Computational Approach for the Firm’s Cost Minimization Problem Using the Selective Infimal Convolution Operator

Abstract

The Infimal Convolution operator is well known in the context of convex analysis. This operator admits a very precise micro-economic interpretation: if several production units produce the same output, the Infimal Convolution of their cost functions represents the joint cost function distributing the production among all of them in the most efficient possible way. The drawback of this operator is that it does not discriminate whether one of some of the production units is not profitable (in the sense that it would be preferable to do without it).This is the motivating idea for the present work, in which we introduce a new operator: the Selective Infimal Convolution. We give not just its definition and basic properties but also an algorithm for its exact computation. Using this, we avoid the combinatorial blowing-up of other classical methods used for solving similar problems. Even more, our approach solves a one-parameter family of problems, not just a single one. We provide an application to the Firm’s Cost Minimization Problem, one of the most important problems in Microeconomics.

Appendix: Proof of the Formula for the SIC (Quadratic Case)

Proposition 5Let$F_{i}(x_{i})=\alpha _{i}+\beta _{i}x_{i}+\gamma _{i}x_{i}^{2}$with domains$[m_{i},M_{i}]$and$\gamma _{i}>0$$\,(i=1,2)$. Let us assume that$F_{1}^{\prime }(m_{1})\le F_{2}^{\prime }(m_{2})$. Define

$$\begin{aligned} l_{1}=\dfrac{(-\beta _{1}+\beta _{2}+2\gamma _{2}m_{2})}{2\gamma _{1}} ;\;l_{2}=\dfrac{(\beta _{1}-\beta _{2}+2\gamma _{1}M_{1})}{2\gamma _{2}} ;\;l_{3}=\dfrac{(-\beta _{1}+\beta _{2}+2\gamma _{2}M_{2})}{2\gamma _{1}} \end{aligned}$$

(40)

and

$$\begin{aligned} F_{12}(\xi )=\alpha _{1}+\alpha _{2}-\frac{(\beta _{1}-\beta _{2})^{2}}{ 4(\gamma _{1}+\gamma _{2})}+\frac{\gamma _{2}\beta _{1}+\gamma _{1}\beta _{2} }{\gamma _{1}+\gamma _{2}}\xi +\frac{\gamma _{1}\gamma _{2}}{\gamma _{1}+\gamma _{2}}\xi ^{2}. \end{aligned}$$

(41)

Then

(A) If $F_{1}^{\prime }(m_{1})\le F_{2}^{\prime }(m_{2})\le F_{1}^{\prime }(M_{1})\le F_{2}^{\prime }(M_{2})$, then:

$$\begin{aligned} (F_{1}\bigodot F_{2}) (\xi ):=\left\{ \begin{array}{lll} F_{1}(\xi -m_{2})+F_{2}(m_{2}) &{} \quad \text { if} &{} \xi \in [m_{1}+m_{2},m_{2}+l_{1}] \\ F_{12}(\xi ) &{} \quad \text { if} &{} \xi \in [m_{2}+l_{1},M_{1}+l_{2}] \\ F_{2}(\xi -M_{1})+F_{1}(M_{1}) &{} \quad \text { if} &{} \xi \in [M_{1}+l_{2},M_{1}+M_{2}] \end{array} \right. \end{aligned}$$

(42)

(B) If $F_{1}^{\prime }(m_{1})\le F_{2}^{\prime }(m_{2})\le F_{2}^{\prime }(M_{2})\le F_{1}^{\prime }(M_{1})$, then:

$$\begin{aligned} (F_{1}\bigodot F_{2}) (\xi ):=\left\{ \begin{array}{lll} F_{1}(\xi -m_{2})+F_{2}(m_{2}) &{} \quad \text { if} &{} \xi \in [m_{1}+m_{2},m_{2}+l_{1}] \\ F_{12}(\xi ) &{} \quad \text { if} &{} \xi \in [m_{2}+l_{1},M_{2}+l_{3}] \\ F_{1}(\xi -M_{2})+F_{2}(M_{2}) &{}\quad \text { if} &{} \xi \in [M_{2}+l_{3},M_{1}+M_{2}] \end{array} \right. \end{aligned}$$

(43)

(C) If $F_{1}^{\prime }(m_{1})\le F_{1}^{\prime }(M_{1})\le F_{2}^{\prime }(m_{2})\le F_{2}^{\prime }(M_{2})$, then:

$$\begin{aligned} (F_{1}\bigodot F_{2}) (\xi ):=\left\{ \begin{array}{lll} F_{1}(\xi -m_{2})+F_{2}(m_{2}) &{} \quad \text { if} &{} \xi \in [m_{1}+m_{2},M_{1}+m_{2}] \\ F_{1}(M_{1})+F_{2}(\xi -M_{1}) &{} \quad \text { if} &{} \xi \in [M_{1}+m_{2},M_{1}+M_{2}] \end{array} \right. \end{aligned}$$

(44)

Proof

The case for n quadratic functions has been studied in Bayón et al. (2010). In this paper we only deal with the case $n=2$.

• (A) Let $(x_{\xi },y_{\xi })$ be the minimum of $F_{1}(x)+F_{2}(y)$ subject to $x+y=\xi$, with $m_{1}\le x_{\xi }\le M_{1}$ and $m_{2}\le y_{\xi }\le M_{2}$.

