Static and dynamic resource allocation models for recovery of interdependent systems: application to the Deepwater Horizon oil spill

Abstract

Determining where and when to invest resources during and after a disruption can challenge policy makers and homeland security officials. Two decision models, one static and one dynamic, are proposed to determine the optimal resource allocation to facilitate the recovery of impacted industries after a disruption where the objective is to minimize the production losses due to the disruption. The paper presents necessary conditions for optimality for the static model and develops an algorithm that finds every possible solution that satisfies those necessary conditions. A deterministic branch-and-bound algorithm solves the dynamic model and relies on a convex relaxation of the dynamic optimization problem. Both models are applied to the Deepwater Horizon oil spill, which adversely impacted several industries in the Gulf region, such as fishing, tourism, real estate, and oil and gas. Results demonstrate the importance of allocating enough resources to stop the oil spill and clean up the oil, which reduces the economic loss across all industries. These models can be applied to different homeland security and disaster response situations to help governments and organizations decide among different resource allocation strategies during and after a disruption.

Appendix

1.1 Proof of Proposition 1

Given a set of $z_i>0$ and $z_i=0$, we seek to prove that (6) has at most three solutions. First, we need to prove the following lemma.

Lemma 1

Given a function $f\left( x\right)$ that is continuously differentiable for $x>0$, if $f'\left( x\right) =0$ has at most two solutions for $x >0$ and $\alpha >0$, then $\mathrm{exp }\left( -\alpha x^2\right) f\left( x\right) =0$ has at most three solutions for $x>0$.

Proof

Assume the function $f\left( x\right)$ is continuously differentiable for $x>0, f'\left( x\right) =0$ has at most two solutions, and $\alpha >0$. If $\mathrm{exp }\left( -\alpha x^2\right) f\left( x\right) =0$ has more than three solutions for $x>0$, then $f\left( x\right) =0$ has more than three solutions for $x>0$ because $\mathrm{exp }\left( -\alpha x^2\right) >0$ for all $x$.

If $f\left( x\right) =0$ has more than three solutions, then $f\left( x\right)$ has at least three local extrema, which means that $f'\left( x\right) =0$ has at least three solutions. This contradicts the premise that $f'\left( x\right) =0$ has at most two solutions. Thus, $\mathrm{exp }\left( -\alpha x^2\right) f\left( x\right) =0$ has at most three solutions for $x>0$. $\square$

If we can show that the first derivative as depicted in (8) has at most two values for $z_0>0$ where the first derivative equals zero, then Lemma 1 applies because (7) is continuously differentiable for $z_0>0$.

The second derivative of (7) with respect to $z_0$ is given in (11).

$$\begin{aligned} F\text {exp}\left( \dfrac{z_0-Z}{\sum _{i:z_i>0}\left. 1\big /k_i\right. }\right) \left[ \left( \dfrac{1}{\sum _{i:z_i>0}\left. 1\big /k_i\right. }\right) ^2-\dfrac{2k_0z_0}{\sum _{i:z_i>0}\left. 1\big /k_i\right. }-4k_0\right] \end{aligned}$$

(11)

Equation (11) has one solution for $z_0$ if (11) equals zero. This implies that the first derivative in (8) has one extreme point and at most two solutions for $z_0$ if (8) equals 0. Because no more than two zeros exist for the first derivative in (8), we can conclude from Lemma 1 that (6) has at most three solutions for $0<z_0<Z$. $\square$

1.2 Proof of Proposition 2

First, we show that if (6) has three real solutions for $z_0$, then each of the three conditions must hold. If (6) has three solutions, then (7) has three solutions because $\mathrm{exp }\left( -k_0z_0^2\right) \ne 0$. Because the second derivative in (11) has exactly one solution for $z_0$ when (11) equals 0, the first derivative in (8) has one local extreme point, which means that at most two solutions exist for $z_0$ when (8) equals 0. This implies that (7) has at most two extreme points. If exactly three solutions exist for $z_0$ when (7) equals 0, then (7) has a local minimum at $z_0=z_-$ where (7) is less than 0 and a local maximum at $z_0=z_+$ where (7) is greater than 0.

Condition 1. The expression in (7) is greater than 0 if $z_0=0$. If (7) has three solutions, then (7) must be less than or equal to 0 when $z_0=Z$. If (7) is less than or equal to 0 when $z_0=Z$, then $F\left( 1-2k_0Z\sum _{i:z_i>0}\left. 1\big /k_i\right. \right) -2Gk_0Z\le 0$. Then it must be true that $Z\ge \left. F\big /\left( 2Gk_0+2Fk_0\sum _{i:z_i>0}\left. 1\big /k_i\right. \right. \right)$, which proves condition 1.

