Chapter 3 Combining. Reminder. Reminder. Shifted Uniform Series. Chapter 3 Combining Factors

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1 Chapter 3 Combining Factors Chapter 3 Combining Factors Shifted Uniform Series Uniform Series and Non-uniform Cash flows Shifted Gradients Shifted Decreasing rithmetic Gradients INEN 33 Sergiy Butenko Industrial & Systems Engineering Texas &M University Chapter 3 Combining Factors Chapter 3 Combining Factors 2 Reminder Reminder P =? F =? 2 n- n 2 n- n Chapter 3 Combining Factors 3 Chapter 3 Combining Factors 4 Consider the following shifted uniform annual series P F Shifted Uniform Series The worth at time zero may be evaluated by several methods: P = (P, i, 4) P = P (P F, i, ), or Series starts at period 2 rather than, and we re looking for its present value (i.e. at time ) F = (F, i, 4) P = F (P F, i, 5) Chapter 3 Combining Factors 5 Chapter 3 Combining Factors 6

2 Example C3.: annually convert the following cash flows to: b. nnual equivalent amount EOY mount EOY mount EOY mount - 4 -$2 8 -$8 -$8 5 -$8 9 -$8 2 -$8 6 -$8 3 -$8 7 -$ Example 3. Cash Flow Chapter 3 Combining Factors 7 Chapter 3 Combining Factors 8 a. P = (-,8)*(P/, 8%, 9) + 6(P/F, 8%, 4) = (-,8)*(6.2469) + 6*(.735) = -,83.42 b. The equivalent series for 9 years is: = P (/P, 8%, 9) = (-,83.42)*(.68) = -,729.4 (has to be between -2 and -8, since is spread over the same no. of periods- 9, if n was fewer than 9 it can be more than 8 and if n was more than 9 it can be less than 2) c. Future worth in year 9 is: F = P (F/P, 8%, 9) = (-,83.42)*(.999) = - 2,596.4 Example C3.2: Using an interest rate of 8% per year compounded annually, convert the following cash flows (see table below for end-of-year amounts) to: b. nnual equivalent amount EOY mount EOY mount EOY mount - 4 -$8 8 -$5 -$8 5 -$5 9 -$5 2 -$8 6 -$5 -$5 3 -$8 7 -$5 -$5 Chapter 3 Combining Factors 9 Chapter 3 Combining Factors Example 3.2 Cash Flow a. P = -,8(P/,8%,4),5(P/,8%,7)(P/F,8%,4) = -,8*3.32-5*5.264*.735 = -$,7.8 or P = -,5(P/,8%,) - 3(P/,8%,4) = -,5*7.39-3*3.32 = - $,72. Chapter 3 Combining Factors Chapter 3 Combining Factors 2

3 Solution (continued): b. = P(/P,8%,) = -,72*.48 = -$,639 c. F = P(F/P,8%,) = -,72*2.336 = -$27,284 Example C3.3 annually, convert the following cash flows to: b. nnual equivalent amount c. Equivalent amount at the end of year 5 d. Future worth at the end of the sequence EOY mount EOY mount EOY mount - 5 -$5 -$6 -$ 6 -$5 -$5 2 -$ 7 -$5 2 -$ $6 3 -$, 4-9 -$6 Chapter 3 Combining Factors 3 Chapter 3 Combining Factors Example 3.3 Cash Flow Example 3.3 Cash Flow a. P = -(P/,8%,2) - 5(P/,8%,8)(P/F,8%,4) - (P/,8%,3)(P/F,8%,7),(P/F,8%,3) = - $ 2,88.3 b. = P(/P,8%,3) = -2,88.3*.2652 = -$355 c. F 5 = P(F/P,8%,5) = -2,88.3*.4693 = -$4,26 d. F 3 = P(F/P,8%,3) = -2,88.3*2.796 = -$7,637 Chapter 3 Combining Factors 5 Chapter 3 Combining Factors 6 Shifted Gradients (reminder) Base mount Gradient (the base is ignored) n- n 2 3 n- n Shifted Gradients Consider the following Gradient Series G G 2G 3G 2G P G F G (n -2)G Gradient series starts in period 2 and goes up to n (n -)G Base amount is ; Gradient series starts in period 3 rather than 2; n = 4 not 3!; P =? Chapter 3 Combining Factors 7 Chapter 3 Combining Factors 8

4 Shifted Gradients The worth at time zero may be calculated by: P = G (P G, i, 4) (P F, i, ), or P = G (F G, i, 4) (P F, i, 5) Example C3.4: b. nnual equivalent amount EOY mount EOY mount EOY mount - 4 $,2 8 $2, $, 5 $,4 9 $2,2 2 $, 6 $,6 $2,4 3 $, 7 $,8 Chapter 3 Combining Factors 9 Chapter 3 Combining Factors 2 Example 3.5 Cash Flow a. P =,(P/,8%,3) + [,2(P/,8%,7) + 2(P/G,8%,7) ](P/F,8%,3) = $9,763 b. = P(/P,8%,) = 9,763*.493 = $,455 c. F = P(F/P,8%,) = 9,763*2.589 = $2, Example C3.5: b. nnual equivalent amount EOY mount EOY mount EOY mount - 4 $,4 8 $8 $2, 5 $,2 9 $7 2 $,8 6 $, $6 3 $,6 7 $9 Chapter 3 Combining Factors 2 Chapter 3 Combining Factors 22 a. P = [ 2,(P/,8%,6) - 2(P/G,8%,6) ] + [ 9(P/,8%,4) - (P/G,8%,4) ](P/F,8%,6) = $8,727 b. = P(/P,8%,) = 8,727*.493 = $,3 c. F = P(F/P,8%,) = 8,727*2.589 = $8, Example 3.6 Cash Flow Example C3.6: b. nnual equivalent amount EOY mount EOY mount EOY mount -$ 4 -$2 8 -$5 -$2 5 -$2 9 -$8 2 -$2 6 -$3 -$ $4 -$, Chapter 3 Combining Factors 23 Chapter 3 Combining Factors 24

5 Example 3.4 Cash Flow P = - - 2(P/,8%,8) + 2(P/F,8%,3) - (P/G,8%,4)(P/F,8%,4) - [ 8(P/,8%,3) + (P/G,8%,3) ](P/F,8%,8) = -$2, b. = P(/P,8%,) = - 2,678.37*.48 = -$375.9 c. F = P(F/P,8%,) = - 2,678.37*2.336 = -$6, Example C3.7 Calculate the Present Value of the following cash flow, i = 2% Chapter 3 Combining Factors 25 Chapter 3 Combining Factors 26 The cash flow may be broken down as follows: P = (P/,.2, 9) = (5.3282) = (from 2 to ) P 2 = (P/G,.2, 3) = (2.228) = 222. (n = 2, 3, 4) P 3 = (n = 5) P 4 = (P/G,.2, 3) = (2.228) = PW,2 = [ ] (P/F,.2, ) = PW 3 = (P/F,.2, 5) = 56.7 PW 4 = 222. (P/F,.2, 7) =.45 P P 3 P4 P 2 PW = = 83.2 Chapter 3 Combining Factors 27 Chapter 3 Combining Factors 28

Solutions to end-of-chapter problems Basics of Engineering Economy, 2 nd edition Leland Blank and Anthony Tarquin

Solutions to end-of-chapter problems Basics of Engineering Economy, 2 nd edition Leland Blank and Anthony Tarquin Chapter 2 Factors: How Time and Interest Affect Money 2.1 (a) (F/P,10%,20) = 6.7275 (b)