Pricing storage of outbound containers in container terminals

Abstract

In container yard of container terminals, a storage charge is imposed to encourage customers to limit the space required for their containers. This study addresses the storage price scheduling problem for the storage of outbound containers. The price schedule consists of the free-time limit, which is the maximum duration for a container to stay in the yard without any charge, and storage charge per day for storing a container past the free-time-limit. Empirical data suggests that the efficiency of loading operations significantly depends on the space utilized by a vessel’s outbound containers. Mathematical models are developed to determine the optimal storage price schedule in such a manner that the terminal’s total profit is maximized or the total system’s cost is minimized. Both single and multi-vessel cases are considered. Properties of the optimal solution are derived from the mathematical models and numerical experiments are conducted to validate solutions.

Appendices

Appendix 1: Proof of Property 1

We will show that for a given value of $T_{S}$, $E\left( {PF(F_{1} ,S_{1} )} \right) > E\left( {PF(0,S_{0} )} \right)$ for an $F_{1} > 0$, where $S_{1} = s_{0} + c_{0} /(T_{S} - F)$ and $S_{0} = s_{0} + c_{0} /T_{S}$, which are obtained from Eq. (5). The objective function $E(PF(F,S)) = E(PF(F)) = \int_{F}^{{T_{S} }} {\left( {s_{0} + \frac{{c_{0} }}{{T_{S} - F}}} \right)(x - F)f(x)dx} - c_{t} a\frac{u}{v}\left\{ {\int_{0}^{{T_{S} }} {xf(x)dx} + f\int_{{T_{S} }}^{\infty } {f(x)dx} } \right\}$.

$\frac{dE(PF(F))}{dF} = \int_{F}^{{T_{S} }} {\frac{{c_{0} }}{{(T_{S} - F)^{2} }}(x - F)f(x)dx} - \int_{F}^{{T_{S} }} {\left( {s_{0} + \frac{{c_{0} }}{{T_{S} - F}}} \right)f(x)dx - } c_{t} a\frac{u}{v}\int_{{T_{S} }}^{\infty } {f(x)dx}$. Note that $\int_{F}^{{T_{S} }} {\frac{{c_{0} }}{{(T_{S} - F)^{2} }}(x - F)f(x)dx \le } \int_{F}^{{T_{S} }} {\left( {\frac{{c_{0} }}{{T_{S} - F}}} \right)f(x)dx}$ because $0 \le \frac{x - F}{{T_{S} - F}} \le 1$ for $F \le x \le T_{S}$. Thus, $\frac{dE(PF(F))}{dF} \le - \int_{F}^{{T_{S} }} {s_{0} f(x)dx - } c_{t} a\frac{u}{v}\int_{{T_{S} }}^{\infty } {f(x)dx} \le 0$.

Hence, all these terms are non-positive. Namely, E(PF(F)) is a non-increasing function of F. Thus, F* = 0.

Appendix 2: Proof of Property 2

Let $f(x) = 1/b, \quad 0 \le x \le b$, which is a uniform distribution. Then, the objective function becomes $E\left({PF(S)} \right) = S\int_{0}^{{\hbox{min} \left\{{\frac{{c_{0}}}{{S - s_{0}}},\,b} \right\}}} {x\frac{1}{b}dx} - c_{t} a\frac{u}{v}\int_{0}^{{\hbox{min} \left\{{\frac{{c_{0}}}{{S - s_{0}}},\,b} \right\}}} {x\frac{1}{b}dx}$.

Case 1

$c_{0}/(S - s_{0}) \le b$ , which can be rewritten as $S \ge c_{0}/b + s_{0}$.

$$E\left({PF(S)} \right) = \frac{1}{2b}\left({S - \frac{{ac_{t} u}}{v}} \right)\left({\frac{{c_{0}}}{{S - s_{0}}}} \right)^{2}$$

$$\frac{{dE\left( {PF(S)} \right)}}{dS} = \frac{1}{2b}\left\{ {\left( {\frac{{c_{0} }}{{S - s_{0} }}} \right)^{2} - 2\left( {S - \frac{{ac_{t} u}}{v}} \right)\frac{{c_{0}^{2} }}{{(S - s_{0} )^{3} }}} \right\} = \frac{{c_{0}^{2} }}{{2b(S - s_{0} )^{3} }}\left( { - S - s_{0} + \frac{{2ac_{t} u}}{v}} \right)$$

Because $S \ge c_{0}/b + s_{0}$, which will be denoted as S₁, ${{c_{0}^{2}} \mathord{\left/{\vphantom {{c_{0}^{2}} {\left({S - s_{0}} \right)^{3}}}} \right. \kern-0pt} {\left({S - s_{0}} \right)^{3}}} \ge 0$. Thus, if $S \ge - s_{0} + \frac{{2ac_{t} u}}{v}$, which we denote as S₂, then $\frac{{dE\left({PF(S)} \right)}}{dS} \le 0$. Namely, E(PF(S)) is a decreasing function for $S \ge S_{2}$, and an increasing function for $S \le S_{2}$.