We first show that the following holds:

• (i) If $F_{1}^{\prime }(m_{1})<F_{2}^{\prime }(m_{2})$ or $(F_{1}^{\prime }(m_{1})=F_{2}^{\prime }(m_{2}))$ then $y_{\xi }>m_{2}\Rightarrow x_{\xi }>m_{1}\$ (or $x_{\xi }=m_{1}\Rightarrow y_{\xi }=m_{2}$).

• (ii) If $F_{2}^{\prime }(m_{2})<F_{1}^{\prime }(M_{1})$ or $(F_{2}^{\prime }(m_{2})=F_{1}^{\prime }(M_{1}))$ then $x_{\xi }=M_{1}\Rightarrow y_{\xi }>m_{2}$ (or $y_{\xi }=m_{2}\Rightarrow x_{\xi }<M_{1}$).

• (iii) If $F_{1}^{\prime }(M_{1})<F_{2}^{\prime }(M_{2})$ or $(F_{1}^{\prime }(M_{1})=F_{2}^{\prime }(M_{2}))$ then $x_{\xi }<M_{1}\Rightarrow y_{\xi }<M_{2}$ (or $y_{\xi }=M_{2}\Rightarrow x_{\xi }=M_{1}$).

We prove just the case (i), the other two follow from a similar reasoning.

(i) Let$F_{1}^{\prime }(m_{1})\le F_{2}^{\prime }(m_{2})$. Assuming that $x_{\xi }=m_{1}$ and $y_{\xi }>m_{2}$ leads to a contradiction. Consider the function:

$$\begin{aligned} \varPhi (\varepsilon )=F_{1}(x_{\xi }+\varepsilon )+F_{2}(y_{\xi }-\varepsilon ) \end{aligned}$$

(45)

Hence $\varPhi ^{\prime }(0)=F_{1}^{\prime }(m_{1})-F_{2}^{\prime }(y_{\xi })<F_{1}^{\prime }(m_{1})-F_{2}^{\prime }(m_{2})\le 0$, which contradicts the minimal nature of $(x_{\xi },y_{\xi }).$

Notice that (i) guarantees that the minimum cannot be obtained for $x_{\xi }=m_{1}$ and $m_{2}<y_{\xi }\le M_{2}$; (ii) guarantees that the minimum cannot be obtained for $y_{\xi }=m_{2}$ and $x_{\xi }=M_{1}$, and finally (iii) guarantees that the minimum cannot be obtained for $y_{\xi }=M_{2}$ and $m_{1}\le x_{\xi }<M_{1}.$

Thus, we have the following possibilities:

• If $y_{\xi }=m_{2}$ and $m_{1}\le x_{\xi }<M_{1}$ then $F_{1}^{\prime }(x_{\xi })\le F_{2}^{\prime }(m_{2})$. As $F_{1}^{\prime }$ is increasing, there must exist some $l_{1}\ge x_{\xi }$ with $F_{1}^{\prime }(l_{1})=F_{2}^{\prime }(m_{2}),$ that is,

  $$\begin{aligned} l_{1}=\dfrac{(-\beta _{1}+\beta _{2}+2\gamma _{2}m_{2})}{2\gamma _{1}} \end{aligned}$$

  (46)

  such that $y_{\xi }=m_{2}$ and $x_{\xi }=\xi -m_{2}\in \left[ m_{1},l_{1} \right]$, from which $\xi \in [m_{1}+m_{2},l_{1}+m_{2}]$ and certainly, in this interval, $(F_{1}\bigodot F_{2})\,(\xi )=F_{1}(\xi -m_{2})+F_{2}(m_{2})$.

• If $x_{\xi }=M_{1}$ and $m_{2}<y_{\xi }\le M_{2}$ then $F_{1}^{\prime }(M_{1})\le F_{2}^{\prime }(y_{\xi })$. As $F_{2}^{\prime }$ is increasing, there must exist some $l_{2}\le y_{\xi }$ with $F_{1}^{\prime }(M_{1})=F_{2}^{\prime }(l_{2}),$ that is,

  $$\begin{aligned} l_{2}=\dfrac{(\beta _{1}-\beta _{2}+2\gamma _{1}M_{1})}{2\gamma _{2}} \end{aligned}$$

  (47)

  such that $x_{\xi }=M_{1}$ and $y_{\xi }=\xi -M_{1}\in \left[ l_{2},M_{2} \right]$, from which $\xi \in [M_{1}+l_{2},M_{1}+M_{2}]$ and certainly, in this interval, $(F_{1}\bigodot F_{2})\,(\xi )=F_{1}(M_{1})+F_{2}(\xi -M_{1})$.

• If $m_{1}<x_{\xi }<M_{1}$ and $m_{2}<y_{\xi }<M_{2}$ then $F_{1}^{\prime }(x_{\xi })=F_{2}^{\prime }(y_{\xi })$. It is clear that, in this case, $\xi \in [l_{1}+m_{2},M_{1}+l_{2}]$ and

  $$\begin{aligned} \underset{y}{\min }\{F_{1}(\xi -y)+F_{2}(y)\}=\alpha _{1}+\alpha _{2}-\frac{ (\beta _{1}-\beta _{2})^{2}}{4(\gamma _{1}+\gamma _{2})}+\frac{\gamma _{2}\beta _{1}+\gamma _{1}\beta _{2}}{\gamma _{1}+\gamma _{2}}\xi +\frac{ \gamma _{1}\gamma _{2}}{\gamma _{1}+\gamma _{2}}\xi ^{2}\nonumber \\ \end{aligned}$$

  (48)

which function we denote $F_{12}(\xi )$.

(B) and (C) are proved using a similar reasoning. $\square$