Condition 2. A local minimum exists at $z_0=z_{-}$, and a local maximum exists at $z_0=z_+$. Since (7) is greater than 0 when $z_0=0$ and (7) is less than 0 when $z_0=Z$, it must be true that $z_-<z_+$. The first derivative in (8) equals 0 when $z_0=z_-$ and $z_0=z_+$, which implies that $2Gk_0=F\text {exp}\left( \frac{z_0-Z}{\sum _{i:z_i>0}\left. 1\big /k_i\right. }\right) \left( \frac{1}{\sum _{i:z_i>0}\left. 1\big /k_i\right. }-2k_0z_0-2k_0\sum _{i:z_i>0}\frac{1}{k_i}\right)$ at those two points. Substituting this expression for $2Gk_0$ into (7) leads to the expression in (12).

$$\begin{aligned} F\text {exp}\left( \dfrac{z_0-Z}{\sum _{i:z_i>0}\left. 1\big /k_i\right. }\right) \left( 2k_0z_0^2-\dfrac{z_0}{\sum _{i:z_i>0}\left. 1\big /k_i\right. }+1\right) \end{aligned}$$

(12)

If (12) equals 0, then $z_0=\left. \left( \frac{1}{\sum _{i:z_i>0}\left. 1\big /k_i\right. }\pm \sqrt{\frac{1}{\left( \sum _{i:z_i>0}\left. 1\big /k_i\right. \right) ^2}-8k_0}\right) \big /\left( 4k_0\right) \right.$, which corresponds to $z^*$ and $z^{**}$ as defined in Proposition 2. The expression in (12) is greater than 0 when $z_0<z^*$ and $z_0>z^{**}$ and less than 0 when $z^*<z_0<z^{**}$. If no solution exists when (12) equals 0, then that would imply that (7) is greater than 0 when $z_0=z_-$ because (12) would always be greater than 0. If that were true, then three solutions would not exist for $z_0$ when (7) equals 0.

Because (7) is less than 0 when $z_0=z_-$, then it must be true that $z^*<z_-$. Because (7) is decreasing for $z_0<z_-$, the first derivative in (8) is less than 0 when $z_0=z^*$, which proves condition 2.

Condition 3. Because (7) is greater than 0 when $z_0=z_+$, it must be true that $z^{**}<z_+<Z$. Because (7) is increasing for $z_-<z_0<z_+$, (8) is greater than 0 when $z_0=z_+$, which proves condition 3.

Second, we show that if the three conditions hold, then (6) has three real solutions for $z_0$. Assume the three conditions hold. Because (8) is less than 0 when $z_0=z^*$, (7) is decreasing over some range. Because (8) is greater than 0 when $z_0=z^{**}$, then (7) is increasing over some range. Since $z^*<z^{**}$, there must be a local minimum at some point $z_0=z_-$ where $z^*<z_-<z^{**}$. The expression in (12) is less than 0 for $z^*<z_0<z^{**}$, which implies that (7) is less than 0 at the local minimum $z_0=z_-$. The expression in (7) must equal 0 for some point $0<z_0<z_-$.

Since (8) is greater than 0 when $z_0=z^{**}$, $2Gk_0<F\text {exp}\left( \frac{z^{**}-Z}{\sum _{i:z_i>0}\left. 1\big /k_i\right. }\right) \left( \frac{1}{\sum _{i:z_i>0}\left. 1\big /k_i\right. }-2k_0z^{**}-2k_0\sum _{i:z_i>0}\dfrac{1}{k_i}\right)$. This means that (7) is greater than (12) when $z_0=z^{**}$. Because (12) equals 0 when $z_0=z^{**}$, (7) is greater than 0. The expression in (7) must equal 0 for some point $z_-<z_0<z^{**}$.

If $Z\ge \frac{F}{2Gk_0+2Fk_0\sum _{i:z_i>0}\left. 1\big /k_i\right. }$ then (7) is less than or equal to 0 when $z_0=Z$. Because $z^{**}<Z$ and (7) is greater than 0 when $z_0=z^{**}$, (7) must equal 0 for some point $z^{**}<z_0<Z$.

This proves there are at least three solutions for $z_0$ when (7) equals 0, which means that (6) has at least three solutions for $z_0$. From Proposition 1, no more than three solutions exist, and we conclude that (6) has three solutions for $z_0$. $\square$