Getting new products manufactured abroad and then delivered to their final desti

Question

Getting new products manufactured abroad and then delivered to their final destination in the USA can be a challenging task. One US Company was tired of getting hit with product delays and associated financial costs/penalties. As a result, they culled through their individual records on (1) their foreign manufacturer, (2) their international delivery company, and (3) their domestic delivery company. All three had contributed to product delays at various times in the past. After looking at historical data on actual cases, they discovered the following delays had been observed (relative to promised times). After looking at historical data on actual cases, they discovered the following cost/financial penalty had been observed ($/day). Obviously, longer delays are increasingly unfavorable, as reflected in the following escalating $-penalty structure. For example, a delay of 6 days incurs a $3, 300 penalty, while a delay of 36 days incurs a $30, 600 penalty. a) What is the expected (i.e., average) total delay, in days, for a new product? b) What is the probability that a new product's total delay is at least 1 day? c) What is the probability that a new product's total delay is at least 5 days? d) What is the probability that a new product's total delay is at least 10 days? e) What is the probability that a new product's total delay is at least 15 days? f) What is the expected (i.e., average) cost/financial penalty for product delays?

Explanation / Answer

E(manufacturer) =0.75*0 + 0.14*1 + 0.05*2 + 0.02*3 + 0.02*4 + 0.01*5 + 0.01*6

= 0.49 weeks = 3.43 days

E(international) = =0.60*0+0.09*1+0.08*2+0.06*(3+4+5) + 0.05*6

= 1.27 days

E(domestic) = 0*0.7+1*0.14+2*0.09+3*0.07

= 0.53 days

Average E(X) = [(3.43+1.27)+(3.43+0.53)]/2

=(4.7 + 3.96 )/ 2

=4.33 days

P(X>1) = 1 – 0.75 = 0.25

P(X>5) = 0.01 + 0.01 = 0.02

P(X>10) = 0.01*(0.06+0.06+0.05) = 0.0017

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