# The strategic interaction between a company and the government surrounding disasters

## Abstract

We analyze the tradeoff between safety and production. The government chooses safety effort and tax rate in the first stage, and then the company strikes a balance between safety effort and production in the second stage. The government, representing the general public, earns taxes on production. Both players’ safety efforts mitigate the negative impact of a disaster. The disaster probability is modeled as a contest between the disaster magnitude and the two players’ safety efforts. Seven propositions are developed. First, as the safety effort of one player approaches infinity, the marginal change in the other player’s safety effort, with respect to the first player’s safety effort, approaches zero. Second, an infinitely large safety effort by any player causes the disaster probability and the negative impact of the disaster to decrease toward a constant. Third, as one player’s safety effort approaches infinity, the other player’s safety effort approaches zero. Fourth, the two players’ safety efforts are strategic substitutes so that an increase in one player’s safety effort decreases the other player’s safety effort. This enables the players to free ride on each other’s safety efforts. Fifth, higher tax rate causes the company to exert higher safety effort. Sixth, with maximum tax rate the company focuses exclusively on safety effort and generates no profit. Seventh, the company’s safety effort is inverse $$U$$ shaped in the disaster magnitude.

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## Notes

1. See http://www.infoplease.com/ipa/A0001451.html for major oil spills since 1967, and http://news.yahoo.com/s/ac/20110317/sc_ac/8079848_worst_oil_spills_in_history for the five worst oil spills in history. The gravity of disasters can be measured according to economic loss, human loss, and symbolic loss. The July 6, 1988 North Sea off Scotland Occidental Petroleum’s Piper Alpha rig disaster killed 167 people, http://en.wikipedia.org/wiki/Piper_Alpha#cite_note-2. The North Sea off Norway March 27, 1980 capsize of the Alexander L. Kielland platform killed 123 people, http://en.wikipedia.org/wiki/Alexander_L._Kielland_(platform), all retrieved August 8, 2014.

2. http://en.wikipedia.org/wiki/Deepwater_Horizon_oil_spill, retrieved August 8, 2014.

3. http://en.wikipedia.org/wiki/Exxon_Valdez_oil_spill, retrieved August 8, 2014.

4. See http://www.infoplease.com/ipa/A0001451.html for major oil spills since 1967, and http://news.yahoo.com/s/ac/20110317/sc_ac/8079848_worst_oil_spills_in_history for the five worst oil spills in history. The gravity of disasters can be measured according to economic loss, human loss, and symbolic loss. The July 6, 1988 North Sea off Scotland Occidental Petroleum’s Piper Alpha rig disaster killed 167 people, http://en.wikipedia.org/wiki/Piper_Alpha#cite_note-2. The North Sea off Norway March 27, 1980 capsize of the Alexander L. Kielland platform killed 123 people, http://en.wikipedia.org/wiki/Alexander_L._Kielland_(platform), all retrieved August 8, 2014.

5. http://www.coso.org/documents/COSOBoardsERM4pager-FINALRELEASEVERSION82409_001.pdf, retrieved August 8, 2014.

6. The scaling function $$F(D,S,s)$$ may be estimated by extrapolating from past data. Examples of factors impacting the scaling function are lawyer’s fees required to keep court cases going to prevent a company from being financially responsible for a disaster, public relations costs to ensure a company’s reputation when suffering from a disaster, and various costs to businesses and society of such disasters that impact the company.

7. Presenting Proposition 7 for general $$F(D,S,s)$$ is technically possible but requires a much more complicated condition than (10).

8. A set of sufficient but not necessary conditions for (10) to be satisfied are $$\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}\ge$$0, $$\frac{\partial ^{3}p(D,S,s)}{\partial S^{3}}\le$$0, $$\left| {\frac{\partial ^{3}p(D,S,s)}{\partial S^{3}}} \right| \ge \left| {\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right|$$, $$\frac{\partial ^{3}p(D,S,s)}{\partial S\partial D^{2}}\ge$$0 and $$\frac{\partial ^{3}p(D,S,s)}{\partial S^{2}\partial D} \le$$0.

