Download Materials Selection - Problem Set 9 | EMA 4714 and more Assignments Materials science in PDF only on Docsity! Problem Set 9 EMA 4714 - Materials Selection and Failure Analysis Due Friday, April 7, 2008 – 8:30 am Dieter Exercise 13.4 A father desires to establish a fund for his new child’s college education. He estimates that the current cost of a year of college education is $12,000 and that the cost will escalate at an annual rate of 4 percent. [a] What amount is needed on the child’s eighteenth, nineteenth, twentieth and twenty-first birthdays to provide for a 4-year college education? ****************************************************** P[F/P,i,n] = 12k/[P/F,4,17] = 12k/0.51337 = $23,400 on 18th birthday = 23.4[1.04]1 = $24,300 on 19th birthday = 23.4[1.04]2 = $25,300 on 20th birthday = 23.4[1.04]3 = $26,300 on 21st birthday [b] If a rich aunt gives $5000 on the day the child is born, how much must be set aside at 10 percent on each of the first through seventeenth birthdays to build up the college fund? ****************************************************** Converting the amounts required on 19th, 20th and 21st birthdays to apresent worth on the 18th birthday [i = 10%]: P(17) = 23.4k+24.3k[P/F,10,1]+25.3k[P/F,10,2]+26.3k[P/F,10,3] P(17) = 23.4k+24.3[0.90909]+25.3k[0.82645]+26.3k[0.75131] P(17) = $86,200 as far as the $5k gift, F(17) = 5k[F/P,i,n] = 5k/[P/F,10,17] = 5k/0.19784 F(17) = $25,300 A = [P(17)-F(17)][A/F,i,n] = 61k/[F/A,10,17] = 61k/40.54470 A = $1500 per year..... Dieter Exercise 13.6 Machine A costs $8500 and has annual operating costs of $4500. Machine B costs $7000 and has an annual operating cost of $4800. Each machine has an economic life of 10 years. If the minimum rate of return is 10 percent, compare the advantages of machine A by [a] the present worth method, [b] the annual cost method and [c] the rate of return on investment. ****************************************************** ** [a] P(A) = 8500+4500[P/A,i,n] = 8500+4500[P/A,10,10] = 8500+ 4500[6.14457] = $36,150 P(B) = 7000+4800[6.14457] = $36,500 [b] A(A) = 8500[A/P,i,n]+4500 = 8500/[P/A,10,10]+4500 A(A) = 8500/6.14457+4500 = $5883 A(B) = 7000[A/P,i,n]+4800 = 7000/6.14457+4800 = $5939 [c] basis for rate of return on investment is to determine an interest rate on the incremental difference in initial capital investment which will equal the differences in annual operating costs.... 8500-7000 = 1500 increment of investment 4800-4500 = 300, savings in annual cost 1500[A/P,i,n] = 300; 1500/[P/A,i,n] = 300; [P/A,i,10] = 5 [10+x]/[20-10] = [6.14457-5]/[6.14457-4.19247] x ~ 5.86, i = 15.86%
