v9 Alternative actions Proportion of Probability Do not develop (42) Develop () customers 3 0.04 0.1 0 44.000 0.08 0.1 O 28.000 0.12 0.2 0 12.000 0.16 0.4 4.000 0 0.20 0.2 20.000 EOL (A) = 0.4 (4.000) +0.2 (20,000) = 5,600, EOL (A) = 0.1 (44,000) + 0.1 (28,000) + 0.2 (12.000) = 9,600, Since A, gives the lower EOL of 5,600, the best decision is not to develop the product. EXAMPLE 9.8-6 A company has recently installed new machinery but has not yet decided on the appropriate number of a certain spare part required for repairs. Spare parts cost 2.000 each but are only available if ordered now. If the plant failed and there was no spare part available, the cost to the business of mending the plant rises to 15,000. The plant has an estimated life of 10 years and the probability distribution of failures during this time, based on the experience with similar plants, is as follows: TABLE 9.42 No. of failures over ten years period 7 2 3 4 5 and over Probability 0.1 0.4 0.3 0.1 0.1 nil Calculate (a) The expected number of failures in the ten-year period. (b) The optimal number of spares that should be purchased now. (c) The cost of the ordering policy chosen. (d) The value of perfect information of the number of failures in ten-year life. [P.T.U. M.Tech. April, 2012: I.C.M.A. (London) May. 1980] . 0