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VIII
INFLATION
Inflation deals with the purchasing power of dollars. Over a period of time, cost of most goods and
services increase. As an example, following data shows approximate prices paid for a typical lunch in a fast
food restaurant over a period of 25 years.
1974
1984
1994
Hamburger
0.30
0.45
0.59
French fries
0.25
0.40
0.55
Soft drink
0.30
0.45
0.69
The increase in the amount paid for the hamburger, 0.30 to 0.59, is due to inflation as we have to pay
$0.29 more to buy the same hamburger in 1994 than in 1974.
As stated earlier, in general, we have to pay more to buy same goods or services but this is not true
for all items. Personal computers are a good example. Prices of computers is actually declining. Same is true
for many electronic goods; color TVs and VCRs etc.
8.1
MEASURING INFLATION
Inflation is measured by changes in the consumer price index, CPI. The consumer price index, and
inflation rates are reported annually by the Federal Government. The consumer price index is calculated
by determining the cost of a basket of same items, approximately one thousand, periodically. The cost of the
basket, relative to the cost of the same basket at some fixed point in time, is the CPI. Table 8.1.1 shows CPI
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values with a base value of 100 in between 1983 and 1984. The cost of this basket at the end of 1988 was
118.3. Producer price index are also reported by the Federal government and can be used but CPI is more
commonly used in determining inflation rates.
Table 8.1.1
CPI and Annual Inflation Rates
Year
CPI
1983
1984
1985
1986
1987
1988
1989
1990
1991
1992
1993
1994
99.6
103.9
107.6
109.6
113.6
118.3
124.0
130.7
136.2
140.3
144.5
148.2
Inflation Rate
3.21
4.32
3.56
1.86
3.65
4.14
4.82
5.40
4.21
3.01
2.99
2.56
Source: Bureau of Labor Statistics, Washington D.C., US Govt. Printing Office
The annual inflation rate, for year (t+1), is calculated as follows:
Inflation Rate for year (t+1) = {CPI(t+1) - CPI(t)}/{CPI(t)}
(8.1.1)
For example, the inflation rate for 1988 is calculated as follows:
Inflation Rate for year 1988 = {CPI(1988) - CPI(1987)}/{CPI(1987)
= {118.3 - 113.6}/{113.6}
= 4.14%
In engineering economy studies, as the life of projects is generally more than one year, average
inflation rate for the life of a project is used. This simplifies calculations. The average inflation rate, f, can
be calculated as follows:
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CPIf = CPIi (F/P, f, N) = CPIi (1+ f )N
(8.1.2)
where
CPIf = Consumer price index at the end of the final year
CPIi = Consumer price index at the start of the initial year
The value of CPIf, in Eq. (8.1.2), is found by using the factor (F/P, i, N). Here, P = CPIi is known
and F = CPIf is to be determined. Appropriate interest rate is the inflation rate, f. As both CPIf and CPIi are
known, Eq.(8.1.2) can be solved for the unknown average inflation rate, f. Average inflation rate for three
years, 1985 to 1986, and 1987 is calculated as follows:
CPIi = CPI(1984) = 103.9
CPIf = CPI(1987) = 113.6
Substituting these values in Eq. (8.1.2), we get
113.4 = 103.9(1 + f )3
or
8.2
f = 3.01%
ACTUAL AND CONSTANT DOLLARS
Actual are constant dollars are defined as follows:
Actual Dollars: These are out of pocket dollars used in daily transactions. Examples are: pay $350
this month on May 15th for car payment; deposit $200 in a savings account today.
Constant Dollars: These are hypothetical dollars which have same buying power at some base
point in time as future actual dollars.
If annual inflation rate is known then actual dollars can be converted to constant dollars and vice
versa. If f is the annual inflation rate (average inflation rate) and elapsed time period between actual and
constant dollars is N years, then constant dollars can be determined by following equation
Constant dollars = Actual dollars (1 + f ) -N
(8.2.1)
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or
Constant dollars = Actual dollars (P/F, f, N)
Eq.(8.2.1) can be solved for finding actual dollars as follows:
Actual dollars = Constant dollars (1 + f )N
or
(8.2.2)
Actual dollars = Constant dollars (F/P, f, N)
Following examples illustrate the meaning and conversion of actual and constant dollars.
Example 8.2.1
Cost of a new standard three bedroom home in a subdivision in 1995 is $96,750.
What was the cost of this home in 1991? Assume annual inflation rate of 5% for such homes in this
subdivision.