## Abbreviations

$$R$$ :

Company’s resource

$$E$$ :

Company’s productive effort

$$S$$ :

Company’s safety effort

$$s$$ :

Government’s safety effort

$$A$$ :

Company’s unit cost of production

$$B$$ :

Company’s unit cost of safety effort

$$H(S)$$ :

Production function

$$b$$ :

Government’s unit cost of safety effort

$$D$$ :

Disaster magnitude

$$p(D,S,s)$$ :

Disaster probability

$$F(D,S,s)$$ :

Scaling function for how the company is negatively affected by disaster

$$f(D,S,s)$$ :

Scaling function for how the government is negatively affected by disaster

$$\tau$$ :

Taxation percentage variable chosen by the government

$$u$$ :

Government’s expected profit

$$U$$ :

Company’s expected profit

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## Author information

Authors

### Corresponding author

Correspondence to Kjell Hausken.

This research was partially supported by the United States Department of Homeland Security (DHS) through the National Center for Risk and Economic Analysis of Terrorism Events (CREATE) under award number 2010-ST-061-RE0001. This research was also partitially supported by the United States National Science Foundation (NSF) award numbers 1334930 and 1200899. However, any opinions, findings, and conclusions or recommendations in this document are those of the authors and do not necessarily reflect views of the DHS, CREATE, or NSF. Author names were listed alphabatically by last name.

## Appendix: Proof of Propositions 1–7

### Proof of Proposition 1

When $$S = 0$$, the limit $$\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{dS(s)}{ds}$$ is always zero so we only consider $$S >0$$. Differentiating the first equation in (5) gives

\begin{aligned}&\frac{d}{ds}\left( {F(D,S,s)\frac{\partial p(D,S,s)}{\partial S}} \right) =\frac{d}{ds}\left( {(1-\tau )\frac{\partial H(S)}{\partial S}-\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) \nonumber \\&\quad =-\frac{d}{ds}\left( {\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) \nonumber \\&\quad \Rightarrow \frac{d}{ds}\left( {F(D,S,s)\frac{\partial p(D,S,s)}{\partial S}} \right) +\frac{d}{ds}\left( {\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) =0\nonumber \\&\quad \Rightarrow \frac{d}{ds}({F(D,S,s)})\frac{\partial p(D,S,s)}{\partial S}+\frac{d}{ds}\left( {\frac{\partial p(D,S,s)}{\partial S}} \right) F(D,S,s)\nonumber \\&\qquad +\frac{d}{ds}\left( {\frac{\partial F(D,S,s)}{\partial S}} \right) p(D,S,s)+\frac{d}{ds}({p(D,S,s)})\frac{\partial F(D,S,s)}{\partial S}=0\nonumber \\&\quad \Rightarrow \left[ {\frac{\partial F(D,S,s)}{\partial s}+\frac{\partial F(D,S,s)}{\partial S}\frac{dS}{ds}} \right] \frac{\partial p(D,S,s)}{\partial S}\nonumber \\&\qquad +\left[ {\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}+\frac{\partial ^{2}p(D,S,s)}{\partial ^{2}S}\frac{dS}{ds}} \right] F(D,S,s)\nonumber \\&\qquad +\left[ {\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}+\frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}\frac{dS}{ds}} \right] p(D,S,s)\nonumber \\&\qquad +\left[ {\frac{\partial p(D,S,s)}{\partial s}+\frac{\partial p(D,S,s)}{\partial S}\frac{dS}{ds}} \right] \frac{\partial F(D,S,s)}{\partial S}=0\\&\quad \Rightarrow \frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial s}+\frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial S}\frac{dS}{ds}\nonumber \\&\qquad +F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}\nonumber \\&\qquad +F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}\frac{dS}{ds}+p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}\nonumber \\&\qquad +p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}\frac{dS}{ds}\nonumber \\&\qquad +\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial s}+\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial S}\frac{dS}{ds}=0\nonumber \\&\quad \Rightarrow \frac{dS}{ds}= \nonumber \\&-\frac{\frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial s}+F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}+p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}+\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial s}}{\frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial S}+F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}+p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}+\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial S}}\nonumber \end{aligned}
(11)

