A multi-objective optimization primary planning model for a POE (Port-of-Entry) inspection

Abstract

Every year, more than 11 million maritime containers and 11 million commercial trucks arrive in the United States, carrying all types of imported goods. As it would be costly to inspect every container, only a fraction of them are inspected before being admitted into the United States. In this paper, a multi-objective optimization (MOO) model is developed for primary allocation of the scarce inspection resources at a POE, especially a land POE (L-POE). This model minimizes the different costs associated with the inspection process, including those associated with delaying the entry of legitimate imports. The resulting model is exploited in two different ways: One is to construct the efficient frontier of the MOO model or a partial efficient frontier with diversity of the solutions maximized; the other is to evaluate a given inspection plan and provide possible suggestions for improvement. The methodologies are described in detail and a simple case study is provided.

Appendices

Appendix A: Variance derivation

Let X be the random variable representing the service time of an entity in the system, then we have:

$$ {\mu_g} = 1/E\left[ X \right] $$

$$ \sigma_g^2 = Var(X) $$

Define an indication variable Y:

$$ Y = j{,}\;{\text{if}}\;{\text{inspection}}\;{\text{routine}}\;j\;{\text{is}}\;{\text{applied}}\;{\text{to}}\;{\text{the}}\;{\text{entity,}} $$

then Pr{Y = j} = p _j , thus

$$ \sum\limits_{{j = 0}}^M {{p_j}} = 1 $$

(A.1)

It is easy to derive that:

$$ E\left[ {X\left| Y \right. = j} \right] = 1/{\mu_j}, $$

and thus

$$ \matrix{ {E\left[ X \right]} \hfill & = \hfill &{E\left[ {E\left[ {X\left| Y \right.} \right]} \right]} \hfill \\ {} \hfill & = \hfill &{\sum\limits_{{j = 0}}^M {{p_j}/{\mu_j}} } \hfill \\ }<!end array> . $$

To calculate Var(X), the following formula is used:

$$ Var\left[ X \right] = E\left[ {Var\left( {X\left| Y \right.} \right)} \right] + Var\left( {E\left[ {X\left| Y \right.} \right]} \right). $$

(A.2)

Assume that the service times of each routine are independent from each other. E[Var(X|Y)] and Var(E[X|Y]) are computed separately.

Since $Var\left( {X|Y = j} \right) = \sigma_j^2$ and Pr{Y = j} = p _j ,

$$ E\left[ {Var\left( {X|Y} \right)} \right] = \sum\limits_{{j = 0}}^M {{p_j}\sigma_j^2} . $$

(A.3)

$$ Var\left( {E\left[ {X\left| Y \right.} \right]} \right) = E\left[ {{{\left( {E\left[ {X\left| Y \right.} \right] - E\left[ {E\left[ {X\left| Y \right.} \right]} \right]} \right)}^2}} \right] $$

Since

$$ E\left[ {E\left[ {X\left| Y \right.} \right]} \right] = \sum\limits_{{j = 0}}^M {\left( {{p_j}/{\mu_j}} \right)}, $$

$$ Var\left( {E\left[ {X\left| Y \right.} \right]} \right) = \sum\limits_{{j = 0}}^M {{p_j}{{\left( {1/{\mu_j} - \sum\limits_{{i = 0}}^M {\left( {{p_i}/{\mu_i}} \right)} } \right)}^2}} . $$

For clarity reason, the index in the inner summation is changed from j to i.

