Extending the Results by Using the Muirhead Inequality

These results can be confirmed and extended when an alternative approach is used, based on the Muirhead inequality (6.5) introduced in Chapter 2, which states that if the sequence (a) is majorising the sequence {b) and хьх2,...,х„ are non-negative, the next inequality holds:

Consider the two non-increasing sequences ax > a2 >,...,> a„ and b > b2 >,...,> b„ of non-negative real numbers. The sequence (a) is said to majorise the sequence [b] if the following conditions are fulfilled:

For any set of non-negative numbers x,x2,...,xn, a symmetric sum is defined by Y ^xf1x2xn‘ which, when expanded, includes n terms. Each term is obtained by a distinct permutation of the elements of the sequence aua2,...,a„. Thus, if (a) = [2,1,0] then

If {a} = [2,0,0], then

Consider now the set of non-negative numbers хьх2,хз and the sequences {a} = [2,0,0] and [b] = [1,1,0]. Clearly, the sequence {a} = [2,0,0] majorises the sequence {b) = [1,1,0] because the conditions in (6.6) are fulfilled:

According to the Muirhead inequality (6.5):

Dividing both sides of the last inequality by the positive constant n (where n=3) leads to

According to inequality (6.7), /?, > p2 therefore, if two components need to be purchased from three suppliers, purchasing both components from a single, randomly selected supplier is the better strategy, resulting in a higher probability that both components will be high-reliability components.

By using the Muirhead general inequality, the results can be extended for n different suppliers.

The probability of purchasing two high-reliability components from a single, randomly selected supplier is pt = Accordingly, the probability of purchasing two high-reliability components from two randomly selected suppliers is

Pi=-afajLtx*i-

Since the sequence |a) = [2,0,0,...,0] (containing n elements) majorises the sequence (/;} = [1,1,0,...,0] (also containing n elements), according to the Muirhead inequality (6.5):

Dividing both sides of the last inequality by the positive number n! yields

The left side of inequality (6.8) is the probability of purchasing two high-reliability components from a randomly selected single supplier while the right side of inequality (6.8) is the probability of purchasing two high-reliability components from two distinct, randomly selected suppliers.

Muirhead’s inequality can also be applied to extend the results for a larger number of purchased components. 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 three different suppliers or (c) purchasing the three components from two randomly selected suppliers. Suppose that the suppliers are characterised by probabilities xu x2 and xз of producing high-reliability components. Because the sequence (a) = [3,0,0] majorises the sequence [(?} = [1,1,1], the next inequality follows immediately from the Muirhead inequality (6.5):

By dividing both sides of (6.9) to n (n = 3), inequality (6.9) transforms into

The left side jq3 / 3 + x /3 + х3 /3 of inequality (6.10) is the probability of purchasing three high-reliability components from a randomly selected supplier. The right side of inequality (6.10) is the probability xx2xy of purchasing three high-reliability components from all three available suppliers.

Since the sequence {a} = [3,0,0] also majorises the sequence [c] = [2,1,0], the following inequality follows immediately from the Muirhead inequality (6.5):

Dividing both sides of (6.11) by 3! (n = 3), gives the inequality which is identical to inequality (6.4).

The meaning of the left and right sides of inequality (6.12) are identical to the meaning of the left and right sides of inequality (6.4). The left side of inequality (6.12) gives the probability of purchasing three high-reliability components from a randomly selected single supplier. The right side of inequality (6.12) gives the probability of purchasing three high-reliability components from two randomly selected suppliers.

Since the sequence (a) = [2,1,0] majorises the sequence (c) = [1,1,1], the following inequality follows immediately from the Muirhead inequality (6.5):

Dividing both sides of (6.13) by n! (n = 3), gives

The left-hand side of inequality (6.14) gives the probability of purchasing three high- reliability components from two randomly selected suppliers. The right-hand side of inequality (6.14) gives the probability of purchasing three high-reliability components from the three available suppliers.

For three suppliers characterised by the probabilities jq=0.9, jc2=0.75 and x3 = 0.25 of selecting a high-reliability component, the Monte Carlo simulation based on 10 million trials resulted in probabilities p2 = 0.26 and p2 = 0.169 of purchasing three high-reliability components from two randomly selected suppliers and from the three available suppliers, correspondingly. The left- and right-hand sides of inequality (6.14) yield p2 = 0.26 and = 0.169 for the same probabilities, which confirms the validity of inequality (6.14).

