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3.47 P = 29,000 + 13,000(P/A,10%,3) + 13,000[7/(1 + 0.10)](P/F,10%,3)
3.48 (a) Find P in year –1 and then move to year 5
3.49 Find P in year 1 for geometric gradient and add to amount in year 1; move total back
to year 0.
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P1 = 22,000 + 22,000[1 – (1.08/1.10)9]/(0.10 – 0.08)
= $189,450
P0 = 189,450(P/F,10%,1)
= 189,450(0.9091)
= $172,229
3.50 (a) Find P in year 4 for the geometric gradient, (b) Spreadsheet
then move all cash flows to future
Shifted Decreasing Gradients
3.51 P = [2000(P/A,10%,6) – 200(P/G,10%,6)](F/P,10%,1)
3.52 First find P in year 1, move to year 0, and then convert to A
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of McGraw-Hill Education.
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P0 = P1(P/F,10%,1)
= 1366.05(0.9091)
= $1241.87
A = P0(A/P,10%,5)
= 1241.87(0.26380)
= $327.61
3.53 P = 1,800,000(P/A,12%,3) + [1,800,000(P/A,12%,7) – 30,000(P/G,12%,7)](P/F,12%,3)
3.54 20,000 = 5000 + 4500(P/A,8%,n) – 500(P/G,8%,n)
3.55 Find present worth of gradient in year 1, move back to year 0, and then set equal to $2500
3.56 P3 = 4,100,000[1 – (0.90/1.06)17]/(0.06 + 0.10)
= $31,141,574
3.57 Find P in year 5, then find future worth of all cash flows
Exercises for Spreadsheets
3.58 (a) P = $298,542 using NPV function shown
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3.60 (a) Amount re-paid is $195,000 for Schedule A, and less at $180,000 for Schedule B.
(b) Annual equivalent loss at 5% over 12 years of $685 is less for Schedule A, even though
3.61 Value of x is $101.74
3.62 (a) F5 = $13,923 using the function = - FV(10%,4,3000)
3.63 Two approaches are: (1) use trial and error in the function = NPV(10%,B5:B10)+B4, and
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of McGraw-Hill Education.
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3.64 (a) Worth today = $10,732,345
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of McGraw-Hill Education.
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3.65 Payment in year 4, x = $2755.89
Extra above arithmetic gradient value of $-1600 is $1155.89
ADDITIONAL PROBLEMS AND FE REVIEW QUESTIONS
3.66 Answer is (b)
3.67 A = 100,000(A/F,10%,4)
3.70 9,500(F/P,20%,5) + x(F/P,20%,3) = 38,000
3.71 A(F/A,20%,5)(F/P,20%,8) = 50,000
3.72 125,000 = 190,000 – G(A/G,20%,5)
3.73 24,000 = 3695(P/A,10%,n)
3.75 10,000 = 2x(P/F,10%,2) + x(P/F,10%,4)
Solution to Case Study, Chapter 3
There are not always definitive answers to case studies. The following are examples only.
Preserving Land for Public Use
Cash flows for purchases at g = –25% start in year 0 at $4 million. Cash flows for parks
development at G = $100,000 start in year 4 at $550,000. All cash flow signs are +.
Cash flow________
Year Land Parks
0 $4,000,000
1 3,000,000
2 2,250,000
3 1,678,000
4 1,265,625 $550,000
5 949,219 650,000
6 750,000
1. Find P. In $1 million units,
P = 4 + 3(P/F,7%,1) + … + 0.750(P/F,7%,6)
2. Find remaining project fund needs in year 3, then find the A for the next 3 years