Since $$p$$ and $$F$$ are bounded and monotonic in s it follows that $$\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial p(D,S,s)}{\partial s}=0$$ and $$\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial F(D,S,s)}{\partial s}=0$$, and thus $$\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}=0$$ and $$\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}=0$$. Hence (11) implies $$\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{dS}{ds}=0$$. Although $$s$$ is determined before $$S$$, in an overall sense the players are interested in $$\mathop {\lim }\nolimits _{S\rightarrow \infty } \frac{ds}{dS}=0$$. In contrast to (9) we thus consider the government’s first order condition

\begin{aligned} \left\{ {{\begin{array}{l} {s>0\Leftrightarrow \frac{\partial u(S,s,\tau )}{\partial s}=-\frac{\partial f(D,S,s)}{\partial s}p(D,S,s)-f(D,S,s)\frac{\partial p(D,S,s)}{\partial s}-b=0}\\ {s=0\Leftrightarrow \frac{\partial u(S,s,\tau )}{\partial s}=-\frac{\partial f(D,S,s)}{\partial s}p(D,S,s)-f(D,S,s)\frac{\partial p(D,S,s)}{\partial s}-b<0}\\ \end{array} }} \right. \end{aligned}
(12)

which is analogous to (5) for the company. Hence analogously, $$\mathop {\lim }\nolimits _{S\rightarrow \infty } \frac{ds}{dS}=0$$. $$\square$$

### Proof of Proposition 2

Follows since $$\frac{\partial p}{\partial s}\le 0$$ ,$$\frac{\partial p}{\partial S}\le 0$$ , $$\frac{\partial f}{\partial s}\le$$0, $$\frac{\partial f}{\partial S}\le$$0, $$\frac{\partial F}{\partial s}\le$$0, $$\frac{\partial F}{\partial S}\le$$0 as stated in Assumption 1, and since $$p, f, F$$ have lower bounds 0. $$\square$$

### Proof of Proposition 3

Inserting $$\mathop {\lim }\nolimits _{S\rightarrow \infty } \frac{\partial p(D,S,s)}{\partial s}=0$$ and $$\mathop {\lim }\nolimits _{S\rightarrow \infty } \frac{\partial f(D,S,s)}{\partial s}=0$$ into (12) implies $$s=0\Leftrightarrow 0>-b$$. Inserting $$\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial p(D,S,s)}{\partial S}=0$$ and $$\mathop {\lim }\nolimits _{s\rightarrow \infty } \frac{\partial F(D,S,s)}{\partial S}=0$$ into the second equation in (5) implies $$S=0\Leftrightarrow 0>(1-\tau )\frac{\partial H(S)}{\partial S}$$. $$\square$$

### Proof of Proposition 4

When $$S = 0$$, $$\frac{dS(s)}{ds}$$ is always zero so we only consider $$S >0$$. Similar to the proof of Proposition 1, we have

\begin{aligned}&\frac{dS}{ds}=\\&-\frac{\frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial s}+F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}+p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}+\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial s}}{\frac{\partial p(D,S,s)}{\partial S}\frac{\partial F(D,S,s)}{\partial S}+F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}+p(D,S,s)\frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}+\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial S}}\nonumber \end{aligned}
(13)

