University:
California State University, Northridge
Course:
MSE 604 | Engineering Economy and Financial Analysis
Academic year:

2024
Views:

487

Pages:

17
Author:

fiziokrat16l

Arithmetic Gradient Factors • An arithmetic (linear) Gradient is a cash flow series that either increases or decreases by a contestant amount over n time periods. •A linear gradient is always comprised of TWO components: •The Gradient component •The base annuity component •The objective is to find a closed form expression for the Present Worth of an arithmetic gradient Linear Gradient Example A1+n-1G A1+n-2G • Assume the following: A1+2G A1+G 0 1 2 3 n-1 N This represents a positive, increasing arithmetic gradient Present Worth: Gradient Component • General CF Diagram – Gradient Part Only (n-1)G 3G 1G (n-2)G 2G 0G We want the PW at time t = 0 (2 periods to the left of 1G) 0 1 2 3 4 ……….. n-1 n To Begin- Derivation of P/G,i%,n P =G ( P / F , i %, 2) + 2G ( P / F , i %, 2) + ... ...+ [(n-2)G](P/F,i,n-1)+[(n-1)G])P/F,i,n) P= G{( P / F , i %, 2) + 2( P / F , i %, 2) + ... ...+ [(n-2)](P/F,i,n-1)+[(n-1)])P/F,i,n)}  1 2 n-2 n-1  + + ... + + P=G  2 3 n-1 n  (1+i) (1+i)   (1+i) (1+i) Multiply both sides by (1+i) Subtracting [1] from [2]…..  1 2 n-2 n-1  + + ... + + P(1+i) =G  1 2 n-2 n-1  2 (1+i) (1+i)   (1+i) (1+i) 1 -  1 2 n-2 n-1  + + ... + + P=G  2 3 n-1 n  (1+i) (1+i)   (1+i) (1+i) G  (1 + i ) − 1 N  − P=  N N  i  i (1 + i ) (1 + i )  N ( P / G, i %, N ) factor 1 The A/G Factor • Convert G to an equivalent A A = G ( P / G, i, n)( A / P, i, n) N  G (1 + i ) − 1 N   i (1 + i ) N  − P=  N N   i  i (1 + i ) (1 + i )   (1 + i ) N − 1  A/G,i,n = 1  n G −  N  i (1 + i ) − 1  Gradient Example $700 $300 $400 $500 $600 $200 $100 0 1 2 3 4 5 6 7 •PW(10%)Base Annuity = $379.08 •PW(10%)Gradient Component= $686.18 •Total PW(10%) = $379.08 + $686.18 •Equals $1065.26 50 Geometric Gradients • An arithmetic (linear) gradient changes by a fixed dollar amount each time period. •A GEOMETRIC gradient changes by a fixed percentage each time period. •We define a UNIFORM RATE OF CHANGE (%) for each time period •Define “g ” as the constant rate of change in decimal form by which amounts increase or decrease from one period to the next Geometric Gradients: Increasing • Typical Geometric Gradient Profile •Let A1 = the first cash flow in the series 0 1 A1 2 3 4 …….. n-1 n A1(1+g) A1(1+g)2 A1(1+g)3 A1(1+g)n-1 Geometric Gradients: Starting • Pg = The Aj’s time the respective (P/F,i,j) factor •Write a general present worth relationship to find Pg…. A1 A1 (1 + g ) A1 (1 + g ) 2 A1 (1 + g ) n −1 = + + + ... + Pg 1 2 3 (1 + i ) (1 + i ) (1 + i ) (1 + i ) n Now, factor out the A1 value and rewrite as.. Geometric Gradients  1 (1 + g )1 (1 + g ) 2 (1 + g ) n −1  Pg A1  = + + + ... + 2 3 n  (1 i ) (1 i ) (1 i ) (1 i ) + + + +   Multuply both sides by (1) (1+g) to create another equation (1+i) (1+g) (1+g)  1 (1 + g )1 (1 + g ) 2 (1 + g ) n −1  (2) = Pg A1 + + + ... +   2 3 (1+i) (1+i)  (1 + i ) (1 + i ) (1 + i ) (1 + i ) n  Subtract (1) from (2) and the result is….. Geometric Gradients n  g 1+g (1 ) 1  +   Pg  = − 1 A1  −  n +1 1+ i   1+i   (1 + i ) Solve for Pg and simplify to yield…. Pg   1 + g n  nA1 P = 1 −    g 1+ i     (1 + i ) A1 g≠i  i−g  For the case i = g     Geometric Gradient: Example •Assume maintenance costs for a particular activity will be $1700 one year from now. •Assume an annual increase of 11% per year over a 6-year time period. •If the interest rate is 8% per year, determine the present worth of the future expenses at time t = 0. •First, draw a cash flow diagram to represent the model. Geometric Gradient Example (+g) •g = +11% per period; A1 = $1700; i = 8%/yr 0 1 $1700 2 3 4 5 6 7 $1700(1.11)1 $1700(1.11)2 $1700(1.11)3 PW(8%) = ?? $1700(1.11)5 Example: i unknown • Assume on can invest $3000 now in a venture in anticipation of gaining $5,000 in five (5) years. •If these amounts are accurate, what interest rate equates these two cash flows? $5,000 0 1 2 3 4 5 •F = P(1+i)n $3,000 •(1+i)5 = 5,000/3000 = 1.6667 •(1+i) = 1.66670.20 •i = 1.1076 – 1 = 0.1076 = 10.76% Unknown Number of Years • Some problems require knowing the number of time periods required given the other parameters •Example: •How long will it take for $1,000 to double in value if the discount rate is 5% per year? Fn = $2000 •Draw the cash flow diagram as…. i = 5%/year; n is unknown! 0 1 P = $1,000 2 ... . . . ……. n Unknown Number of Years Fn = $2000 • Solving we have….. 0 1 2 ... . . . ……. n P = $1,000 •(1.05)x = 2000/1000 •Xln(1.05) =ln(2.000) •X = ln(1.05)/ln(2.000) •X = 0.6931/0.0488 = 14.2057 yrs •With discrete compounding it will take 15 years