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    Lecture 5: Inflation: Engineering Economy

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    Lecture 5 discusses inflation and its impact on economic analyses. Inflation decreases the purchasing power of currency over time. There are two main ways to account for inflation: 1) convert cash flows to constant value dollars using the inflation rate and real interest rate, or 2) use the market interest rate, which includes inflation. Different interest rates must be used depending on whether the analysis considers constant dollars, future dollars, or real rates of return.

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    Lecture 5: Inflation: Engineering Economy

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    Qed Vio
    83 views37 pages
    Lecture 5 discusses inflation and its impact on economic analyses. Inflation decreases the purchasing power of currency over time. There are two main ways to account for inflation: 1) convert cash flows to constant value dollars using the inflation rate and real interest rate, or 2) use the market interest rate, which includes inflation. Different interest rates must be used depending on whether the analysis considers constant dollars, future dollars, or real rates of return.

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    Lecture 5 2016_Ramzan

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    Lecture 5 discusses inflation and its impact on economic analyses. Inflation decreases the purchasing power of currency over time. There are two main ways to account for inflation: 1) convert cash flows to constant value dollars using the inflation rate and real interest rate, or 2) use the market interest rate, which includes inflation. Different interest rates must be used depending on whether the analysis considers constant dollars, future dollars, or real rates of return.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    83 views37 pages

    Lecture 5: Inflation: Engineering Economy

    Uploaded by

    Qed Vio
    Lecture 5 discusses inflation and its impact on economic analyses. Inflation decreases the purchasing power of currency over time. There are two main ways to account for inflation: 1) convert cash flows to constant value dollars using the inflation rate and real interest rate, or 2) use the market interest rate, which includes inflation. Different interest rates must be used depending on whether the analysis considers constant dollars, future dollars, or real rates of return.

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    © All Rights Reserved
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    of 37

    Lecture 5 : Inflation

    Lecture slides to accompany

    Engineering Economy
    7th edition

    Leland Blank
    Anthony Tarquin
    LEARNING OUTCOMES

    • Relationship between Inflation and Purchasing Power

    • Constant Value Dollar and Purchasing Power

    • 3 Different Ways of Stating an Interest Rate


    • Calculate PW, FW and AW of cashflows with inflation
    considered.
    Understanding Inflation
    Inflation: Increase in amount of money needed to purchase same amount
    of goods or services. Inflation results in a decrease in purchasing power,
    i.e., one unit of money buys less goods or services

    • Purchasing power is a good indicator for the real value of money (real
    money)
    • Increase in money supply (related very much with the interest rate dictated by
    the central bank) - As the value of money decreases (because of its
    abundancy), it takes more money for the same amount of goods or services.
    • Reduce or stagnant productivity i.e. constant amount of money, less goods
    Constant Value Dollar
    The money that maintain its value over time i.e represents the same
    purchasing power over time.
    Constant Value (Real Dollar) Actual Dollar
    Year RM
    0 RM 100 RM 100
    1 RM 100 RM 100 (1+0.05)
    2 RM 100 RM 100 (1+0.05)2

    Say we have an inflation of 5% Give more to compensate for the


    inflation (decrease value of
    money)
    Constant Value Dollar Definition
    An adjusted value of currency used to compare dollar values from one
    period to another. Due to inflation, the purchasing power of the dollar
    changes over time, so in order to compare dollar values from one year to
    another, they need to be converted from nominal (current) dollar values to
    constant dollar values.

    Constant dollar value may also be referred to as real dollar or today’s


    dollar (present dollar).
    Money Time Value due to the Inflation

    Money in one period of time t1, can be brought to the same value as
    money in another period of time t2 by using;

    𝑎𝑚𝑜𝑢𝑛𝑡 𝑖𝑛 𝑝𝑒𝑟𝑖𝑜𝑑 𝑡2
    𝐴𝑚𝑜𝑢𝑛𝑡 𝑖𝑛 𝑝𝑒𝑟𝑖𝑜𝑑 𝑡1 = (𝐄𝐪. 𝟏)
    𝑖𝑛𝑓𝑙𝑎𝑡𝑖𝑜𝑛 𝑟𝑎𝑡𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡1 𝑎𝑛𝑑 𝑡2

    Lets use dollar as the currency.


