1 / 15

Inflation / Deflation

Inflation / Deflation. Inflation is an increase over time in the price of a good or service with a constant value A gallon of 87 octane gasoline increases in price (but still only gets Dr. J. 32 miles down the road)

malcolm-gamble
Download Presentation

Inflation / Deflation

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Inflation / Deflation • Inflation is an increase over time in the price of a good or service with a constant value • A gallon of 87 octane gasoline increases in price • (but still only gets Dr. J. 32 miles down the road) • Deflation is a decrease over time in the price of a good or service of constant value. • A 2 GHz computer has decreased in price • (but still does the same number of computations/min.)

  2. Other Examples: • Inflationary • Tuition • Fees • Books • Industry Salaries • Cars • Gas • Deflationary • CPU Memory • Computers • Constant • Bread • Milk

  3. Food & Drink • Housing • Clothing • Transportation • Medical Care • Entertainment • Personal Care • Other Goods / Services Average Inflation Rate (f) • In most engineering econ problems, different items will have different inflation rates • Average Inflation Rate is based on a market basket of goods • – CPI or Consumer Price Index • Simplifies cash flow in an analysis!

  4. Deriving Equation for Inflation P = 50,000 f = 10% increase annually F1 = 50,000 + 50,000 (.10) = 50,000 (1 + .10) = 55,000 F2 = 55,000 + 55,000 (.10) = 55,000 (1 + .10) = 50,000 (1 + .10)(1 + .10) = 50,000 (1 + .10)2 = 60,500 F3 = 60,500 + 60,500 (.10) = 50,000 (1 + .10)2(1 + .10) = 50,000 (1 + .10)3 = 66,550 Generally: Fn = P (1 + f )n

  5. Constant Dollars vs. Actual Dollars • Constant Dollars – represent constant purchasing power independent of the passage of time. • Actual Dollars – an estimate of a future cash flow for Year n that take into account any anticipated changes in amount caused by inflationary or deflationary effects.

  6. $A = $C (1+ f )n $C = $A (1+ f )– n 1980 2012 ACTUAL $ $785 / QTR Tuition & Fees $785 (1+ f )32 / QTR Tuition & Fees Constant Dollars vs. Actual Dollars 1980 2012 CONSTANT $ $785 / QTR Tuition & Fees $785 / QTR Tuition & Fees f = Average Inflation Rate

  7. Incorporating Inflation • Inflation can be accounted for as an additional component on top of the interest rate: • d = i + f + i•f(d replaces i in tables/equations) • where: • i is the effective interest rate • f is the constant inflation rate

  8. Real Life Examples  OSU research proposal budgets were supposed to contain a 4% cost increase factor for each successive year. e.g. If I have a grad student helping with my research, for each year I employ her, I will need to figure a 4% increase in her stipend, her tuition waiver, her insurance contribution, … What happens if we use a 4% average inflation rate for some common student-oriented prices?

  9. 1979 2012 $0.799 / gal $0.799 (1+ 0.04 )33 = $2.92 / gal F33 = $0.799 (F/P, 4%, 30) (F/P, 4%, 3) = $0.799 (3.2434) (1.1249) = $2.92 / gal Example 1 Dr. J. used a motorcycle to get to college. At the Fall of 1979, the cost of a gallon of gas was 79.9 ¢. What should the cost have been in 2012 at a 4% average annual rate of inflation? Find: F33 (19792012) Given: P = $0.799 / gal f = 4% $3.35 / gal Hardrockn Depot (cash price) July 11, 2012 Actual average inflation rate was 4.4 % / YR

