Engineering Economy 8th, Summaries of Engineering Economy

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Lecture slides to accompany Engineering Economy 8 th^ edition Leland Blank Anthony Tarquin

Chapter 3

Combining

Factors and

Spreadsheet

Functions

LEARNING OUTCOMES

**1. Shifted uniform series

1. Shifted series and single** **cash flows
2. Shifted gradients**

3 - 4 Example Using P/A Factor: Shifted Uniform Series P 0 =? A = $10, 0 1 2 3 4 5 6 i = 10% The present worth of the cash flow shown below at i = 10% is: (a) $25,304 (b) $29,562 (c) $34,462 (d) $37,

Solution : (1) Use P/A factor with n = 5 (for 5 arrows) to get P 1 in year 1

(2) Use P/F factor with n = 1 to move P 1 back for P 0 in year 0 P 0 = P 1 (P/F,10%,1) = A(P/A,10%,5)(P/F,10%,1) = 10,000(3.7908)(0.9091) = $34, Answer is___ 0 1 2 3 4 5

P 1 =?

Actual year Series year

How much money would be available in year 10 if $8000 is deposited each year in years 3 through 10 at an interest rate of 10% per year? 0 1 2 3 4 5 6 7 8 9 10

FA =?

A = $ i = 10% Solution : Re-number diagram to determine n = ___(number of arrows) 0 1 2 3 4 5 6 7 8 Cash flow diagram is: FA = 8000(F/A,10%,8) = 8000(11.4359) = ____________ Example Using F/A Factor: Shifted Uniform Series Actual year Series year

Example: Series and Random Single Amounts

Find the present worth in year 0 for the cash flows

shown using an interest rate of 10% per year.

0 1 2 3 4 5 6 7 8 9 10 PT =? A = $ i = 10% First, re-number cash flow diagram to get n for uniform series: n =__ $ 0 1 2 3 4 5 6 7 8 9 10 PT =? A = $ i = 10% $ 0 1 2 3 4 5 6 7 8 Solution: Actual year Series year

A = $ i = 10% 0 1 2 3 4 5 6 7 8 Actual year Series year 0 1 2 3 4 5 6 7 8 9 10 $ Use P/A to get PA in year 2: PA = 5000(P/A,10%,8) = 5000(5.3349) = $26, Move PA back to year 0 using P/F: P 0 = 26,675(P/F,10%,2) = 26,675(0.8264) = $22, Move $2000 single amount back to year 0: P 2000 = 2000(P/F,10%,8) = 2000(0.4665) = $ Now, add P 0 and P 2000 to get PT: PT =_____________________ Example: Series and Random Single Amounts PT =? PA

Example: Series and Random Amounts Convert the cash flows shown below (black arrows) into an equivalent annual worth A in years 1 through 8 (red arrows) at i = 10% per year. 0 1 2 3 4 5 6 7 8 A = $ i = 10% $ 0 1 2 3 4 5 A =? Solution:

1. Convert all cash flows into P in year 0 and use A/P with n = 8 2. Find F in year 8 and use A/F with n = 8 Solve for F: F = 3000(F/A,10%,5) + 1000(F/P,10%,1) = 3000(6.1051) + 1000(1.1000) = ____________ Find A: A = 19,415(A/F,10%,8) = 19,415(0.08744) = _________ Approaches:

Shifted Arithmetic Gradients Shifted gradient begins at a time other than between periods 1 and 2 Present worth PG is located ____________ gradient starts Must use multiple factors to find PT in actual year 0 To find equivalent A series, find PT at actual time 0 and apply (___, __, __)

Shifted Geometric Gradients

Shifted gradient begins at a time other than between periods 1 and 2 Equation yields Pg for all cash flows (base amount A 1 is included) For negative gradient, change signs on both g values Pg = A 1 {1 - [(1+g)/(1+i)] n

Equation (i ≠ g): /(i-g)}

There are ___ ________ for geometric gradient factors

Example: Shifted Geometric Gradient Weirton Steel signed a 5-year contract to purchase water treatment chemicals from a local distributor for $7000 per year. When the contract ends, the cost of the chemicals is expected to increase by 12% per year for the next 8 years. If an initial investment in storage tanks is $35,000, determine the equivalent present worth in year 0 of all of the cash flows at i = 15% per year.

Negative Shifted Gradients For negative arithmetic gradients, change sign on G term from + to - For negative geometric gradients, change signs on both g values All other procedures are the same as for positive gradients General equation for determining P: P = present worth of base amount - PG Pg = A 1 {1-[(1-g)/(1+i)] n /(i+g)} Changed from + to - Changed from + to - Changed from - to +

Example: Negative Shifted Arithmetic Gradient For the cash flows shown, find the future worth in year 7 at i = 10% per year F =? 0 1 2 3 4 5 6 7 700 650 500 450 550 600 G = $- 50 Solution:^ Gradient G first occurs between actual years 2 and 3; these are gradient years 1 and 2 0 1 2 3 4 5 6 Actual years Gradient years PG is located in gradient year 0 (actual year 1); base amount of $700 is in gradient years 1- 6 PG = 700(P/A,10%,6) – 50(P/G,10%,6) = 700(4.3553) – 50(9.6842) = ______ F = PG(F/P,10%,6) = 2565(1.7716) = ________ i = 10% PG =?

Summary of Important Points

P for shifted uniform series is _______________ of first A; n is equal to number of A values F for shifted uniform series is in _________________ as last A; n is equal to number of A values For gradients, first change equal to G or g occurs between gradient years 1 and 2 For ________arithmetic gradients, change sign on G from + to - For negative _________gradients, change sign on g from + to -

HOMEWORK

1. Please solve every Examples in your textbook. You do not have to **submit your works.

1. You don’t have to submit these problems, however, I recommend you** solve them for the tests. ① 3. ② 3. ③ 3. ④ 3. ⑤ 3. ⑥ 3. ⑦ 3. ⑧ 3.