Thus,

$$ \begin{array}{*{20}{c}} \hfill {Var(X) = \sum\limits_{{j = 0}}^{M} {{{p}_{j}}\sigma _{j}^{2}} + \sum\limits_{{j = 0}}^{M} {{{p}_{j}}{{{\left( {1/{{\mu }_{j}} - \sum\limits_{{i = 0}}^{M} {\left( {{{p}_{i}}/{{\mu }_{i}}} \right)} } \right)}}^{2}}} } \\ \hfill { = \sum\limits_{{j = 0}}^{M} {{{p}_{j}}\sigma _{j}^{2}} + \sum\limits_{{j = 0}}^{M} {{{p}_{j}}{{{\left( {1/{{\mu }_{j}}} \right)}}^{2}}} + \sum\limits_{{j = 0}}^{M} {{{p}_{j}}{{{\left( {\sum\limits_{{i = 0}}^{M} {\left( {{{p}_{i}}/{{\mu }_{i}}} \right)} } \right)}}^{2}}} - 2\sum\limits_{{j = 0}}^{M} {\left( {\left( {{{p}_{j}}/{{\mu }_{j}}} \right)\sum\limits_{{i = 0}}^{M} {\left( {{{p}_{i}}/{{\mu }_{i}}} \right)} } \right)} } \\ \hfill { = \sum\limits_{{j = 0}}^{M} {{{p}_{j}}\sigma _{j}^{2}} + \sum\limits_{{j = 0}}^{M} {{{p}_{j}}{{{\left( {1/{{\mu }_{j}}} \right)}}^{2}}} - {{{\left( {\sum\limits_{{j = 0}}^{M} {\left( {{{p}_{j}}/{{\mu }_{j}}} \right)} } \right)}}^{2}}} \\ \end{array} $$

(A.4)

The last step is valid because:

• 1. $\sum\limits_{{j = 0}}^M {{p_j}{{\left( {\sum\limits_{{i = 0}}^M {\left( {{p_i}/{\mu_i}} \right)} } \right)}^2} = } {\left( {\sum\limits_{{i = 0}}^M {\left( {{p_i}/{\mu_i}} \right)} } \right)^2}$, due to (A.1);

• 2. $\sum\limits_{{j = 0}}^M {\left( {\left( {{p_j}/{\mu_j}} \right)\sum\limits_{{i = 0}}^M {\left( {{p_i}/{\mu_i}} \right)} } \right)} = {\left( {\sum\limits_{{j = 0}}^M {\left( {{p_j}/{\mu_j}} \right)} } \right)^2}$

Thus,

$$ \sum\limits_{{j = 0}}^M {{p_j}{{\left( {\sum\limits_{{i = 0}}^M {\left( {{p_i}/{\mu_i}} \right)} } \right)}^2}} - 2\sum\limits_{{j = 0}}^M {\left( {\left( {{p_j}/{\mu_j}} \right)\sum\limits_{{i = 0}}^M {\left( {{p_i}/{\mu_i}} \right)} } \right)} = - {\left( {\sum\limits_{{j = 0}}^M {\left( {{p_j}/{\mu_j}} \right)} } \right)^2} $$

Note that for conformity, indices are converted from i back to j.

Since we assume that 1/μ ₀ = 0 and $\sigma_0^2 = 0$, Var(X) can also be rewritten as follows:

$$ Var(X) = \sum\limits_{{j = 1}}^M {{p_j}\sigma_j^2} + \sum\limits_{{j = 1}}^M {{p_j}{{\left( {1/{\mu_j}} \right)}^2}} - {\left( {\sum\limits_{{j = 1}}^M {\left( {{p_j}/{\mu_j}} \right)} } \right)^2} $$

Appendix B: Validation of function separation

Let f(x,y) be a function of x and y in the form of

$$ f\left( {x,y} \right) = \frac{{ax}}{{1 - by}}, $$

(B.1)

where a and b are positive real constants, x,y are non-negative real numbers and y < 1/b.

It can be shown that

$$ \mathop{{\min }}\limits_{{x,y}} f\left( {x,y} \right) $$

(B.2)

is equivalent to

$$ \min x\;{\text{and}}\;\min y\;{\text{simultaneously}}. $$

(B.3)

The partial derivatives can be derived as following.

$$ \frac{{\partial f}}{{\partial x}} = \frac{1}{{1 - by}} > 0. $$

(B.4)

This is because y < 1/b, and thus 1 − by > 0.

$$ \frac{{\partial f}}{{\partial y}} = \frac{{abx}}{{{{\left( {1 - by} \right)}^2}}} > 0, $$

(B.5)

since a,b and x are non-negative.

Thus, f(x,y) decreases monotonically as x or y decreases within the domain in which the function is defined. Thus, minimizing f(x,y) is equivalent to minimizing x and y simultaneously.