Satisficing is to accept the first-seen option that met certain thresholds, known as aspirations. Assume options are presented as vectors $$\vec{x}=(x^1,x^2,\cdots,x^k)$$, and the aspiration level as $$\vec{\theta}=(\theta^1,\theta^2,\cdots,\theta^k)$$. There is an indicator function for accept,

$\sigma=\mathbb{1}\{x^i \ge \theta^1 \ \forall i\},$

and a payoff function of the option, $$\phi(\vec{x})\in\mathbb{R}$$. The option is assumed to be stochastic with density function $$f(\vec{x})$$.

The probability of an option is acceptable is

$\Pr[\sigma=1 \mid \theta] = \int_\theta^\infty f(\vec{x})d\vec{x}$

the expected payoff of any option is

$E[\phi(\vec{x})] = \int_\Omega f(\vec{x})\phi(\vec{x})d\vec{x}$

and the expected payoff of an acceptable option is

$E[\phi(\vec{x}) \mid \sigma=1] = \frac{\int_\theta^\infty f(\vec{x})\phi(\vec{x})d\vec{x}}{\int_\theta^\infty f(\vec{x})d\vec{x}}.$

Consider $$N$$ options presented in tandem, the decision maker must take one of them. Thus the last one must be accepted if presented. The policy would be on setting the aspiration $$\vec{\theta}$$ to maximize the expected payoff. Let the value of a policy at state $$n$$ to be $$V_n(\vec{\theta})$$. Then we have

$V_N(\vec{\theta}) = E[\phi(\vec{x})]$

and

$V_n(\vec{\theta}) = E[\phi(\vec{x}) \mid \sigma=1]\Pr[\sigma=1 \mid \vec{\theta}] + V_{n+1}(\vec{\theta})\Pr[\sigma=0 \mid \vec{\theta}]$

for $$n=1,\cdots,N-1$$.

The goal of optimal satisficing is to find $$\arg\max_{\theta\in\Omega} V_n(\theta)$$. As $$E[\phi(\vec{x})] < E[\phi(\vec{x}) \mid \sigma=1]$$ for positive function $$\phi(\vec{x})$$, the optimal $$\theta$$ usually decreases as $$n$$ increases.

Heuristic satisficing is to use a fixed $$\theta$$ for all $$n$$. That is, with

$V_H(\theta) = \sum_{n=1}^{N-1}\Big((1-P_\sigma)^{n-1}P_\sigma E[\phi(\vec{x}) \mid \sigma=1]\Big) + (1-P_\sigma)^{N-1}E[\phi(\vec{x})]$

to find $$\arg\max_{\theta\in\Omega} V_H(\theta)$$. It is found that, the value of heuristic satisficing $$V_H$$ has only a slight decrease from the optimal satisficing $$V_1$$.

Satisficing can be converted into infinite horizon: Introduce a time cost of $$c$$ per unit time in heuristic satisficing. The value is then

$V_\infty = \int_0^\infty (1-P_\sigma)^t \Big(P_\sigma E[\phi(\vec{x}) \mid \sigma=1]-c\Big) dt$

It is found that, as $$c$$ increases, $$\theta$$ decreases. This reflects that the higher the time cost, the more relaxed the aspiration have to be.

## Bibliographic data

@incollection{
title = "On Optimal Satisficing: How Simple Policies can Achieve Excellent Results",
author = "J. N. Bearden and T. Connolly",
booktitle = "Decision Modelling in Uncertain and Complex Environments",
editor = "T. Kugler and J. C. Smith and T. Connolly and Y. J. Son",
}