Developed By:Dr. Don Smith, P.E.Department of Industrial EngineeringTexas A&M UniversityCollege Station, TexasExecutive Summary Version

Chapter 14 Effects of Inflation

LEARNING OBJECTIVESImpact of inflationPW with inflation consideredFW with inflation consideredAW with inflation consideredSomeone once said: Inflation is too much money chasing too few goods and services.

Sct 14.1 Understanding the Impact of Inflation Inflation is: an increase in the amount of money necessary to obtain the same amount of product or service before the inflated price was present. Social phenomena where too much money chases too few goods/services Impact because the value of the currency changes downward in value

Deflation Opposite of Inflation Deflation: when the value of the currency increases over time and produces increased value; In the future, smaller amounts of the currency can purchase the same amount of goods and services than it can presently. When deflation occurs nationally and for longer periods of time, money becomes less available (tighter). It has a very negative effect on the economy.

Equating ValueMoney in time period t1 can be related to money in time period t2 by the following:

The Inflation rate -- f The inflation rate, f, is a percent per time period; Stated in a manner similar to interest rates. Example: If f = 5.0% per year inflation, $100 today requires $105 to buy the same amount next year. Inflation is on top of the interest rate

The Basic Inflation Relationship Let n represent the amount of time between t1 and t2 then . . . .Future Dollars = Todays dollars(1+f)n Dollars in period t1 are termed:Constant-value dollars or todays dollars Dollars in time period t2 are termed:Future dollars or then-current dollars.

Example of InflationAssume a firm desires to purchase an asset that costs $209,000 in todays dollars. Assume a reasonable inflation rate of, say, 4% per year In 10 years, that same piece of equipment would cost:$209,000(1.04)10 = $309,371Does not count an interest rate or rate of return consideration.

Inflation can be Significant From the previous example we see that even at a modest 4% rate of inflation, the future impact on cost can and is significant! The previous example does not consider the time value of money. A proper engineering economy analysis should consider both inflation and the time value of money.

Three Important Rates Real or inflation-free interest rateDenoted as i. Inflation-adjusted interest rate Denoted as if Inflation rateDenoted as f In this chapter, i, if, and f are used extensively

Real or Inflation-free Interest Rate - i Rate at which interest is earned; Effects of any inflation have been removed; Represents the actual or real gain received/charged on investments or borrowed funds.

Sct 14.2 Present Worth Calculations Adjusted for Inflation In prior chapters, present worth was calculated assuming that all cash flows were in constant value dollars; Study Table 14-1 for an example of $5,000 inflated at 4% per year with a discount rate of 10% per year.

Comparison of $$ Values from Table 14-1

Derivation of a Combined Interest Rate We now derive if the inflation-adjusted interest rate. Start with:

Assume i is the real interest rate

Derivation of if - continued Assume F is a future dollar amount with inflation built in.P is:

Defining ifif is equal to:

if = (i + f + if)Then,

Sct 14.3 Future Worth Calculation Adjusted for Inflation This section deals with solving:F = f(i,if,n)There are four different interpretations for future worth calculations.1. Actual future $$;2. Purchasing power of future $$ stated in constant-value $$;3. Future $$ required at t = n to maintain t = 0 purchasing power;4. $$ at t = n to maintain purchasing power and earn a stated interest rate of i% per time period.

Case 1: Actual $$ at time t = n Solve: F = P(1+if)n = P(F/P,if,n) Apply the following equation: if = (i + f + if)

Case 1: Example P = $1,000 now and the market rate of interest is 10% per year. (if = 4%/year) Remember, the market rate of interest includes both the inflation rate and the discount rate.If n = 7, what is the future value of the $1,000 now?F = $1,000(F/P,10%,7) = $1,948.

Case 2: Purchasing Power in Constant-Value $$Calculate Fn using the market rate of interest.F7 = $1000(F/P,10%,7) =$1948.Deflate the F7 dollars at the inflation rate.F = $1948/(1.04)7= 1948/1.3159 = $1481$1481 is 24% less that the $1948;Inflation reduces the purchasing power by 24% over the 7-year period.Purchasing power in constant-value $$

Case 2: Equations Constant-value with purchasing power

Finding the REAL Interest Rate Given a market rate of interest; and, the inflation rate, f; Find the real interest rate.We know:if = i + f + ifSolve for i

Meaning of the Real Interest Rate The real rate, i Is the rate at which current $$ expand with their same purchasing power; Into equivalent future $$. If f > market rate if a negative real interest rate will result

Given if = 10% and f = 4% Find the real interest rate for this case. Find the real interest rate for a market rate of 10% and inflation at 4%. Solve:F = $1,000(F/P,5.77%,7) = $1481Inflation of 4%/year has reduced to real rate to less than 6% per year!