Consider purchasing four components from three suppliers. The alternative of purchasing all four components from a randomly selected supplier is compared with the alternative of purchasing the four components from all three suppliers.

Since the sequence [a] = [4,0,0] majorises the sequencejc) = [2,1,1], the following inequality follows immediately from the Muirhead inequality (6.5):

Dividing both sides of (6.15) by n! (n = 3), gives

The left side of inequality (6.16) yields the probability of purchasing four high- reliability components from a single, randomly selected supplier. The right side of inequality (6.16) yields the probability of purchasing four high-reliability components from the three available suppliers.

For high-reliability fractions of 0.9, 0.4 and 0.3 characterising the suppliers, the Monte Carlo simulation yielded the value 0.23 for the left side of inequality (6.16) and the value 0.0576 for the right side of inequality (6.16). The same values are calculated by substituting the values jr, = 0.9, x2 = 0.4 and x} = 0.3 in inequality (6.16).

Consider now the alternative of purchasing all four components from a randomly selected supplier compared with the alternative of purchasing three components from a randomly selected supplier and another component from another randomly selected supplier.

Since the sequence (a) = [4,0,0] majorises the sequence (c) = [3,1,0], the following inequality follows immediately from the Muirhead inequality (6.5):

Dividing both sides of (6.17) by n (n = 3), gives

The left side of inequality (6.18) yields the probability of purchasing four high- reliability components from a randomly selected supplier. The right side of inequality (6.18) gives the probability of purchasing four high-reliability components from two suppliers by purchasing three components from one randomly selected supplier and the other component from another randomly selected supplier.

For high-reliability fractions of 0.9, 0.4 and 0.3 characterising the suppliers, the left side of inequality (6.18) yields the value 0.23 for the probability of four high- reliability components while the right side of (6.18) yields the probability of 0.1.

Finally, consider now the alternative of purchasing all four components from a randomly selected supplier compared with the alternative of purchasing two components from a randomly selected supplier followed by purchasing another two components from another randomly selected supplier.

Since the sequence [a] = [4,0,0] majorises the sequence [c] = [2,2,0], the following inequality follows immediately from the Muirhead’s inequality (6.5):

Dividing both sides of (6.19) by n! (n = 3), gives

The left side of inequality (6.20) gives the probability of purchasing four high-reli- ability components from a randomly selected supplier. The right side of inequality

(6.20) gives the probability of purchasing four high-reliability components from two randomly selected suppliers by purchasing two components from each supplier.

For high-reliability component fractions of 0.9,0.4 and 0.3 characterising the suppliers, the left side of inequality (6.20) yields the value 0.23 for the probability of four high-reliability components while the right side of (6.20) yields the probability 0.072.

Consider purchasing three components from four suppliers. The alternative of purchasing all three components from a randomly selected supplier is compared with the alternative of purchasing the three components from two different suppliers.

Since the sequence (a) = [3,0,0,0] majorises the sequence (c) = [2,1,0,0], the following inequality follows immediately from the Muirhead inequality (6.5):

Dividing both sides of (6.21) by n (n = 4), gives

The left side of inequality (6.22) gives the probability of purchasing three high- reliability components from a single, randomly selected supplier. The right side of inequality (6.22) gives the probability of purchasing three high-reliability components from two randomly selected suppliers.

For high-reliability fractions of 0.85,0.55,0.35 and 0.25 characterising the suppliers, the Monte Carlo simulation yielded the value 0.21 for the left side of inequality (6.22) and the value 0.13 for the right side of inequality (6.22). The same values are calculated by substituting the values x, = 0.85, x2 = 0.55, x3 = 0.35 and x4 = 0.25 in the left and the right parts of inequality (6.22). The same reasoning can be applied for any number of purchased components and any number of suppliers.

In summary, if no information is available about the fractions of high-reliability components characterising the separate suppliers, the best strategy is to purchase the components from a single supplier or from the smallest possible number of suppliers. Despite the total lack of knowledge related to the proportions of high-reliability components characterising the separate batches/suppliers and existing dependencies (correlations) among the proportions of high-reliability components from the separate suppliers, the use of inequalities reduced the epistemic uncertainty and allowed an appropriate choice to be made, associated w'ith a reduced risk of not selecting high-reliability components.

These highly counter-intuitive results fly in the face of the conventional practice always advocating diversification as a risk reduction measure and expose the dangers of blindly following conventional wisdom in risk reduction.

 
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