    Dollars in period t1are called constant value dollar (present dollar).
    Dollars in period t2 are called future dollars which have taken
    inflation into account.
    If f is the inflation rate per period, and t is the number of period between
    time t1 and t2.

    Equation (Eq. 1) can be rewrite as;

    𝐹𝑢𝑡𝑢𝑟𝑒 𝐷𝑜𝑙𝑙𝑎𝑟𝑠
    𝐶𝑜𝑛𝑠𝑡𝑎𝑛𝑡 𝑉𝑎𝑙𝑢𝑒 𝐷𝑜𝑙𝑙𝑎𝑟𝑠 = 𝑡
    (𝐄𝐪. 𝟐)
    (1 + 𝑓)

    𝐹𝑢𝑡𝑢𝑟𝑒 𝐷𝑜𝑙𝑙𝑎𝑟𝑠 = Constant Value Dollars (1 + 𝑓)𝑡 (𝐄𝐪. 𝟑)

    𝑭𝑽 = 𝑷𝑽(𝟏 + 𝒓)𝒕
    Three Different Interest Rates
    ► Real accumulation of wealth or interest rate i – Rate at which
    interest is earned when effects of inflation are removed; i represents
    the real increase in purchasing power

    ►Inflation rate f – Rate of change in value of currency

    ►Market or inflation-adjusted rate if – Rate that takes inflation into


    account. Commonly stated rate everyday

    if = i + f + (i)(f) Please memorize this equation


    Example: Constant Value Dollars
    How much would be required today to purchase an item that increased in cost
    by exactly the inflation rate? The cost 30 years ago was $1000 and inflation has
    consistently averaged 4% per year.
    Solution: Solve for future dollars

    Future dollars = constant value dollars(1 + f)t


    = 1000(1 + 0.04)30
    = $3243

    Note: This calculation only accounts for the decreased purchasing power
    of the currency. It does not take into account the time value of money
    (to be discussed)
    Example: Market vs. Real Rate
    Money in a medium-risk investment makes a guaranteed 8% per
    year. Inflation rate has averaged 5.5% per year. What is the real
    rate of return on the investment?

    Solution
    Solve for the real rate i in relation for if
    if = i + f + (i)(f)
    𝑖𝑓 − 𝑓 0.08 − 0.055
    𝑖= =
    1+𝑓 1 + 0.055

    𝑖 = 0.024 = 2.4%
    Two GENERAL ways to work problems when considering
    inflation:

    (1) Convert to constant value (CV) dollars, then use real rate i.
    If f = inflation rate (% per year), the equation is:
    Constant-value dollars = future dollars = then-current dollars
    (1+ f)t (1+ f)t

    (2) Leave money amounts as is and use market rate (interest rate
    adjusted for inflation), if
    if = i + f + (i)(f)
    PW Calculations with Inflation
    Two ways to account for inflation in PW calculations

    (1) Convert cash flow into constant-value (CV) dollars and use regular i
    where: CV = future dollars/(1 + f)t = then-current dollars/(1 + f)t
    f = inflation rate
    (Note: Calculations up to now have assumed constant-value dollars)

    (2) Express cash flow in future (then-current ) dollars and use market
    interest rate where if = i + f + (i)(f)
    Example: PW with Inflation
    A honing machine will have a cost of $25,000 (future cost) six years from
    now. Find the PW of the machine, if the real interest rate is 10% per year
    and the inflation rate is 5% per year using (a) constant-value dollars,
    and (b) future dollars.

