Inflation

The loss of value of money over time.

                      Now                @10% inflation            In 1 year 
Wages         $10/hour   --->   x       1.1          --->  $11/year
Product       $1/Each     --->   x       1.1          --->  $1.1/Each

If the wages keep pace with inflation, the buying power remins the same. You can still buy 10 items for 1 hour's wage.
But $1 is now  worth 10% less if wages stay steady.
Normally, we want to be ahead of inflation.

i = Market interest rate (Rate bank shows)
f = Inflation Rate

What if we invest $1000 for 1 year.

F = 1000(1.10) = $1100 (Called Constant value, Actual or today's $)

But since everything is now 4% more expensive, actual purchase power is 4% less.

1100/1.04 = $1057.69 (Can only buy this much product)

Thus, the real i = Rate adjusted for inflation = i_f = [(1+i)/(1+f)] - 1

         Actual Rate obtained = (1.10/1.04) -1 = 0.0577 = 5.77%

What is the rate required to keep pace with the inflation, to get 4% more buying power in 1 year?


Required i_f = i_fR = [(1+i)(1+f)] - 1
Hence i_fR = [(1.10)(1.04)] - 1 = 0.1440 = 14.4%
and F=1000(1.1440)= $1144 (Future$ required to equal buying power +10%)
                                               (Called future or current $)

Thus to obtain now / actual $ in future use :

i_f = [(1+i)/(1+f)] - 1  (Gives lower rate requiring more $ invested now)

To obtain current / future $ required in future use :

i_fR = [(1+i)(1+f)] - 1  (Gives higher rate requiring less investment  now, since using future $)

Example: i = 10%,  f = 5%. How much money do you need to invest to buy a stereo equipment in 3 years, if
a) The equipment costs $2000 now.
b) You anticipate that the equipment will cost $2000 in 3 years.

a) This is a now $ in future case use :

            i_f = [(1+i)(1+f)] - 1 = (1.10/1.05) -  1 = 4.76%
     2000 = P(1.0476)³ . Hence P = $1739.57
Since, you actually need 2000(1.05)³ = $2313.25 actual $ in year 3, you need to invest $1739.57 now @ 10%.

b) This is a future $ in future case use :

        i_fR = [(1.05)(1.10)] - 1 = 15.5%
    2000 = P(1.155)². Hence P = $1298.03

Since, you need $2000 future / current $ in year 3, you need to only invest $1298.03 now @ 10%.

Note: Without inflation f = 0
  2000 = P(1.10)³. Hence P = $1502.63

The required amount to be invested now can be summarized as follows :
1739.57(Actual $) > 1502.67(No inflation) > 1298.03 (Future $)

Effect on Cash Flows ?

Example: What is the affect of F = 5% on the PW of the following cash flow @ i = 10% ?

Without Inflation :
PW = 2200 - 600(P/A,10,4)
        = 2200 - 600(3.170)
       = $298

With 5% inflation using future (Current $)

Year     CF         Inflation Factor     Future($)      (P/F,10,N)          PW     

  0       2200                ---                2200.00             ---             2200.00
  1         600    /        (1.05)¹               571.43           0.9091          519.49
  2         600    /        (1.05)²               544.21           0.8264          449.74
  3         600    /        (1.05)³                    518.30           0.7513          389.39
  4         600    /        (1.05)⁴               493.62           0.6830          337.14   
                                                                                                   $504.24

Or when using future $ :
i_ER = (1.10)(1.05) - 1 = 15.5%
PW = 2200 - 600(P/A,15.5,4)
       = 2200 - 600(2.8263)
       = $504.24

How is inflation rate determined?

• Cost indexes track relative costs of products and services for "Projects" over time.
• Helps to update costs over several years when inflation is not  uniform .
• Index is a dimensionless number to update a Cost/Price/Value.

The following equation is used to update a cost with an inflation rate : 
             C_t = C_o(I_t/I_o )
   C_t = Cost estimated @ present time t
   C_o = Original cost @ time t = 0
   I_t = Index value @ time t
   I_o = Index value @ time t = 0

Example: What is the current (2000) cost of a school renovation if the cost for the project was estimated to be $2,225,000 in 1998 ?

                 C_t = 2,225,000(6258.59 / 5920.49) = $2,352,434

Added to the Web: May 20, 2002. Last updated: April 26, 2002. Web page design by Dan Solarek.	http://cset.sp.utoledo.edu/engt3600/