(a) If $S_{{^{1}}} < S_{{^{2}}}$, then the maximum objective value is obtained at S = S ₂. (b) If $S_{1} \ge S_{2}$, then in the range of $S \ge S_{1}$, E(PF(S)) decreases as the value of S increases. Thus, the maximum objective value is obtained at S = S ₁. From constraint (6) and $\frac{u}{b}\int_{0}^{{\frac{{c_{0}}}{{S - s_{0}}}}} {xdx} = \frac{u}{2b}\left({\frac{{c_{0}}}{{S - s_{0}}}} \right)^{2} = \frac{{uc_{0}^{2}}}{{2b(S - s_{0})^{2}}}$, we obtain $\frac{{uc_{0}^{2}}}{{2b(S - s_{0})^{2}}} \le v$, which is equivalent to $S \le s_{0} - c_{0} \sqrt {u/(2bv)}$ or $S \ge s_{0} + c_{0} \sqrt {u/(2bv)}$. However, because $S \ge s_{0} , (c) S \ge s_{0} + c_{0} \sqrt {u/(2bv)}$, which will be denoted as S₃. From (a), (b), and (c), (d) the maximum objective value is obtained at S = max {S₁, S₂, and S₃}.

Case 2

$S < c_{0}/b + s_{0}$.

In this case, $E\left({PF(S)} \right) = S\int_{0}^{b} {x\frac{1}{b}dx} - c_{t} a\frac{u}{v}\int_{0}^{b} {x\frac{1}{b}dx} = \frac{b}{2}\left({S - \frac{{ac_{t} u}}{v}} \right)$. Namely, $E\left({PF(S)} \right)$ is an increasing function of S. Now, consider the constraint in (6). Following the analysis in case 1, we know that $S \ge S_{2}$. If $S_{3} \le S_{1}$, then clearly E(PF(S)) is maximized at S₁, If S₃ > S₁, then there is no feasible solution. Thus, in case 2, if there is a feasible solution, then E(PF(S)) is maximized at S = S₁.

Considering cases 1 and 2 simultaneously, E(PF(S)) is maximized at S = max {S₁, S₂, and S₃}.

Appendix 3: Proof of Property 3

Suppose that $f(x) = 1/b,\;0 \le x \le b$, Then, the objective function can be expressed as

$$E\left({TC(F,S)} \right) = \left({c_{t} + \frac{{c_{v}}}{g}} \right)\frac{{au\int_{0}^{{T_{S}}} {C_{R} (y)dy}}}{v} + \int_{{T_{S}}}^{\infty} {\{c_{0} + s_{0} (x - F)\} f(x)dx}= \left({c_{t} + \frac{{c_{v}}}{g}} \right)\frac{au}{v}\left({\int_{0}^{{\hbox{min} \left\{{\frac{{c_{0}}}{{S - s_{0}}} + F,b} \right\}}} {\frac{x}{b}dx} + F\int_{{\hbox{min} \left\{{\frac{{c_{0}}}{{S - s_{0}}} + F,b} \right\}}}^{b} {\frac{1}{b}dx}} \right) + \int_{{\hbox{min} \left\{{\frac{{c_{0}}}{{S - s_{0}}} + F,b} \right\}}}^{b} {(c_{0} + s_{0} (x - F))\frac{1}{b}dx}$$

Case 1

${{c_{0}} \mathord{\left/{\vphantom {{c_{0}} {(S -}}} \right. \kern-0pt} {(S -}}s_{0}) + F \ge b$. Then, the objective function becomes

$$E\left({(TC(F,S)} \right) = \left({c_{t} + \frac{{c_{v}}}{g}} \right)\frac{au}{v}\left({\int_{0}^{b} {\frac{x}{b}dx} + F\int_{b}^{b} {\frac{1}{b}dx}} \right) + \int_{b}^{b} {\left\{{c_{0} + s_{0} (x - F)} \right\}\frac{1}{b}dx} = \frac{abu}{2v}\left({c_{t} + \frac{{c_{v}}}{g}} \right)$$

which remains constant for all values of F and S.