Since we assume $$\frac{\partial p}{\partial S}\le 0$$,$$\frac{\partial p}{\partial s}\le 0$$, $$\frac{\partial F}{\partial S}\le 0$$, $$\frac{\partial F}{\partial s}\le 0$$, $$\frac{\partial ^{2}p}{\partial S^{2}}\ge 0$$, and $$\frac{\partial ^{2}F}{\partial S^{2}}\ge 0$$, the denominator of (13) is always positive. So we have $$\frac{dS}{ds}\le 0$$ if and only if the numerator of (13) is positive. One sufficient condition to ensure this is: $$\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}\ge 0$$ and $$\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}\ge 0$$. In summary we have shown $$\frac{dS}{ds}\le 0$$ if $$\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}\ge 0$$ and $$\frac{\partial ^{2}F(D,S,s)}{\partial S\partial s}\ge 0$$. Analogously, similar to the proof of Proposition 1 and using the first order condition (11), we can show $$\frac{ds}{dS}\le 0$$ if $$\frac{\partial ^{2}p(D,S,s)}{\partial S\partial s}\ge 0$$ and $$\frac{\partial ^{2}f(D,S,s)}{\partial S\partial s}\ge 0$$.

### Proof of Proposition 5

When $$S = 0$$, $$\frac{dS(\tau )}{d\tau }=0$$ so we only consider $$S > 0$$. Differentiating the first equation in (5) gives

\begin{aligned}&\frac{d}{d\tau }\left( {F(D,S,s)\frac{\partial p(D,S,s)}{\partial S}} \right) =\frac{d}{d\tau }\left( {(1-\tau )\frac{\partial H(S)}{\partial S}-\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) \nonumber \\&\Rightarrow \frac{\partial }{\partial S}\left( {F(D,S,s)\frac{\partial p(D,S,s)}{\partial S}} \right) \frac{dS}{d\tau }=-\frac{\partial H(S)}{\partial S}+(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}\frac{dS}{d\tau }\\&-\frac{\partial }{\partial S}\left( {\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) \frac{dS}{d\tau }\nonumber \\&\Rightarrow \frac{dS}{d\tau }=\frac{\frac{\partial H(S)}{\partial S}}{(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-\frac{\partial }{\partial S}\left( {\frac{\partial F(D,S,s)}{\partial S}p(D,S,s)} \right) -\frac{\partial }{\partial S}\left( {F(D,S,s)\frac{\partial p(D,S,s)}{\partial S}} \right) }\nonumber \\&=\frac{\frac{\partial H(S)}{\partial S}}{(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-\frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}p(D,S,s)-2\frac{\partial F(D,S,s)}{\partial S}\frac{\partial p(D,S,s)}{\partial S}-F(D,S,s)\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}}\nonumber \end{aligned}
(14)

Because of $$\frac{\partial H}{\partial S}<0, \frac{\partial F}{\partial S}\le 0,\frac{\partial ^{2}p(D,S,s)}{\partial ^{2}S}\ge 0, \frac{\partial ^{2}F(D,S,s)}{\partial S^{2}}\ge 0$$, and $$\frac{\partial ^{2}H(S)}{\partial S^{2}}\le 0$$ as stated in Assumptions 1 and 2, we have $$\frac{dS(\tau )}{d\tau }\ge 0$$. $$\square$$

### Proof of Proposition 6

Inserting tax rate $$\tau =1$$ into (2) gives $$U(S,s,\tau =1)=-F(D,S,s)p(D,S,s)$$. Since we assume $$\frac{\partial p}{\partial s}\le 0$$ and $$\frac{\partial F}{\partial S}\le$$0 in Assumption 1, S needs to be maximum to maximize $$U$$. Hence from (1) we have $$S(\tau =1,s)=\frac{R}{B}$$. $$\square$$

