What is Value? Engineering Economics: Session 2. Page 1

Transcription

1 Engineering Economics: Session 2 Engineering Economic Analysis: Slide 26 What is Value? Engineering Economic Analysis: Slide 27 Page 1

2 Review: Cash Flow Equivalence Type otation Formula Excel Single Uniform Series Compound Amount F = P (1 + i) (F/P,i,) Present Worth P = F /(1 + i) (P/F,i,) Compound Amount (1 + i) 1 F = A (F/A, i, ) i Sinking Fund i A= F (A/F, i, ) (1 + i ) 1 Present Worth (1 + i) 1 P= A (P/A, i, ) i (1 + i ) Capital Recovery i (1 + i) A= P (A/P, i, ) (1 + i ) 1 Engineering Economic Analysis: Slide 28 Single Payment Example Finding P given F An investor can purchase land that will be worth $k in 6 years If the investor s discount rate is 8%, what is the max they should pay today? P= F( P / F, i, ) = $, 000 = ( ) 6 = $, = $6, 300 F (1+ i) Engineering Economic Analysis: Slide 29 Page 2

3 Single Payment Example Solving for i or What rate of return will you need to double your investment in years? F = P( F / P, i, ) = P (1 + ) i P (1 + i ) = 2 P (1 + i ) = 2 ln(1 + i ) = ln(2) i ln(1 + i ) = ln(2) e ln( 2) ln(1 + i ) = e 1 + i= e ln( 2) ln( 2) i = e 1 i = 7.2% Engineering Economic Analysis: Slide 30 Single Payment Example Solving for i or How many years must elapse for an investment to double at a rate of return of 6%? F = P( F/ P, i, ) = P (1 + ) i P ( ) = 2 P (1.08) = 2 ln(1.08) = ln(2) i ln(1.08) = ln(2) ln(2) = ln(1.08) = 9 Engineering Economic Analysis: Slide 31 Page 3

4 Discount Rate Approximation: Rule of 70 or 72 To approximate effect of discounting: Rule of 72 or Rule of 70 umber of years to double = 70 / Interest rate (in percent) P (1 + i) = 2 P for small x (1 + i) = 2 ln(1 + x) x ln(1 + i ) = ln(2) ln(2) 0.69 i ln(1 + i) = ln(2) i i ln(2) = 70 ln(1 + i ) i % Engineering Economic Analysis: Slide 32 Discount Rate Approximation: Rule of 70 or 72 To approximate effect of discounting: Rule of 72 or Rule of 70 umber of years to double = 70 / Interest rate (in percent) Examples When would $00 invested at % double? Rule Æ 7.2 years Actual Æ What is the value of $00 in 8 years, at 9%? Rule Æ $2,000 Actual Æ $1,993 Engineering Economic Analysis: Slide 33 Page 4

5 Discount Rate Approximation: Rule of 70 or 72 Period Interest Rate 12% 11% % 9% 8% 7% 6% 5% 4% Mapping Interest Rate and Periods to Double Rule Actual umber of Periods Engineering Economic Analysis: Slide 34 Review: Finite Series of Equal Payments a) Future Value (F) i = A (1 + r ) i [(1 + r ) - 1] = A r b) Payment (A) P A A A A A [(1+r) ] = P r [(1 + r) -1] = P (crf) 0-1 crf = Capital Recovery Factor Engineering Economic Analysis: Slide 35 Page 5

6 Using the Compound Amount Factor: Finding F, Given i, A, Suppose: You put $3k into savings for years of ea. yr) Your savings account earns 7% What is your account worth after years? F = A( F / A, i, ) (1 + i) 1 = A i (1.07) 1 = $ = $300(13.82) = $4,145 Engineering Economic Analysis: Slide 36 Using the Capital Recovery Factor: Finding A, given P Suppose: Your firm purchases lab equipment for $250k The loan s interest rate is 8% What payment will repay the loan? A= P( A/ P, i, ) i(1 + i) = P (1 + i) 1 6 = $250, (1.08) (1.08) 6 1 = $250, 000(0.2163) = $54, 075 Engineering Economic Analysis: Slide 37 Page 6

7 Other Special Cases: Linear Gradient of Cash Flows Series of cash flows changing by uniform amount per period. (n-2)g (n-1)g o P1 G 2G 3G n-2 n-1 n Engineering Economic Analysis: Slide 38 Deriving Equivalence for a Linear Gradient of Payments G 2 G ( 1) G P = (1 + i ) (1 + i ) (1 + i ) P = n ( n 1) G n (1 + i ) Let x = 1/(1 + i ) 2 3 P = 0 + ax + 2 ax ( 1) ax 2 1 ( ( 1) ) P = ax + x + x + + x x + ( 1) x 0 + x + 2 x ( 1) x = x 2 (1 x ) (1 + i ) i 1 P = G 2 i (1 + i ) Engineering Economic Analysis: Slide 39 Page 7

8 Other Special Cases: Geometric Series Payment grows from an amount A 1 by g% per period A n = A 1 (1+g) n-1 If g>0, series grows If g<0, series shrinks Engineering Economic Analysis: Slide 40 Geometric Gradient of Payments G 2G ( 1) G P = (1 + i) +... (1+ i) 3 (1+ i) (1 + g) P = n 1 A 1 n=1 (1 + i) n 1 (1 A 1 +g ) (1+ i),if i g i g P = A 1,if i = g (1 + i) Engineering Economic Analysis: Slide 41 Page 8

9 Example Problem: Geometric Series Facility has aging cooling system which currently runs 70% of the time the plant is open Pump will only last 5 more years. As it deteriorates, the pump run time is expected to increase 7% per year ew cooling system would only run 50% of the time Assumptions Either pump uses 250 kwh, Electricity cost $0.05/KWh Plant runs 250 days per year, 24 hours per day Firm s discount rate is 12% What is the value of replacing the pump? Engineering Economic Analysis: Slide 42 Example Problem: Geometric Series Current pump power cost = 70% x 250kWh x $0.05/kWh x 250 days x 24 hrs/day = $52,500 ew pump power Cost = $37,500 P Old P ew P ew (1.07) (1.12) = $52, = $214, 360 = $37, 500(P / A,12%, 5) = $37, 500(3.605) = $135, 200 Value = P Old P ew = $79,160 Engineering Economic Analysis: Slide 43 Page 9