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Inflation Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years, Price = 1000(1.1) 5 = $1, But we still only have 1 ton of copper $1,610 5 years from now buys the same as $1,000 now
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IENG 217Cost Estimating for Engineers
Inflation
InflationSuppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years,
Price = 1000(1.1)5 = $1,610.51
But we still only have 1 ton of copper
InflationSuppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years,
Price = 1000(1.1)5 = $1,610.51
But we still only have 1 ton of copper
$1,610 5 years from now buys the same as $1,000 now
InflationSuppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years,
Price = 1000(1.1)5 = $1,610.51
But we still only have 1 ton of copper
$1,610 5 years from now buys the same as $1,000 now
10% inflation
InflationSuppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years,
Price = 1000(1.1)5 = $1,610.51
But we still only have 1 ton of copper
$1,610 5 years from now buys the same as $1,000 now
10% inflation (deflation = neg. inflation)
Combined Interest Rate
Suppose inflation equals 5% per year. Then $1 today is the same as $1.05 in 1 year
Suppose we earn 10%. Then $1 invested yields $1.10 in 1 year.
Combined Interest Rate
Suppose inflation equals 5% per year. Then $1 today is the same as $1.05 in 1 year
Suppose we earn 10%. Then $1 invested yields $1.10 in 1 year.
In today’s dollars$1.00 $1.10
$1.05 $1.10
Combined Interest RateThat is
$1.10 = 1.05 (1+d)1
1(1+.10) = 1(1+.05)(1+d)
Combined Interest RateThat is
$1.10 = 1.05 (1+d)1
1(1+.10) = 1(1+.05)(1+d)
1+i = (1+f)(1+d)
i = d + f + df
Combined Interest RateThat is
$1.10 = 1.05 (1+d)1
1(1+.10) = 1(1+.05)(1+d)
1+i = (1+f)(1+d)
i = d + f + df
i = interest rate (combined)f = inflation rated = real interest rate (after inflation rate)
Solving for d, the real interest earned after inflation,
wherei = interest rate (combined)f = inflation rated = real interest rate (after inflation
rate)
Combined Interest Rate
ffid
1
Example
Suppose we place $10,000 into a retirement account which earns 10% per year. How much will we have after 20 years?
Example
Suppose we place $10,000 into a retirement account which earns 10% per year. How much will we have after 20 years?
Solution: F = 10,000(1+.1)20
= $67,275
Example (cont.)
How much is $67,275 20 years from now worth if the inflation rate is 3%?
Example (cont.)
How much is $67,275 20 years from now worth if the inflation rate is 3%?
Solution: FT = 67,275(P/F,3,20)
= 67,275(1.03)-20
= $37,248
Alternate: Recall
= (.1 - .03)/(1+.03) = .068
Example (cont.)
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1
Alternate: Recall
= (.1 - .03)/(1+.03) = .068
FR = 10,000(1+d)20
= 10,000(1.068)20
= $37,248
Example (cont.)
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1
Alternate: Recall
= (.1 - .03)/(1+.03) = .068
FR = 10,000(1+d)20
= 10,000(1.068)20
= $37,248
Note: This formula will not work with annuities.
Example (cont.)
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Where, Dr = real dollars at some point in time
Da = actual dollars at the time it occurs f = inflation rate k = base time to determine real dollar
Relationship Real to Actualkn
ar fDD
1
1 knra fDD )1(
Relationship Real to ActualK-Corp is considering building a new ceramic mug line. Estimated values for construction, manufacturing and sales revenue are considered real (today’s dollars). Inflation is 3%.
Inflation 3.0%Year n Dr Factor Da
0 (750,000) 1.00 (750,000)1 200,000 1.03 206,0002 300,000 1.06 318,2703 400,000 1.09 437,0914 500,000 1.13 562,7545 600,000 1.16 695,564
Corporate Tax
Net Sales $ 10,000,000Less Expenses $ 5,000,000Gross Income $ 5,000,000Depreciation - 800,000Interest - 1,500,000Taxable Income $ 2,700,000
Tax = $ 113,900 + .35(2,700,000 - 335,000)= $ 941,650
Corporate Tax
Net Sales $ 10,000,000Less Expenses $ 5,000,000Depreciation - 800,000Interest - 1,500,000Taxable Income $ 2,700,000
Tax = $ 113,900 + .35(2,700,000 - 335,000)= $ 941,650
C
CDc
G
Cash Flow
Tax = (G – C – DC) x t , t = tax rate
FC = Cash Flow
= Net Income – tax + Depreciation = (G – C – DC) – tax + DC
= G – Gt – C + Ct + DCt
= (G – C)(1 – t) + tDC
Cash Flow Sale Price $2.50 COGS 30.0%
Depr 20.0% Admin 10.0% Tax 34.0%K-Corp Ceramic Mug Line
Yr 0 Yr 1 Yr 2 Yr 3 Yr 4 Yr 5Construction (600,000)Start Up Expense (150,000)Production 80,000 120,000 160,000 200,000 240,000Net Sales 200,000 300,000 400,000 500,000 600,000 COGS 60,000 90,000 120,000 150,000 180,000 Admin & Sales 20,000 30,000 40,000 50,000 60,000 Depreciation 120,000 120,000 120,000 120,000 120,000Income before Tax 0 60,000 120,000 180,000 240,000 Tax 0 20,400 40,800 61,200 81,600Net Income 0 39,600 79,200 118,800 158,400Add Depr 120,000 120,000 120,000 120,000 120,000Net Cash Flow (750,000) 120,000 159,600 199,200 238,800 278,400Cum Cash Flow (750,000) (630,000) (470,400) (271,200) (32,400) 246,000
Conversion to Cash Flow