  10. 1980 2012 $66.50 / Cr Hr $66.50 (1+ .04 )32 = $233 / Cr Hr F32 = $66.50 (F/P, 4%, 30) (F/P, 4%, 2) = $66.50 (3.2434) (1.0816) = $233.29 / Cr Hr Example 2 Dr. J. went to Iowa State University for a B.S. in Computer Engineering. In Fall 1980, his fee card said he had to pay $665 for 15 credit hours of quarterly credit (which converts to $66.50 / sem cr hr). What would that cost be this year at a 4% average annual rate of inflation? Find: F32 (1980 2012) Given: P = $66.50 / Cr Hr f = 4% $114.30 / Cr Hr +$62.40 / Cr Hr (Engineering Fee) = $176.70 / Cr Hr 233.85 / Cr Hr (non-resident) SDSMT Catalog 2011-2012 Actual average inflation rate was 3.2 % / YR at SDSMT, and 5.5 % / YR at ISU

  11. 1980 2012 $254.20 (1+ .04 )– 32= $72.46 / QTR F32 = $254.20 (P/F, 4%, 30) (P/F, 4%, 2) = $254.20 (.3083) (.9246) = $72.46 / QTR (2.4% inflation rate) $254.20 / QTR (381.30 / sem) SDSMT Fee Descriptions 2011/2012 Example 3 At Iowa State University in Fall 1980, Dr. J’s fee card showed a $120 technology fee. SDSMT’s tablet fee was $381.30 for academic year 2011-2012 (or $254.20 / QTR). What might the SDSMT fee have been back in 1980, using a 4% average annual rate of inflation? Find: P(2010 1980) Given: F32 = $254.20 / QTR f = 4% $120 / QTR paid for programmable TI calculators that were bolted to the tables at ISU… Looks like 4% inflation rate is roughly right for prices, but maybe pay rates don’t keep up…

  12. $25 000 / YR 1986 2012 $25 000 (1+ 0.04 )26 = $69 312/ YR F26 = $25 000 (F/P, 4%, 25)(F/P, 4%,1) = $25 000 (2.6658)(1.0400) = $69 311/ YR Example 4 Starting industry salary for Dr. J. as a computer engineer in 1986 was $25 000. Assuming a 4% average annual rate of inflation, what should the starting salary have been at graduation in 2012? $65 370/ YR BLS low end for SD CpE May 2011 Find: F26 (19862012) Given: P = $25 000 / YR f = 4% Computer Engineering salaries increased at a 3.92% inflation rate…

  13. $5.50 / HR 1981 $5.50 (1+ 0.04 )31 = $18.55 / HR F31 = $5.50 (F/P, 4%, 30)(F/P, 4%,1) = $5.50 (3.2434)(1.0400) = $18.55 / HR 2012 Example 5 In 1981 Dr. J. worked as a rodman / chainman for a land surveyor. Intern-type pay was $5.50 / HR then. Again, using the 4% average annual rate of inflation, what would the equivalent intern wage have been in the Summer of 2012? $19.67 / HR average pay for Interns Summer 2011 Find: F31 (1981 2012) Given: P = $5.50 / HR f = 4% … and Engineering Intern salaries beat the 4% inflation rate … 4.2% average rate

  14. $3.35 / HR 1981 2012 $3.35 (1+ 0.04 )31 = $11.30 / HR F31 = $3.35 (F/P, 4%, 30) (F/P, 4%, 1) = $3.35 (3.2434)(1.0400) = $11.30 / HR Example 6 In 1976 Dr. J. got his first job sweeping floors at an aerial photography firm. The National Minimum Wage rose to $3.35 / HR in 1981, when he quit. Using a 4% average annual rate of inflation, what should the equivalent minimum wage be in the Summer of 2012? National Min. is $7.25 / HR 01 Jan 2012 $9.04 / HR Washington Min. Wage Law for 2012 (Highest Nationally) Find: F31 (1981 2012) Given: P = $3.35 / HR f = 4% Actual average inflation rate was 2.52% / YR

  15. Concluding Thoughts • Perhaps it isn’t that the cost of gas, tuition, fees, or engineering salaries have risen outrageously since • “the good old days”… • Our analysis shows one factor is that the high school degree, entry-level wages don’t keep up. • Important to finish your degree quickly!

More Related