### Proof of Proposition 7

When $$S = 0$$, $$\frac{dS}{dD}=\frac{d^{2}S}{dD^{2}}=0$$ is always zero so we only consider $$S > 0$$. From the first equation in (5) we have

\begin{aligned} F\frac{\partial p(D,S,s)}{\partial S}&= (1-\tau )\frac{\partial H(S)}{\partial S}\Rightarrow \nonumber \\ \frac{d}{dD}\left( {F\frac{\partial p(D,S,s)}{\partial S}} \right)&= F\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}+F\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}\frac{dS}{dD}=\frac{d}{dD}\left( {(1-\tau )\frac{\partial H(S)}{\partial S}} \right) \nonumber \\&= (1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}\frac{dS}{dD}\nonumber \\ \Rightarrow \frac{dS}{dD}&= \frac{F\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}}{(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-F\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \end{aligned}
(15)

where $$H(S)$$ does not depend on $$D$$ according to (1). Since we assume $$\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}\ge 0$$ and $$\frac{\partial ^{2}H(S)}{\partial S^{2}}\le 0$$ in Assumption 2 and $$F\ge 0$$ in (2), we have $$\frac{dS}{dD}\ge 0$$ if and only if $$\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}\le 0$$.

From (15) we have

\begin{aligned} \begin{array}{l} \frac{d^{2}S}{dD^{2}}=\frac{d}{dD}\left( {\frac{F\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}}{(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-F\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}}} \right) \\ =\frac{\left( {F\frac{\partial ^{3}p(D,S,s)}{\partial S\partial D^{2}}+F\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}\frac{dS}{dD}} \right) \left( {(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-F\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right) +\left( {F\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}} \right) \left( {(1-\tau )\frac{\partial ^{3}H(S)}{\partial S^{3}}\frac{dS}{dD}-F\frac{\partial ^{3}p(D,S,s)}{\partial S^{3}}\frac{dS}{dD}-F\frac{\partial ^{3}p(D,S,s)}{\partial S^{2}\partial D}} \right) }{\left( {(1-\tau )\frac{\partial ^{2}H(S)}{\partial S^{2}}-F\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right) ^{2}}\\ \end{array}\quad \end{aligned}
(16)

For the special case $$\frac{\partial ^{2}H(S)}{\partial S^{2}}=0$$ (linear production), (15) and (16) become

\begin{aligned} \frac{dS}{dD}&= -\frac{\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}}{\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}}\end{aligned}
(17)
\begin{aligned} \frac{d^{2}S}{dD^{2}}&= -\frac{\left( {\frac{\partial ^{3}p(D,S,s)}{\partial S\partial D^{2}}+\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}\frac{dS}{dD}} \right) \left( {\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right) +\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}} \right) \left( {\frac{\partial ^{3}p(D,S,s)}{\partial S^{3}}\frac{dS}{dD}+\frac{\partial ^{3}p(D,S,s)}{\partial S^{2}\partial D}} \right) }{\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right) ^{2}} \end{aligned}
(18)

Substituting (17) into (18), we have

\begin{aligned} \frac{d^{2}S}{dD^{2}}=-\frac{\frac{\partial ^{3}p(D,S,s)}{\partial S\partial D^{2}}\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}-\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}} \right) ^{2}-\frac{\partial ^{3}p(D,S,s)}{\partial S^{3}}\frac{\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}} \right) ^{2}}{\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}}+\frac{\partial ^{3}p(D,S,s)}{\partial S^{2}\partial D}\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S\partial D}} \right) }{\left( {\frac{\partial ^{2}p(D,S,s)}{\partial S^{2}}} \right) ^{2}} \end{aligned}
(19)

So we have $$\frac{d^{2}S}{dD^{2}}\le 0$$ if and only if (10) is satisfied. For the second part, when $$D$$ goes to infinity, we must have $$\frac{dS}{dD}\le 0$$. Since $$S$$ must be non-negative, the limit exists. $$\square$$

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Hausken, K., Zhuang, J. The strategic interaction between a company and the government surrounding disasters. Ann Oper Res 237, 27–40 (2016). https://doi.org/10.1007/s10479-014-1684-5