Solution: We need to find constant dollar equivalent of $96,750 with base in 1991. The cost of home
in 1995 is in actual dollars and the elapsed time between actual dollars (1995) and constant dollars (1991)
is four years. Using Eq.(8.2.1) we get
Constant dollars = Actual dollars (1 + f)-N
= 96,750(1 + 0.05) -4
= $79,596.23
The cost of purchasing this home in 1991 was $79,596.23 and it is lower. The increase to $96,750
in 1995, four years, is due to inflation at the rate of 5% per year. In other words, it takes only 79,596.23
constant dollars or (1991) dollars to buy the home which costs $96,750 in 1995 (now).
Example 8.2.2 A person is planning to a buy a new car in two years. The cost of the car now
(1992) is $12,500. How much the person will expect to pay for this same car in two years if annual inflation
rate is 4%?
Solution: We need to find the actual dollar equivalent of the cost of the new car. Using Eq.(8.2.2)
Actual dollars = Constant dollars (1 + f)N
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= Constant dollars (F/P, f, N)
= 12,500 (F/P, 4, 2)
= 12,500(1.082)
= $13,525.00
The person will have to pay $13,525.0 (actual dollars) in two years to buy the same car.
It should be pointed out that the constant dollar amount is same as long as the base point remains the
same. For Example 8.2.2, the base point is 1992. The cost of car is $12,500 in 1992. In 1992, actual and
constant dollars both are $12,500. The actual and constant dollar (base 1992) cost of this car for several
years is shown below:
Year
Constant Dollar
Actual Dollar
1992
$12,500
$12,500
1993
12,500
12,500(F/P, 4, 1) = 13,000.00
1994
12,500
12,500(F/P, 4, 2) = 13,525.00
1995
12,500
12,500(F/P, 4, 3) = 14,062.50
Note that the constant dollar cost of this car is $12,500 if the base point is 1992. However, if the base
point changes, the cost of car in constant dollars will also change. For example, the constant dollar cost of the
car, if base point is 1993, is $13,000.
8.3
EQUIVALENCE CALCULATIONS
Equivalence calculations can be done using either actual or constant dollars. If actual dollars are used
interest rate, i, is used. However, if constant dollars are used then an inflation free interest rate, i', should be
used. An expression for i' is derived in Section 8.3.2. In using any of these methods, all the cash flows must
215
be same type, all actual or all constant dollars.
8.3.1
Equivalence Using Actual Dollars
The first step is to convert all the cash flows to actual dollars. After this, factors or interest formulas
with an interest rate of i are used in equivalence calculations.
Example 8.3.1 A family of four is planning to take a vacation in five years. The cost of vacation
is $2,000 now. They are planning to save for it by making four equal annual deposits (A) with first deposit
at t = 1 (end of the 1st year). The bank pays 6% compounded yearly. If annual inflation rate is estimated to
be 5%, how much should be deposited every year to provide for this vacation?
Solution: The amount of equal annual deposit (A) can be obtained by equating the cost of vacation
to the value of the deposits at the end of the 5th year.
The first step is to convert all cash flows to actual dollars. The four deposits will be made at the end
of the 1st, 2nd, 3rd, and 4th years. These are all actual cash flows as the deposits are made from out of
pocket dollars. The cost of vacation, $2,000, is at t = 0 or in constant dollars. Using t = 0 as the base, cost of
vacation at t = 5 (actual dollars) can be calculated using Eq. (8.2.2)
Actual dollar cost of vacation at the end of the 5th year = Constant $ (F/P, 5, 5)
= 2,000(1.276)
= $2,552
Now all the cash flows, deposits as well as the cost of vacation are in actual dollars. A cash flow
diagram is shown in Fig. 8.3.1. We equate the values of deposits and cost of vacation at t = 5 using i = 6%
compounded yearly and solve for A.
2,552 = A(F/A, 6, 4)(F/P, 6, 1)
= A(4.375)(1.06)
or
A
= $550.30
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Annual deposits of $550.30 for four years in the bank will provide enough money for the vacation
which will be taken at the end of the 5th year.
2,552
1
2
3
4
A
8.3.2
5
Figure 8.3.1 Actual Dollar
Cash Flows for Example 8.3.1
A
Using Constant Dollars
In using constant dollars, the first step is convert all actual dollars to constant dollars. This can be
done by using Eq.(8.2.1). For carrying out equivalence calculations, inflation free interest rate, i', is used. The
derivation of a formula for determining i' follows:
As stated earlier, there are two different ways for calculating equivalent cash flows. One is to work
in actual dollars and use i as the interest rate and the other is to use constant dollars and i' as the interest rate.
To derive an expression for i', the inflation free interest rate, we find the value of F at t = N for a given
amount P using both the methods . The following terminology will be used.
