\beta,q} E_{\theta,x} \pi _M =-\frac{l}{2}(1-\frac{1}{2}\beta)q_0 ^2+\left[ {w+l\left( {1-\frac{1}{2}\beta } \right)} \right]q_0 -\frac{l}{2}\left( {1-\frac{1}{2}\beta } \right)-\frac{1}{2}\beta ^2 $$

(A-10)

The first order conditions are:

$$ \frac{\partial \pi _M }{\partial \beta }=\frac{l}{4}q_0 ^2-\frac{l}{2}q_0 +\frac{l}{4}-\beta $$

(A-11)

Therefore, the optimal solutions are: $\beta _2^\ast =\frac{l(1-q_0)^2}{4}, q_2^\ast =q_0$, and $w_2^\ast =\frac{p}{2}$. □

1.3 Proof of Proposition 1a & 1b

$\frac{\partial \beta _2^\ast}{\partial l}=\frac{(1-q_0)^2}{4} > 0,$ and $\frac{\partial \beta _2^\ast }{\partial q_0 }=\frac{-l(1-q_0 )}{2} < 0$. □

1.4 Proof of Proposition 2a

Since q ₀ < 1, $\beta _2^\ast =\frac{l(1-q_0)^2}{2} > \beta _1^\ast =0$. □

1.5 Proof of Proposition 2b

$$ \pi _{R,1}^\ast =-\frac{1}{2}p+\left(p-\frac{p}{2}\right)=-\frac{1}{2}p+\frac{1}{2}p=0 $$

$$ \pi _{R,2}^\ast =\frac{p}{2}(q_0 -q_0 ^2)+\frac{l^2(1-q_0)^4}{64} $$

Since q ₀ < 1, q ₀ > q ₀ ², we have $\pi _{R,2}^\ast > \pi _{R,1}^\ast$. □

1.6 Proof of Proposition 2c

$$ \pi _{M,1}^\ast =\frac{p}{2} $$

$$ \pi _{M,2}^\ast =\frac{l^2(1-q_0)^4}{32}-\frac{l}{2}(q_0 -1)^2+\frac{p}{2}q_0 $$

$$ \pi _{M,2}^\ast -\pi _{M,1}^\ast =\frac{l^2(1-q_0 )^4}{32}-\frac{l}{2}(q_0 -1)^2+\frac{p}{2}q_0 -\frac{p}{2} $$

Hence, we can show that when $l < \frac{4}{(1-q_0)^2}\left({2+\sqrt {4+(1-q_0)p} } \right), \pi _{M,2}^\ast -\pi _{M,1}^\ast < 0$ (i.e., $\pi _{M,2}^\ast < \pi _{M,1}^\ast)$ will be true, and when $l\ge \frac{4}{(1-q_0 )^2}\left({2+\sqrt {4+(1-q_0)p} } \right)$, $\pi _{M,2}^\ast -\pi _{M,1}^\ast \ge 0$(i.e., $\pi _{M,2}^\ast \ge \pi _{M,1}^\ast)$ will also be true. □

1.7 Proof of Proposition 3

Since $\pi _{R,2}^\ast =\frac{p}{2}(q_0 -q_0 ^2)+\frac{l^2(1-q_0 )^4}{64}$, we can take first and second partial derivative of the retailer’s optimal profit with respect to the order limit quantity q ₀:

$$ \frac{\partial \pi _{R,2}^\ast }{\partial q_0 }=\frac{p}{2}(1-2q_0 )-\frac{l^2(1-q_0)^3}{16} $$

$$ \frac{\partial ^2\pi _{R,2}^\ast }{\partial q_0 ^2}=-p+\frac{3l^2(1-q_0)^2}{16} $$

In order to have a maximum, the second partial derivative needs to be negative. By setting it to be less than zero, we obtain the sufficient condition under which there will be an optimal order limit quantity $q_{0: }q_0 > 1-\frac{4\sqrt p }{l\sqrt 3 }$.

Setting the first derivative to 0, we obtain the optimal order limit quantity q ₀ after excluding two complex roots as follows:

$$ q_0^\ast =1+\frac{8\left({\frac{2}{3}} \right)^{1/3}p}{\left( {9l^4p+\sqrt 3 \sqrt {27l^8p^2-256l^6p^3} } \right)^{1/3}}+\frac{\left({\frac{2}{3}} \right)^{2/3}\left( {9l^4p+\sqrt 3 \sqrt {27l^8p^2-256l^6p^3} } \right)^{1/3}}{l^2} $$

□