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# sol10 - AMS 341 Spring 2011 Homework Set 10 — Solution...

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Unformatted text preview: AMS 341 Spring, 2011 Homework Set # 10 — Solution Notes #1, 13.4: Deﬁne fi (c) = xi (c) = Then maximum revenue in sites i, i + 1..., 3 millions of dollars invested in site i in order to get fi (c) f3 (0) = 3, x3 (0) = 0, f3 (1) = 7, x3 (1) = 1, f3 (2) = 8, x3 (2) = 2, f3 (3) = 13, x3 (3) = 3, f3 (4) = 15, x3 (4) = 4, f2 (0) = 3 + f3 (0) = 6, x2 (0) = 0, 3 + f3 (1) = 10 f2 (1) = max , x2 (1) = 0, 6 + f3 (0) = 9 3 + f3 (2) = 11 f2 (2) = max 6 + f3 (1) = 13 , x2 (2) = 1 or 2, 10 + f3 (0) = 13 3 + f3 (3) = 16 6 + f (2) = 14 3 , x2 (3) = 2, f2 (3) = max 10 + f3 (1) = 17 12 + f3 (0) = 15 3 + f3 (4) = 18 6 + f3 (3) = 19 f2 (4) = max 10 + f3 (2) = 18 , x2 (4) = 1 or 3, 12 + f3 (1) = 19 14 + f3 (0) = 17 4 + f2 (4) = 23 7 + f2 (3) = 24 f1 (4) = max 8 + f2 (2) = 21 , x1 (4) = 1, 9 + f2 (1) = 19 11 + f (0) = 17 2 Optimal allocations of investments is: 1 million to site 1, 2 million to site 2 and 1 million to site 3, for a revenue total of 24 million. #6,Ch.13 Review: Let pi be the proﬁt from the begining of the (i + 1)-th year with a new machine to the end of the 6th year. The proﬁt for using a machine for j years before buying a new machine is: j = 1 : 4500 + 3000 − 500 − 5000 = \$2000 j = 2 : 4500 + 3000 + 1800 − 500 − 700 − 5000 = \$3100 j = 3 : 4500 + 3000 + 1500 + 500 − 500 − 700 − 1100 − 5000 = \$2200 Therefore, p6 p4 p2 p0 = \$0, = max(3100 + p6 , 2000 + p5 ) = \$4000, = max(2200 + p5 , 3100 + p4 , 2000 + p3 ) = \$8000, = max(2200 + p3 , 3100 + p2 , 2000 + p1 ) = \$12000 p5 = 2000 + p6 = \$2000, p3 = max(2200 + p6 , 3100 + p5 , 2000 + p4 ) = \$6000, p1 = max(2200 + p4 , 3100 + p3 , 2000 + p2 ) = \$10000, Optimal solution: Buy a new machine at the begining of each year and use it for one year. The maximum proﬁt after 6 years is \$12000. Since you start year 1 with a machine, the proﬁt is 12,000+5000=17,000. ...
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