    Solution: (a Determine constant-value dollars and use i in PW equation

    CV = 25,000 / (1 + 0.05)6 = $18,655


    PW = 18,655(P/F,10%,6)
    = $10,530

    (b) Leave as future dollars and use if in PW equation


    if = 0.10 + 0.05 + (0.10)(0.05) = 15.5%
    PW = 25,000(P/F,15.5%,6)
    = $10,530
    FW Calculations with Inflation

    FW values can have four different interpretations


    (1) The actual amount accumulated
     Use if in FW equation FW = PW(F/P, if, n)
    (2) The purchasing power in terms of CV dollars of the future amount
     Use if in FW equation and divide by (1+f)t or use real i
    where real i = (if – f)/(1 + f) FW = PW(F/P,i,n)
    (3) The number of future dollars required to have the same purchasing
    power as a dollar today with no time value of money considered
     Use f instead of i in F/P factor FW = PW(F/P,f,n)
    (4) The amount required to maintain the purchasing power of the present
    sum and earn a stated real rate of return
     Use if in FW equation FW = PW(F/P, if, n)
    Example: FW with Inflation
    An engineer invests $15,000 in a savings account that pays interest at a
    real 8% per year. If the inflation rate is 5% per year, determine (a) the amount
    of money that will be accumulated in 10 years, (b) the purchasing power of
    the accumulated amount (in terms of today’s dollars), (c) the number of
    future dollars that will have the same purchasing power as the
    $15,000 today, and (d) the amount to maintain purchasing power and earn a
    real 8% per year return.

    Solution:
    (a) The amount accumulated is a function of the market interest rate, if
    if = 0.08 + 0.05 + (0.08)(0.05) = 13.4%

    Amount Accumulated = 15,000(F/P,13.4%,10)


    = $52,750
    Example: FW with Inflation (cont’d)
    (b) To find the purchasing power of the accumulated amount deflate
    the inflated dollars

    Purchasing power = 15,000(F/P,13.4%,10) / (1 + 0.05)10


    = $32,384
    (c) The number of future dollars required to purchase goods that cost
    $15,000 now is the inflated cost of the goods

    Number of future dollars = 15,000(F/P,5%,10)


    = $24,434

    (d) In order to maintain purchasing power and earn a real return, money must
    grow by the inflation rate and the interest rate, or if = 13.4%, as in part (a)

    FW = 15,000(F/P,13.4%,10)
    = $52,750
    AW (Capital Recovery) with Inflation

    If a small company invests $150,000 in a new production line machine,


    how much must it receive each year to recover the investment in 5
    years? The real interest rate is 10% and the inflation rate is 4% per year.

    The A/P and A/F factors require the use of if when inflation is considered

    Solution: Capital recovery (CR) is the AW value


    if = 0.10 + 0.04 + (0.10)(0.04) = 14.4%
    CR = AW = 150,000(A/P,14.4%,5)
    = $44,115 per year
    Summary of Important Points
    Inflation occurs because value of currency has changed

    Inflation reduces purchasing power; one unit buys less


    goods ort services
    Two ways to account for inflation in economic analyses:
    (1) Convert all cash flows into constant-value dollars and use i
    (2) Leave cash flows as inflated dollars and use if

    During deflation, purchasing power of money is greater in future than


    at present
    Future worth values can have four different interpretations,
    requiring different interest rates to find FW
    Use if in calculations involving A/P or A/F when inflation is
    considered
    14-18
    14.1 (a) There is no difference.
    (b) Today’s dollars are inflated compared to dollars of 2 years ago.
    Therefore, in order
    for the dollars to have the same value (i.e., constant-value dollars) as 2
    years ago,
    divide today’s dollars by (1 + f)2.

    14.2 (a) During periods of inflation


    (b) During periods of deflation
    (c) When inflation is zero

    14.3 0.10 = 0.04 + f + 0.04f


    1.04f = 0.06
    f = 0.0577 or 5.77% per year

    14.4 if = 0.20 + 0.05 + (0.20)(0.05)


    = 0.26 or 26%
    14.5 if per month = 0.30/12 + 0.015 + (0.30/12)(0.015)
    = 0.040375 or 4.0375% per month
    Nominal if per year = 12(4.0375)
    = 48.45% per year

    14.6 0.35 = 0.25 + f + 0.25f


    1.25f = 0.10
    f = 0.08 or 8% per year

    14.7 if = 0.04 + 0.01 + (0.04)(0.01)


    = 0.0504 or 5.04% per quarter
    14.8 if per month = 18/12 = 1.5%
    Use inflation-adjusted interest rate equation to solve for i.
    0.015 = i + 0.005 + (i)(0.005)
    1.005i = 0.01
    i = 0.00995 or 0.995% per month

