Engineering Economy

IEN255 Chapter 11 - Inflation

Inflation vs Deflation

Inflation - a loss in purchasing power of money over time

Deflation - prices decrease of time

Measuring inflation - Consumer Price Index (CPI)

Survey of Current Business - for industrial products

Engineering Economy

Average inflation rate (f)

Step 1 - simple compounding $100(1 + 0.04)(1 + 0.08) = $112.32

Step 2 - equiv eqn $100(1 + f)2 = $112.32 solve for f

Engineering Economy

General inflation rate vs Specific

General Inflation Rate ( ) - average inflation rate calculated based on the CPI

Specific Inflation Rate (fj) - inflation rate calculated based on a specific segment j of the economy.(if just one item, drop the subscript j)

equations 11.1 and 11.2 with defs

f

Engineering Economy

Example 11.2

Engineering Economy

Actual vs Constant Dollars

Actual (current) dollars - estimates of future cash flows for year n that take into account any anticipated changed in amount due to inflation

Constant (real) dollars - constant purchasing power independent of time (adjusted using an accepted general inflation rate)

To convert use 11.4 where A’n = constant dollar expression for the cash

flow where An = actual dollar expression for the cash

flow

),,/(')1(' nfPFAfAA nn

nn

Engineering Economy

Example 11.3

Engineering Economy

Figure 11.1

Engineering Economy

Conversion from Actual to Constant Dollars

(11.5)

example 11.4

),,/()1(

'' nfFPA

f

AA nn

nn

Engineering Economy

Equivalence calculations under inflation

Two TypesMarket Interest Rate (i) (typically inflation-

adjusted MARR)Inflation-free Interest Rate (i’)- real interest

rate (prev chaps)

Case 1: All in constant dollars Case 2: All in actual dollars Case 3: Mix (convert all to either case 1 or 2)

Engineering Economy

Constant Dollar Analysis

Example 11.5

Engineering Economy

Actual Dollar Analysis

Deflation method

copy page 573

Engineering Economy

Actual Dollar Analysis cont

Adjusted discount method

example 11.7

figure 11.3

Engineering Economy

Mixed Dollar Analysis

Page 577 and 578

Engineering Economy

Effects of Inflation on Project Cash Flows

Depreciation Expenses under Inflation

page 581

Engineering Economy

Multiple Inflation Rates

Table 11.3

Engineering Economy

Multiple Inflation Rates

Solution on page 583

Engineering Economy

Effects of Borrowed Funds under Inflation

Solution on page 584

Engineering Economy

Effects of Borrowed Funds under Inflation

Page 585

Engineering Economy

Rate of Return Analysis Under Inflation

Table 11.5

Engineering Economy

Rate of Return Analysis Under Inflation

RoR without inflation PW(i’) = -30,000+12,700(P/F,i’,1) + 14,033

(P/F,i’,2)+ 10,478 (P/F,i’,3) + 9,589 (P/F,i’,4)

Solve for IRR’ = i’* = 21.88%

RoR under inflation PW(i’) = -30,000+12,336(P/F,i’,1) + 13,108

(P/F,i’,2)+ 10,036 (P/F,i’,3) + 9,307 (P/F,i’,4)

Solve for IRR’ = i’* = 19.40%

Engineering Economy

Rate of Return Analysis Under Inflation

put table 11.5 and 11.6 on the same page next to each other (shrink if necessary)

Engineering Economy

Effect of Inflation on Working Capital

Figure 11.6

Engineering Economy

Table 11.7

Engineering Economy

IEN255 Summer’99 Chapter 10,11,13 HW#4

Homework Assignment:Chapter 10#’s 10.4; 10.6; 10.24Chapter 11#’s 11.1; 11.6; 11.13Chapter 13#’s 13.9

Due together (Tues July 27)