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Econ. Lecture 4

Econ. Lecture 4

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Econ. Lecture 4

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  1. Econ. Lecture 4 Read 70-84 HW Problems 2.18, 2.19, 2.21, 2.27, 2.28, 2.29

  2. Finishing interest formula’s (Uniform payment, Linear and Geometric Gradients)We finished with:

  3. Solving this equation for A: Known as the Uniform Sinking Fund Factor, (A/F, i, n) Finding A, given F, i, n

  4. Since F = P(1+i)n, we can substitute this for F in the previous equation The Uniform Series Capital Recovery Factor, (A/P, i, n)

  5. Solving this equation for P, The Uniform Series Present Worth Factor, (P/A, i, n)

  6. All of these equations can be solved using interest tables, Appendix C. • Note, if the table doesn’t exist for a given interest rate, you have to use the equations.

  7. Example: • $100 deposited at the end of each year in savings account, 6%, for five years, what do you have? • A = $100 • F = ? • n = 5 years • i = 6% • F = A(F/A,i,n) = 100(F/A,6%,5) = 100(5.637) $563.70

  8. Or Notice that the tables can make life a lot easier.

  9. Recall on the first day, $100 a month for 40 years (work from 25-65) at an 8% rate. • Bond = 8%, Stock Index Fund = 12% • F=? • A = $1200 /year • i = 8%

  10. F = $1200(F/A,40,8%) = 1200 (259.0565) = $310,867.80 • F = $1200(F/A,40,12%) = 1200(767.0914) = $920509.68 • Just for kicks, what if at age 20 you decided to save $100/month • F = $1200(F/A,45,12%) = 1200(1358.23) = $1,629,876

  11. Example: • You wish to make uniform deposits every three months to have $10,000 at the end of 10 years. At 6% annual, how much would you have to deposit?  • F = $10,000 • A = ? • n = 40 quarterly deposits • i= 1.5% per quarter • A = F(A/F,i,n) = F(A/F,1.5%,40) = $10,000(0.184) = $184

  12. Example: • Buying a car. Total price is $6200. The down payment is $1240, with 48 monthly payments, with 1% per month. Payments are due at the end of each month. What is the monthly payment?

  13. P = 6200 – 1240 = $4960 • A = ? • n = 48 monthly payments • i = 1% • A = P(A/P,i,n) = 4960(A/P,1%,48) = 4960(0.0263) = $130.45

  14. Example: • Selling a house. Mortgage of $232.50 for 10 years. Selling mortgage to a bank with 1% interest per month. How much will the bank pay for mortgage. • A = $232.50 • n = 120 months • i = 1% • P = ?  • P = A(P/A,i,n) = $232.50(P/A,1%,120) = 232.509*(69.701) = $16,205.48

  15. Linear Gradient

  16. Sometimes cash flow will increase by a constant amount. 100 125 150 175 200 n=5 P

  17. This can be broken into two components: 100 100 100 100 100 0 G 2G 3G (n-1) G + P* = P’

  18. G = $25, n = 5 • We can compute P by adding P’ with P*. We have an equation for P’, • P* = G(P/G,i,n)

  19. Gradient series present worth factor, (P/G,i,N)

  20. Example: • Maintenance on a machine is expected to be $155 after one year and increase $35 each year for seven years. How much should be set aside now for the upcoming eight years. i = 6% • P = 155(P/A,6%,8) + 35(P/G,6%,8) = 155(6.210) + 35(19.841) = $1656.99

  21. Gradient to equal payment

  22. (A/G,i,n) Converting a gradient into an annual payment scheme. • Future worth given G can be calculated by substituting, (A/G,i,n) into (F/A,i,n).

  23. (F/G,i,N)