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ME 443

ME 443 . TIME VALUE of MONEY OPERATIONS (PART1) Prof. Dr. Mustafa Gökler. INTRODUCTION.

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ME 443

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  1. ME 443 TIME VALUE of MONEY OPERATIONS (PART1) Prof. Dr. Mustafa Gökler

  2. INTRODUCTION • Since the cash flow profiles are usually quite different among the several design alternatives, in order to compare the economic performances of the alternatives, one must compensate for the differences in the timing of cash flows. • To compare alternative investment projects at a common point in the time scale, the TIME VALUE OF MONEY must be considered.

  3. INTRODUCTION • The value, economic worth, or purchasing power of moneychanges over time during periods of inflation and deflation • Money has a time value even in the absence of inflation or deflation • People prefer to receive $X today rather than 1 year later even when there is no inflation or deflation. One reason for this preference is the opportunity cost of money.

  4. CASH FLOW DIAGRAMS End-of-period cash flows are assumed, unless otherwise noted

  5. INTEREST CALCULATIONS If a single sum of money has a current or presentvalue of P, its value in nyears would be equal to Fn = P + In where Fnis the accumulated value of P over nyears, or the future value of P, and Inis the increase in the value of P over nyears. Inis referred to as the accumulated interest.

  6. SIMPLE INTEREST Inis a linear function of time In= P i n where, i is annual interest rate (or interest rate per interest period ) n is number of years (or number of periods) Fn = P + In Fn = P (1 + i n)

  7. COMPOUND INTEREST In computing the value of In, i is interpreted as the rate of change in the accumulated value of money In= i Fn-1 Fn = Fn-1 + In Fn = Fn-1 (1+i) Since practically all monetary transactions are based currently on compound interest rates instead of on simple interest rates, it will be assumed that compounding occurs unless otherwise stated.

  8. PRESENT/FUTURE WORTH

  9. FUTURE WORTH F = P(1 + i)n P= F(1 + i)-n P = the equivalent value of an amount of money at time zero, or present worth F = the equivalent value of an amount of money at time n, or future worth i = the interest rate per interest period n = the number of interest periods

  10. FUTURE WORTH F = P(F |P i,n) where (F |P i,n) and (1 + i)nare referred to as thesingle sumFuture Worth (Value) Factor (F |P i,n) = (1 + i)n

  11. PRESENT WORTH P= F(1 + i)-n P=F(P| F i,n) where (1 + i)-nand (P |F i,n) are referred to asthe single sum, present worth factor. (P |F i,n) = (1 + i)-n

  12. NOTATION (P |F i,n) (Present from Future) i = 5% and n=3 (P/F 5,3) (P |F i,n) = (1 + 0.05)-3 (F|P i,n) (Future from Present) i=8% and n=6 (F |P 8,6) (F |P i,n) = (1 + 0.08)6

  13. NOTATION In some text books (P |F i,n) (Present from Future) (P |F i,n) = PVSP i%,,n Sp : Single payment (single sum) (F|P i,n) (Future from Present) (F |P i,n) = FVSP i%,,n

  14. PRESENT/FUTURE WORTH

  15. SERIES OF CASH FLOWS P = A1(1 + i) -1+ A2 (1 + i) -2+ ..... + An-1(1 + i) -(n-1)+ An(1 + i) -n F = A1(1 + i) + A2 (1 + i) 2+ ..... + An-1(1 + i) n-1+ An(1 + i)n

  16. UNIFORM SERIES OF CASH FLOWS

  17. UNIFORM SERIES of C.F. PRESENT WORTH P = A [(1 + i)n -1]/[i(1 + i)n] P = A (P |A i,n) (P |A i,n) = [(1 + i)n -1]/[i(1 + i)n] referred to as the uniform series, present worth factor.

  18. UNIFORM SERIES of C.F. ANNUAL EQUIVALENT WORTH A = P [i(1 + i)n] /[(1 + i)n -1] A = P (A |P i,n) (A |P i,n) = [i(1 + i)n] /[(1 + i)n -1] referred to as capital recovery factor.

  19. UNIFORM SERIES of C.F. FUTURE WORTH F = A [(1 + i)n -1]/i F = A (F |A i,n) (F |A i,n) = [(1 + i)n -1]/i referred to as the uniform series, future worth factor.

  20. UNIFORM SERIES of C.F. ANNUAL EQUIVALENT WORTH A = F (i) /[(1 + i)n -1] A = F (A |F i,n) (A|F i,n) = (i) /[(1 + i)n -1] referred to as sinking fund factor

  21. EXAMPLE • An individual wishes to deposit a single sum of money in a savings account so that five equal annual withdrawals of $2000 can be made before depleting the fund. If the first withdrawal will not occur until 3 years after the deposit and the fund pays interest at a rate of 12% compounded annually, how much should be deposited?

  22. EXAMPLE

  23. EXAMPLE P = A(P|A 12,5) (P|A 12,5) = [(1 + 0.12)5 -1]/[0.12 (1 + i)5] (P|A 12,5)= 3.6048 P = $2000(3.6048) P= $7209.60

  24. EXAMPLE

  25. EXAMPLE (P|F 12,2) = (1 + 0.12)-2 = 0.7972 P = A(P|A 12,5)(P|F 12,2) = $2000(3.6048)(0.7972) = $5747.49

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