1. IENG 217 Cost Estimating for Engineers Inflation

2. Suppose the price of copper is $1,000/ton and price rises by 10% per year. In 5 years, Price = 1000(1.1) 5 = $1,610.51 But we still only have 1 ton of copper

3. 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 = $1,610.51 But we still only have 1 ton of copper $1,610 5 years from now buys the same as $1,000 now

4. 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 = $1,610.51 But we still only have 1 ton of copper $1,610 5 years from now buys the same as $1,000 now 10% inflation

5. 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 = $1,610.51 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)

6. 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.

7. 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

8. Combined Interest Rate That is $1.10 = 1.05 (1+d) 1 1(1+.10) = 1(1+.05)(1+d)

9. Combined Interest Rate That 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

10. Combined Interest Rate That 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 rate d = real interest rate (after inflation rate)

11. Solving for d, the real interest earned after inflation, where i = interest rate (combined) f = inflation rate d = real interest rate (after inflation rate) Combined Interest Rate f fi d    1

12. Example Suppose we place $10,000 into a retirement account which earns 10% per year. How much will we have after 20 years?

13. 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

14. Example (cont.) How much is $67,275 20 years from now worth if the inflation rate is 3%?

15. Example (cont.) How much is $67,275 20 years from now worth if the inflation rate is 3%? Solution: F T = 67,275(P/F,3,20) = 67,275(1.03) -20 = $37,248

16. Alternate: Recall = (.1 -.03)/(1+.03) =.068 Example (cont.) j ji d    1

17. Alternate: Recall = (.1 -.03)/(1+.03) =.068 F R = 10,000(1+d) 20 = 10,000(1.068) 20 = $37,248 Example (cont.) j ji d    1

18. Alternate: Recall = (.1 -.03)/(1+.03) =.068 F R = 10,000(1+d) 20 = 10,000(1.068) 20 = $37,248 Note: This formula will not work with annuities. Example (cont.) j ji d    1

19. Where, D r = real dollars at some point in time D a = actual dollars at the time it occurs f = inflation rate k = base time to determine real dollar Relationship Real to Actual

20. K-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%.

21. Corporate Tax Net Sales $ 10,000,000 Less Expenses$ 5,000,000 Gross Income$ 5,000,000 Depreciation- 800,000 Interest - 1,500,000 Taxable Income$ 2,700,000 Tax= $ 113,900 +.35(2,700,000 - 335,000) = $ 941,650

22. Corporate Tax Net Sales $ 10,000,000 Less Expenses$ 5,000,000 Depreciation- 800,000 Interest - 1,500,000 Taxable Income$ 2,700,000 Tax= $ 113,900 +.35(2,700,000 - 335,000) = $ 941,650 C C DcDc G

23. Cash Flow Tax = (G – C – D C ) x t, t = tax rate F C = Cash Flow = Net Income – tax + Depreciation = (G – C – D C ) – tax + D C = G – Gt – C + Ct + D C t = (G – C)(1 – t) + tD C

24. Cash Flow

25. Conversion to Cash Flow