    14.9 Let CV = constant-value dollars


    CV1 = 45,000/(1 + 0.05)1 = $42,857
    CV2 = 45,000/(1 + 0.05)2 = $40,816
    CV3 = 45,000/(1 + 0.05)3 = $38,873
    CV4 = 45,000/(1 + 0.05)4 = $37,022
    14.10 Future, inflated dollars = 10,000(1 + 0.05)10 = $16,289
    14.11 Number of future dollars required = 1,500,000(1 + 0.04)30
    = $4,865,096
    14.12 Assume C1 is the cost today
    2C1 = C1(1 + 0.07)n
    (1 + 0.07)n = 2.000
    n log 1.07 = log 2.000
    n = 10.2 years
    14.13 0.28 = i + 0.06 + i(0.06)
    1.06i = 0.22
    i = 0.2075 or 20.75%
    14.15 Buying power = 250,000/(1 + 0.04)5
    = $205,482
    14.16 (a) Constant-value dollars have to increase by only the real
    interest rate of 5% per year.
    CV5 = 30,000(F/P,5%,5)
    = 30,000(1.2763)
    = $38,289
    (b) if = 0.05 + 0.04 + (0.05)(0.04)
    = 9.2%
    F = 30,000(F/P,9.2%,5) = 30,000(1.55279)
    = $46,584
    14.17 Find f using F/P or P/F factor
    5400 = 4050(F/P,f,5)
    (F/P,f,5) = 1.3333
    By factor equation
    (1 + f)5 = 1.3333
    1 + f = 1.33330.2
    1 + f = 1.0592
    f = 0.0592 or 5.92% per year
    14.24 The two ways to account for inflation in PW calculations are:
    (1) Convert all cash flow amounts into constant-value (CV) dollars,
    and
    (2) Change the interest rate to consider inflation, that is, to account for
    the changing
    currency value.
    14.25 if = 0.10 + 0.04 + (0.10)(0.04)
    = 14.4%
    PW = 50,000(P/F,14.4%,2)
    = 50,000[1/(1.144)2]
    = $38,205
    14.26 if = 0.10 + 0.04 + (0.10)(0.04)
    = 14.4%
    PW = 125,000(P/F,14.4%,3)
    = 125,000(0.66792)
    = $83,490
    14.35 if = 0.10 + 0.06 + (0.10)(0.06)
    = 16.6% per year
    F = 10,000(F/P,16.6%,10)
    = 10,000(1 + 0.166)10
    = $46,450
    14.36 Find F in future dollars using f = -3.0%
    F = 50,000(1 – 0.03)5
    = 50,000(0.85873)
    = $42,937
    14.37 Purchasing power = 100,000(F/P,10%,15)/(1 - 0.01)15
    = 100,000(4.1772)/0.86006
    = $485,687

    14.38 Buying power = 60,000(F/A,10%,5)/(1 + 0.04)5


    = 60,000(6.1051)/
    14.47 Calculate amount needed at 5% inflation rate
    and then find A using market rate.
    F = 72,000(1 + 0.05)3
    = 72,000(1.1576)
    = $83,347
    A = 83,347(A/F,12%,3)
    = 83,347(0.29635)
    = $24,700 per year
    14.50 if = 0.12 + 0.03 + (0.12)(0.03)
    = 15.36%
    A = -3,700,000(A/P,15.36%,5)
    = -3,700,000(0.30086)
    = $-1,113,182 per year
    14.51 if = 0.10 + 0.04 + (0.10)(0.04)
    = 14.4% per year
    A = -40,000(A/P,14.4%,3) - 24,000 + 6000(A/F,14.4%,3)
    = -40,000(0.43363) - 24,000 + 6000(0.28963)
    = $-39,607 per year

    14.52 if = 0.09 + 0.03 + (0.09)(0.03)


    = 12.27% per year

    A = -180,000(A/P,12.27%,5) - 70,000(P/F,12.27%,3)(A/P,12.27%,5)
    = -180,000(0.27927) - 70,000(0.70666)(0.27927)
    = $-64,083 per year

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