An inventory modeling analysis of hydrogen fueling station capacity considering an outside option

Abstract

A number of issues have to be addressed in order for a hydrogen economy to become a reality, including the need to expand hydrogen infrastructure to accommodate vehicular needs. The design of a hydrogen station is a complex problem that entails the determination of numerous operating parameters including the overall hydrogen fueling station capacity. In this paper, a newsvendor formulation is selected to form the basis of capacity determination, in particular, the classical formulation with an outside option that reflects the added value that a consumer might give to hydrogen. This study demonstrates the behavior of optimal capacity decisions for hydrogen stations related to demand distribution, consumer behavior, and economics of hydrogen fuel. Moreover, through assessing the consequences of adding extra capacity for stimulating demand, this paper suggests opportunities for additional profits. The results from our study highlight the effect that consumer beliefs have on demand and optimal station size.

Appendix

1.1 Proof of Lemma 1:

The capital investment, $f(C)$, is a function of the capacity, $C$. We define the first order derivative of $f$ with respect to $C$, $df/dC$, by $f^\prime$. According to Eq. (3), $C$ is equal to the product of the normalized capacity, $s$, and the cumulative distribution function, $O(\hat{u})$; that is, $C=sO(\hat{u})$. Therefore, $f(C)$ is a function of $s$ and $\hat{u}$; moreover, $f_s\triangleq \frac{\partial f}{\partial s}=\frac{d f}{dC}\frac{\partial C}{\partial s}= f^\prime O(\hat{u})$, $f_O\triangleq \frac{\partial f}{\partial O} =\frac{d f}{dC} \frac{\partial C}{\partial O}=f^\prime s$, and $f_{\hat{u}}\triangleq \frac{\partial f}{\partial {\hat{u}}} = \frac{df}{dC}\frac{\partial C}{\partial O}\frac{dO(\hat{u})}{du} = f^\prime s o(\hat{u})$ where $o(u)$ is the probability density function of the outside option’s utility, $u$. Therefore, $f_{\hat{u}}/o(\hat{u})=f^\prime s = f_O$ and $sf_s=O(\hat{u})f_O=f^\prime sO(\hat{u})=f^\prime C$. $\square$

1.2 Proof of Lemma 2:

Equation (6) shows that the neutral utility, $\hat{u}$, is equal to zero if the customers’ utility of purchasing hydrogen, $v-p$, is equal to zero. Under this condition, $\hat{u}$ is not affected by the choice of standardized capacity, $s$. Moreover, Eq. (5) indicates that the service rate, $r(s)$, is not affected by the change in $s$ if $s$ is outside the range of $[x^-, x^+]$. Therefore, if $v-p>0$ and $x^-\le s\le x^+$, the choice of capacity, $s$, can influence the service rate, $r(s)$, which in turn impacts the neutral utility, $\hat{u}$. $\square$

1.3 Proof of Proposition 1:

When $s\ge x^+$, $\frac{\partial E[\pi ]}{\partial s}=-O(\hat{u})(c_O+h-l+A^\prime )<0$ because $c_O>l$, $h>0$ and $A^\prime \ge 0$. Therefore, if there exists a value of $s$, less than $x^+$, such that $p-c_O-(p+h-l)F(s)-A^\prime >0$, then $\frac{\partial E[\pi ]}{\partial s}>0$ at that $s$ value. $A^\prime$ is continuous at any value of $s$ under Assumption 1; therefore, $\frac{\partial E[\pi ]}{\partial s}=O(\hat{u})\{(p-c_O)-(p+h-l)F(s)-A^\prime \}$ is also continuous at any value of $s$. This indicates that there must exist $\tilde{s}$ that satisfies $\tilde{s}<x^+$ and $\frac{\partial E[\pi ]}{\partial s}\left|_{s=\tilde{s}}\right.=0.$