(i)
PA
= Amount at t = 0 in actual dollars
PC
= Amount at t = 0 in constant dollars
FA
= Amount at t = N in actual dollars
FC
= Amount at t = N in constant dollars or constant $ equivalent of FA .
Using actual dollars and i as the interest rate we find F given P.
FA = PA (F/P, i, N)
(8.3.1)
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(ii)
Using constant dollars and i', we find F given P as follows:
FC = PC(F/P, i', N)
(8.3.2)
The constant dollar equivalent of FA is FC. Using Eq.(8.2.1) and annual inflation rate, f, the value of
FC is
FC = FA (P/F, f, N)
(8.3.3)
The two values of FC from Eqs.(8.3.3) and (8.3.2) can be equated.
FA (P/F, f, N) = PC(F/P, i', N)
or
FA
= PC(F/P, i', N)(1.0/(P/F, f, N))
(8.3.4)
Equating the two values of FA from Eqs.(8.3.4) and (8.3.1), we get
P C(F/P, i', N)(1.0/(P/F, f, N)) = PA (F/P, i, N)
(8.3.5)
It should be noted that P A is equal to P C as at the base point, t = 0, actual dollars and constant dollars
are same. Solving for (F/P, i', N) from Eq.(8.3.5), we get
(F/P, i', N) = (F/P, i, N)(P/F, f, N)
Substituting the interest formulas for all the factors
(1 + i')N
= (1 + i)N (1 + f)-N
or
(1 + i')
= (1 + i)/(1 + f)
or
i'
= {(1 + i)/(1 + f)} - 1
(8.3.6)
Eq.(8.3.6) allows us to calculate inflation free interest rate, i' when i and f are known.
Example 8.3.2
Using constant dollar analysis, determine the amount of equal annual deposit in
Example 8.3.1.
Solution: The first step is to convert all actual dollars to constant dollars. Note that all four deposits
are in actual dollars and should be converted to constant dollars using Eq.(8.2.1). The cost of vacation, $2,000,
at t = 5 is already in constant dollars.
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Constant $ equivalent of first unknown A = Actual dollar(P/F, f, 1)
= A(P/F, 5,1)
= (0.9524)A
Similarly, for remaining three equal deposits
Constant $ equivalent of 2nd A = A(P/F, 5, 2) = (0.9070)A
Constant $ equivalent of 3rd A = A(P/F, 5, 3) = (0.8638)A
Constant $ equivalent of 4th A = A(P/F, 5, 4) = (0.8227)A
A cash flow diagram with all cash flows (dollars) is shown in Fig. 8.3.2.
2,000
1 2
3
4
5
0.8227 A
0.9524 A
Next, equate the value of deposits to the cost of vacation at t = 5 using inflation free interest rate, i'.
Inflation free interest rate, i' ,is determined using Eq. (8.3.6).
i' = {(1 + i))/(1 + f)} - 1
= {(1+0.06)/(1+0.05)} - 1
= 0.0095 or 0.95%
Equating the value of deposits to the cost of vacation at t = 5, we get
2,000 = [(0.9524)A(F/P, 0.95,4) + (0.9070)A(F/P,0.95,3) + (0.8638)A(F/P,0.95,2)
+(0.8227)A(F/P,0.95,1)]
For each of the factor, the interest formula is used to find its value as the value of the factor for
0.95% is not tabulated. Solving for A, we get
A
• $502
219
Note that this answer is approximately the same as obtained using actual dollar analysis. However,
the amount of calculations needed in the constant dollar analysis are more than the actual dollar analysis.
Example 8.3.3
A 40 year old man is trying to save for his retirement. He has estimated that he
will need $35,000 per year for 5 years, in today’s dollars, starting at the age of 61 years. The money for
retirement will be saved by making twenty equal annual deposits with first deposit at t =1 or at the age of
41 years. If inflation rate is 4% and bank will pay 10% interest compounded annually, how much money
should be deposited each year so that the man will meet his retirement needs.
Solution:
This problem, like equivalence problems without inflation, can be solved in many
different ways.
1st Method:
Use actual dollars and interest rate i: A cash flow diagram for this problem is
shown in Fig. 8.3.3.
