# Using Inequalities for Reducing Epistemic Uncertainty and Ranking Decision Alternatives

## Selection from Sources with Unknown Proportions of High-Reliability Components

Consider a real-world example featuring a market where three suppliers *A _{h} A_{2}* and Дз produce high-reliability components of the same type, with proportions

*x*and *з, which are unknown. In the case of suspension automotive springs, for example, this means that only a fraction

_{h}x_{2}*x*= 1, 2, 3 of the manufactured suspension springs can last for more than a specified number of cycles if tested on a specially designed test rig and the rest of the springs fail significantly below the specified limit.

_{h}iIf two components are to be purchased and installed in a device, the question of interest is which strategy maximises the probability that both components will be highly reliable: (i) purchasing the two components from the same supplier or (ii) purchasing the two components from different suppliers.

The same problem can be formulated for machine centres (instead of suppliers) producing high-reliability components with unknown proportions.

This problem is mathematically equivalent to the following problem:

Three boxes contain high-reliability components in unknown proportions *x _{u} x_{2 }*and а'з, characterising the boxes, correspondingly (Figure 6.1).

Two components are to be installed in a device. Which strategy maximises the probability that both components will be highly reliable?

- 1. Taking the two components from the same, randomly selected box, or
- 2. Taking the two components from two different, randomly selected boxes.

At first glance, it seems that either of these strategies could be selected because the proportions *x, x _{2}* and of high-reliability components characterising the boxes/ suppliers are unknown. Surprisingly, this common-sense conclusion is incorrect. The probability of selecting two high-reliability components from the same box is

This is composed of the probabilities of three mutually exclusive events: (i) the probability (l/3)jc_{(}^{2} that box 1 will be selected and both components taken from box 1 will

FIGURE 6.1 Three boxes/suppliers containing high-reliability components with unknown proportions.

be highly reliable; (ii) the probability (1 / 3)xf that box 2 will be selected and both components taken from box 2 will be highly reliable and (iii) the probability (l/3)xf that box 3 will be selected and both components taken from box 3 will be highly reliable.

Accordingly, the probability of selecting two high-reliability components from two randomly selected boxes is

This is composed of the probabilities of three mutually exclusive events: (i) the probability (1 / 3)x,x_{2} that boxes 1 and 2 will be selected and both components taken from boxes 1 and 2 will be highly reliable; (ii) the probability (l/3)x,x_{3} that boxes 1 and 3 will be selected and both components taken from boxes 1 and 3 will be highly reliable and (iii) the probability (l/3)x_{2}x_{3} that boxes 2 and 3 will be selected and both components taken from boxes 2 and 3 will be highly reliable.

The question of interest can be answered by comparing the probabilities /?, and *p _{2},* which is equivalent to proving the conjectured non-trivial algebraic inequality

This inequality can be proved rigorously by using a number of different techniques: by a direct manipulation, by using segmentation, by using the rearrangement inequality or by reducing it to a standard inequality (e.g., to the Muirhead inequality).

The segmentation technique is a powerful technique whose essence is to segment (split) the original inequality into simple parts and express each part as an inequality by using some of the standard inequalities. Adding the separate parts assembles the original inequality and completes the proof. Proving inequality (6.1) is equivalent to proving the inequality

because inequality (6.1) can be obtained from inequality (6.2) by dividing both sides of (6.2) by the positive constant '3'. Inequality (6.2) can be segmented into three parts by using the standard arithmetic mean - geometric mean (AM-GM) inequality:

Adding the three inequalities gives the original inequality (6.2) and completes the proof.

This is a surprising and highly counter-intuitive result. After all, the proportions of high-reliability components characterising the boxes are unknown. Despite the total lack of knowledge related to the proportions of high-reliability components characterising the separate batches/suppliers and regarding existing interdependences (correlations) among the proportions of high-reliability components in the boxes, inequality (6.1) still holds. The reduced epistemic uncertainty allows an appropriate choice to be made, associated with a reduced risk of not selecting two high-reliability components.

The same approach can be generalised for a larger number of selected components. Consider an analogous case of purchasing components from suppliers whose proportions of high-reliability components are unknown. If, for example, three components are to be purchased from three suppliers *(n* = 3) and installed in an assembly, the question of interest is to choose between several competing strategies: (a) purchasing the three components from a single, randomly selected supplier; (b) purchasing the three components from the three available suppliers or (c) purchasing the three components from two randomly selected suppliers. Suppose that three suppliers are characterised by unknown proportions x,_{;} *x _{2}* and x

_{3}of high-reliability components and unknown interdependences of the proportions of high-reliability components among the boxes. Proving that strategy

*‘a’*is better than strategy

*'If*reduces to proving the inequality

The left side of inequality (6.3) is the probability of purchasing three high-reliability components from a randomly selected supplier. The right side of inequality (6.3) is the probability of purchasing three high-reliability components from the three available suppliers.

Inequality (6.3) can be proved by a direct application of the AM-GM inequality:

Proving that variant ‘a’ is better than variant *‘c’* reduces to proving the inequality

The left side of inequality (6.4) is the probability of purchasing three high-reliability components from a randomly selected single supplier. The right side of inequality (6.4) is the probability of purchasing three high-reliability components from two randomly selected suppliers.

Applying segmentation by using the AM-GM inequality six times yields the following inequalities:

Adding the six inequalities and dividing both sides by the positive constant 3 gives inequality (6.4) and completes the proof.