Case 2

${{c_{0}} \mathord{\left/{\vphantom {{c_{0}} {(S -}}} \right. \kern-0pt} {(S -}}s_{0}) + F < b$ the objective function becomes

$$E\left({TC(F,S)} \right) = \left({c_{t} + \frac{{c_{v}}}{g}} \right)\frac{au}{v}\left({\int_{0}^{{\frac{{c_{0}}}{{S - s_{0}}} + F}} {\frac{x}{b}dx} + F\int_{{\frac{{c_{0}}}{{S - s_{0}}} + F}}^{b} {\frac{1}{b}dx}} \right) + \int_{{\frac{{c_{0}}}{{S - s_{0}}} + F}}^{b} {\{c_{0} + s_{0} (x - F)\} \frac{1}{b}dx}$$

$$= \frac{au}{2bv}(c_{t} + \frac{{c_{v}}}{g})\left\{{\frac{{c_{0}^{2}}}{{(S - s_{0})^{2}}} - F^{2} + 2Fb} \right\} - \frac{{s_{0} c_{0}^{2}}}{{2b(S - s_{0})^{2}}} - \frac{{c_{0}^{2}}}{{b(S - s_{0})}} + \frac{{s_{0} F^{2}}}{2b} - \frac{{c_{0} F}}{b} + \frac{{s_{0} b}}{2} + c_{0} - Fs_{0}.$$

$$\frac{{\partial E\left({TC(S,F)} \right)}}{\partial F} = F\left\{{\frac{{s_{0}}}{b} - \frac{au}{bv}\left({c_{t} + \frac{{c_{v}}}{g}} \right)} \right\} + \frac{au}{v}\left({c_{t} + \frac{{c_{v}}}{g}} \right) - \frac{{c_{0}}}{b} - s_{0}.$$

From $\frac{{\partial E\left({TC(S,F)} \right)}}{\partial F} = 0$, $F = \frac{{vg(c_{0} + bs_{0}) - abu(gc_{t} + c_{v})}}{{vgs_{0} - au(gc_{t} + c_{v})}}$. Let this value of F be denoted as F₁. Note that the first term of E(TC(F,S)), $\left({c_{t} + \frac{{c_{v}}}{g}} \right)\frac{{au\int_{0}^{{T_{S}}} {C_{R} (y)dy}}}{v}$, represents the cost at the PCT and $\int_{0}^{{T_{s}}} {C_{R} (y)dy}$ indicates the expected DTY of a container at the terminal. Thus, $\frac{au}{v}\left({c_{t} + \frac{{c_{v}}}{g}} \right)$ represents the additional cost for a container to stay at the terminal one more unit time. In addition, s₀ represents the additional storage cost for a container to stay at an ODCY one more unit time. If s₀ is greater than $\frac{au}{v}\left({c_{t} + \frac{{c_{v}}}{g}} \right)$, then no container may have to be stored at an ODCY (consider that c₀ needs to be additionally paid for the storage at an ODCY). Thus, $s_{0} < \frac{au}{v}\left({c_{t} + \frac{{c_{v}}}{g}} \right)$. Thus, for a given value of S, E(TC(S,F)) monotonically increases when F < F₁ and monotonically decreases when F ≥ F₁. Thus, F* = 0 or b. Note that $\min_{S} E\left({TC(S,b)} \right) = E\left({TC(0,0)} \right)$, because in both cases, no container visits an ODCY. However, in general, $E\left({TC(0,0)} \right) \ge \min_{S} E\left({TC(0,S)} \right)$. Thus, F* = 0.

Next, $\frac{{\partial E\left({TC(S,F)} \right)}}{\partial S} = \frac{{c_{0}^{2}}}{{b(S - s_{0})^{3}}}\left\{{S - \frac{au}{v}\left({c_{t} + \frac{{c_{v}}}{g}} \right)} \right\}.$

Considering S > s₀, E(TC(S,F)) decreases until S reaches $\frac{au}{v}\left({c_{t} + \frac{{c_{v}}}{g}} \right)$ and then increases. Note also that this function is valid in the range satisfying $\frac{{c_{0}}}{{S - s_{0}}} + F(= 0) \le b$, which is equivalent to $S_{1} = c_{0}/b + s_{0}$. Let, $S_{1} = c_{0}/b + s_{0}$ and $S_{4} = au\left({c_{t} + c_{v}/g} \right)/v$. From constraint (6), we obtain $\frac{{uc_{0}^{2}}}{{2b(S - s_{0})^{2}}} \le v$, which can be converted to $S \ge s_{0} + c_{0} \sqrt {\frac{u}{2bv}}$. Let $S_{3} = s_{0} + c_{0} \sqrt {u/(2bv)}$. Therefore, TC(S,F) is minimized at $S = Max\{S_{1},\,\,S_{3},S_{4} \}$. If $S^{*} = S_{1}$, then E(TC(S*,0)) is the same as the objective value of case 1, which is a constant for all the values of (S,F) and thus all the values of (S,F) satisfying $c_{0}/(S - s_{0}) + F \ge b$ become the optimal solutions.