35,000
1 2
21
A
A
35,000
25
Figure 8.3.3 Cash Flow
Diagram for Example 8.3.3
Note that deposits, A, are actual dollars while withdrawals of $35,000 are constant dollars with
base at t = 0. These constant dollars are converted to actual dollars using Eq. (8.2.2).As the values of
factor for 21, 22, 23 and 24 years are not tabulated, we use either the interest formulas or product of two
factors from the table. Using the product of two factors,
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At t = 21; Actual dollars
= 35,000 (F/P, 4, 21)
= 35,000(F/P, 4, 20)(F/P,4,1)
= 79,752.4
Similarly;
t = 22; Actual dollars = 35,000(F/P, 4, 22) = 82,973.2
t = 23; Actual dollars = 35,000(F/P, 4, 23) = 86,270.6
t = 24; Actual dollars = 35,000(F/P, 4, 24) = 89,721.5
t = 25; Actual dollars = 35,000(F/P, 4, 25) = 93,310.0
The meaning of the actual dollars of $93,325.6 at t = 25 is that at the age of 65 years, the man will
need $93,325.6 to purchase same goods/services as he could for $35,000 at t = 0 or at the age of 40
years.
Now both the deposits and withdrawals are in actual dollars. Equating the values of deposits and
withdrawals at t = 20 , we get
A(F/A, 10, 20) = 79,752.4(P/F, 10, 1) +........ + 93,325.6(P/F, 10, 5)
Solving for A
A = $5,676.30
The man will need to deposit $5,676.3 for 20 years to meet his retirement needs.
2nd Method: Use constant dollars and i': Withdrawals of $35,000 are constant dollars. We need
to convert the 20 annual deposits(A) to constant dollars. Eq. (8.2.1) is used for each A. For example
First A at t = 1; Constant dollar = A(P/F, 4, 1) = 0.9615A
2nd A at t = 2; Constant dollar = A(P/F, 4, 2) = ).9246A
Similarly, we find constant dollar equivalent for all A's.
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Next, find inflation free interest rate, i', using Eq. (8.3.6)
i' = {(1+i)/(1+f)} - 1
= {(1+0.10)/(1+0.04)} - 1
= 0.0576
We equate the value of deposits and withdrawals at t = 20 as follows:
0.9615A(F/P, 5.76, 19) + 0.9246(F/P, 5.76, 18) +....... = 35,000(P/A, 5.76, 5)
Solving for A, we get
A = $5,676.3
3rd Method:
Use a combination of actual and constant dollars: In some cases, this may provide
a better solution.
We first use constant dollars, $35,000, and find their equivalent value at t = 20. Note that as these
are constant dollars for equivalence i' should be used. Let this value we PC20 (present value in constant
dollars at t =20)
PC20 = 35,000(P/A, 5.76, 5)
= 149,100
Next we convert this constant dollar equivalent to actual dollars at t = 20.
Actual dollars at t = 20; Actual PA20 = PC20(F/P, 4, 20)
= 149,100(2.191)
= $326,678.1
It should be pointed out that N = 20 is used because base for all the constant dollars is at
t = 0.
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Next we equate value of deposits, A, and withdrawal of $326,678.1 at t = 20 using i = 10%.
A(F/A, 10, 20) = 326,678.1
or
A = $5,703.6
This value of A is little bit different because the value of the factor for i' = 5.76% is not tabulated
and interpolation was used.
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PROBLEMS
8.1
Using Table 8.2.1 determine the following:
(a) Average inflation rate for three years, 1988 to 1990, using CPI values. (4.79%)
(b) Average inflation rate for four years, 1990 to 1993, using CPI values.
8.2
The cost of a standard lawn mower in 1992 was $650. The average inflation rate is estimated to
be 5%. Determine the cost of this lawn mower in 1994 and 1995. (716.53; 752.46)
8.3
The cost of a new motorcycle is $24,600 in 1997. If annual inflation rate is 4.5%, determine the
constant dollar prices of this motorcycle with bases in 1992 and 1994.
8.4
Determine inflation free interest rates for following:
(i) Inflation rate = 8% and interest rate = 10%
(ii) Inflation rate = 5% and interest rate = 7%
8.5
A new college graduate is planning to save a fixed percentage of her salary to buy a new car five
years from now (t=5). The car presently costs $9,000 and its price is expected to increase at the rate of
inflation (8% per year). The first deposit will be made one year from now and her present salary of
$30,000 per year will be assumed to occur at t=0. She expects her salary to increase at the rate of inflation.
If she can earn 14% per year compounded annually, what fixed % of her salary must be saved each year
(for 5 years) for purchase of the car. Solve using:
(i) Actual dollar analysis. (5.37%)
(ii) Constant dollar analysis.
8.6
A young couple with a 9 year old son (now t=0) wants to save for their son's college expenses in
advance. He will start college at the age of 18 years ( t = 18). It is estimated that $6,000 per year in terms
of today's dollars (constant $) will be required to support his college expenses for four years. Inflation rate
will be 8% per year and savings can earn 12% per year compounded annually. Determine the equal
amount this couple must save each year for the college expenses. (First deposit at t=0, last deposit t=9, and
first withdrawal at t=9).
224
APPENDIX A
INTEREST FACTORS