Case 3: Future Amount with No Interest With inflation, prices and costs increase over time; Future $$ are worth less out in time; At t = n, more $$ are needed; We only need to apply the inflation rate to the present sum;F = P(1+f)n For the $1000 example,F = $1,000(1.04)7 = 1,000(F/P,4%,7) = $1316

Case 4: Inflation and Real InterestHave an established and required MARR Objective:Maintain purchasing power, andAccount for time value of money Approach:Calculate if and apply the required equivalence formulas at the if rate.

Case 4: Apply the if interest rate Given P = $1,000 f = 4% Real interest rate of 5.77%; Calculate if as;if = 0.0577 + 0.04 + 0.0577(0.04) = 0.10 Then:F = $1000(F/P,10%,7) = $1948

Case 4: Exampleif = 0.0577 + 0.04 + 0.0577(0.04) = 0.10 Then:F = $1000(F/P,10%,7) = $1948

$1948 seven years out is equivalent to $1,000 now with a real return of 5.77% per year and inflation at 4% per year.

Setting the MARR Rate From Sct 10.2 and Sct 10.5Firms should set their MARR rate to:Cover the cost of capital;Cover or buffer the inflationary aspects perceived to exist;Account for risk.

Inflation-adjusted MARR Let MARRf = the inflation adjusted MARR Then define the MARRf as:MARRf = i + f + i(f) Thus;F = P(1 + MARRf)n = P(F/P,MARRf ,n)Simply another use of the time value of money relations applied to the MARR

Importance of Inflation Impacts Most countries inflation is from 2% to 8% per year; Some countries with weak currencies, political instability, poor balance of payments can have hyperinflation (as high as 100% per year).

When Hyperinflation is Present Spend money almost immediately; Money looses value quickly; Very difficult to perform engineering economy calculations in a hyper-inflated economy; Future values are unreliable and, Future availability of investment capital is very uncertain.

Sct 14.4 Capital Recovery Calculations Adjusted for Inflation With inflation present:Current dollars invested in a productive asset must be recovered over time with future inflated dollars.With loss of future purchasing power, future dollars will have less buying power than current dollars;More dollars will be required to recover a present investment is a productive asset.

Capital Recovery Analysis - Inflation Assume an investment of $1000 today in a productive asset. Assume inflation is 8% per year and a real interest rate of 10% is required. Assume a 5-year recovery period and S = $0Use if formula to determine inflation-adjusted interest rate of 18.8% per year

Capital recovery amount for AW analysis:A = 1000(A/P,18.8%,5) = $325.59 per year

More on Capital Recovery Analysis with InflationA = 1000(A/P,18.8%,5) = $325.59 per year (in future dollars). Annual equivalent of $1000 five years from now at 18.8% (inflation considered) is:A = 1000(A/F,18.8%,5) = $137.59 per year To accumulate F = $1000 at a real rate of 10% (without inflation) is:1000(A/F, 10%,5) = $163.80 per year

More on Capital Recovery Analysis with Inflation - continued Conclusion: For a fixed F value Uniformly distributed future costs should be spread over as long a period of time as possible;The leveraging effect of inflation will reduce the payment to $137.59 vs. $163.80.

Chapter Summary Inflation, when treated like an interest rate, makes the cost of the same product or service increase over time; This is due to the decreasing purchasing power of the currency when inflation is in effect.View in terms of:Todays dollars (constant-value); Future dollars (then-current);Important relationships:Inflated interest rateif = i + f + ifReal interest ratei = (if f)/(1+f)

Summary - continued PW of F with inflationP = F(P/F,if ,n)Future worth of P in constant-value dollars with the same purchasing powerF = P(F/P,i,n) F to cover a current amount with no interestF = P(F/P,f,n)

Summary continued F to cover a current amount with interestF = P(F/P,if ,n) Annual equivalent A of a future dollar amountA = F(A/F,if ,n) Annual equivalent A of a present amount in future dollarsA = P(A/P,if ,n)

Chapter 14End of Set

At this point:1. Introduce yourself - your students are likely to want to know something about your qualifications and interests - overall, where you are coming from.2. Have students introduce themselves. Ask why they are taking this class. If you are fortunate enough to have a Polaroid camera, take pictures of each student for later posting on a class board so both they and you get to know each other.3. Discuss both choice of textbook and development of syllabus.4. If you are expecting students to work in teams, at east introduce the choice of team members. If at all possible, have students participate in a team building or team study exercise. It works wonders. Most student have been told to work in teams in prior classes, but have never examined exactly what a team is and how it works. One hour spent in a team building/examination exercise saves many hours and avoids many problems later on.