Assume $s_1$ and $s_2$ are two different values of $s$ on $[x^-, x^+]$ and they both solve $\frac{\partial E[\pi ]}{\partial s}=0$. Without loss of generality, we assume $s_1<s_2$. Let $g(s,\hat{u})=p-c_O-(p+h-l)F(s)-A^\prime (s,\hat{u})$, then $g(s_1,\hat{u}_1)=g(s_2,\hat{u}_2)=0$. $A^\prime (s_1,\hat{u}_1)>A^\prime (s_2,\hat{u}_2)$ because $F(s_1)<F(s_2)$, which contradicts to $\frac{dA^\prime }{ds}=\frac{dC}{ds}A^{\prime \prime }\ge 0$. Therefore, the solution to $\frac{\partial E[\pi ]}{\partial s}=0$ must be unique. $\square$

1.4 Proof of Proposition 2:

When $s=\tilde{s}$, $\frac{dE[\pi ]}{ds}=\frac{\tilde{o}}{\tilde{O}^2}\frac{v-p}{\bar{s}}[1-\tilde{F}][E[\tilde{\pi }]-(\tilde{A}^\prime \tilde{C}-\tilde{A})]>0$ according to Lemma 3. When $s\ge x^+$, $\frac{dE[\pi ]}{ds}=-O(\hat{u})(c_O+h-l+A^\prime )<0$. $\frac{dE[\pi ]}{ds}$ is continuous at any value of $s$. Therefore, there must exist $s^*$ that satisfies $\tilde{s}\le s^*\le x^+$ and $\frac{dE[\pi ]}{ds}\left|_{s=s^*}\right.=0.$

We take the second order derivative of the expected profit to $s$ at $s^*$:

$$\begin{aligned} \left.\frac{d^2E[\pi (s, \hat{u})]}{ds^2}\right|_{s=s^*}\!=\! O(\hat{u}) \left.\left\{ \begin{array}{lllll}&-(p+h-l)f(s)\\&+\frac{d(o(\hat{u})/O(\hat{u})^2)}{d\hat{u}}\frac{(v-p)^2}{\bar{x}^2}[1\!-\!F(s)]^2(E[\pi ]\!+\!A\!-\!CA^\prime )\\&-\frac{o(\hat{u})}{O(\hat{u})^2}\frac{v-p}{\bar{x}}f(s)(E[\pi ]+A-CA^\prime )\\&-\frac{o(\hat{u})}{O(\hat{u})^2}\frac{v-p}{\bar{x}}[1-F(s)]\left(C\frac{dA^{\prime }}{ds}\right)\\&-\frac{dA^\prime }{ds}\\ \end{array} \right\} \right| \begin{array}{llllll}&\\&\\&\\&\\&\\&_{s=s^*} \end{array} \end{aligned}$$

The first term in the bracket is nonpositive since $p+h-l>0$ and $f(s)\le 0$ for any value of $s$. Lemma 4 has shown that $E[\pi ^*]+A^*-C{A^\prime }^*$ is nonnegative; therefore the second term in the bracket is nonpositive if $\frac{d(o(\hat{u})/O(\hat{u})^2)}{d\hat{u}}\le 0$. Let $R(\hat{u})=o(\hat{u})/O(\hat{u})$, then $\frac{d(o(\hat{u})/O(\hat{u})^2)}{d\hat{u}} = \frac{d(R/O)}{d\hat{u}} = \frac{R^\prime O-Ro}{O^2}=(R^\prime -R^2)/O$. $\frac{d(o(\hat{u})/O(\hat{u})^2)}{d\hat{u}}\le 0$ if $R^\prime \le R^2$. Most probability distributions support $R^\prime \le R^2$. Therefore, we can say that the second term in the bracket is also nonpositive. The third term is nonpositive according to Lemma 4. The last two terms in the bracket are nonpositive if $A$ is convex since $\frac{dA^{\prime }}{ds}=\frac{dC}{ds}A^{\prime \prime }\ge 0$. The convexity of $A$ ensures that $\frac{d^2E[\pi (s, \hat{u})]}{ds^2}\left|_{s=s^*}\right.\le 0$, which suggests that the global optimum occurs at $s^*$. $\square$