Cash Flows & Compound Interest

Cash Flow - Inflow & Outflow of Money

Examples of Cash Inflows - Revenues, cost reductions, salvage receipts, Receipts of loan principal.

Examples of Cash Outflows - Purchase of assets, operating costs, maintenance costs,Interest                                               payments, Repayment of  loan principle.

Cash Flow is not an obligation or accrual of unpaid receipts.

Point of View

Cash flows are classified dependent upon the point of view of which party is paying and which is receiving payment. Let's look at several examples.

$2000 loan :                 You    =   +$2000       Inflow
                                   Bank     =    -$2000       Outflow

$50 Telephone bill:         You   =       -$50       Outflow
                                  Verizon  =       +$50       Inflow

Both happening in        You     =     $1950        Net Inflow
   one period               Bank   =    -$2000        Outflow         
                              Verizon    =         $50        Inflow      

Using our first example from Lesson 1:      P = $1000      i = 10%(Simple)      N = 5 yrs
However, let's pay I yearly.

We would have:

Year           CashFlow         Principal               Interest

  0                 $1000              $1000                  --
  1                 -$100                      0                -$100
  2                 -$100                      0                -$100
  3                 -$100                      0                -$100
  4                 -$100                      0                -$100
  5               -$1100            -$1000                -$100

The latter two columns would be considered to be an auxiliary table, breaking down the cash flow into a more explanatory fashion. Cash flows can be broken into convenient groupings, but auxiliary totals must equal the total cash flow.

End of Year Convention

For practicality (sacrificing a bit of accuracy) transactions are assumed to occur at the end of a period, no matter where they occur during each year.
Note in the example above: Start at year 0 = Begin year 1 = Start of time. Years 1 - 5 signify the end of year values.

Cash Flow Diagrams 
Graphical representation of Cash Flow

Cash flow diagrams help to visualize the exchange of funds. We will use them in many of our problems. Every Cash flow diagram contains the following components:

Time line -- with discrete periods
Cash flow vectors -- Up (+) = Inflow = Benefit  
                                                  or
                       -- Down (-) = Outflow = Cost
Interest rate

From our Simple Interest Example:

Just as with auxiliary tables, Cash flow diagrams can be split into separate equivalent diagrams. Vectors are additive.

Determining cash flows and drawing diagrams is part of every engineering economics problem.

Compound Interest

Interest (i) applies to total amount (P + sum of all I) during each period.

Consider the following Cash flow:
A $1000 deposit for 5 years at 10% / year would result in:

                  Amount accrued
Year     Begin Year        Interest      End year

   1     P = 1000               100        F1 = 1100          F1 = P + Pi
   2           1100                110        F2 = 1210         F2 = P + Pi+ (P + Pi) i
   3           1210                121        F3 = 1331         F3 = P + Pi+ (P + Pi) i + (P + Pi+ (P + Pi) i) i
   4           1331                133                1464              = P + Pi+ Pi+Pi2 + Pi+Pi2 + Pi2 + Pi3
   5           1464                146                1610              = P + 3Pi + 3Pi 2 + Pi3
                                                                                     = P(1+3i +3i2 + i3) = P(1 + i)3
  
Hence F = P(1 + i)N   where (1 + i)N  is called Compound Amount Factor.  

Solving Problems

It is now easy to solve problems regarding the Equivalence of Present (P) and Future (F) values over time (N) with interest (i). The following steps will help:

1.    Identify cash flows.
2.    Identify P, F, i & N.
3.    Determine the missing value.
4.    Solve for missing value using equation.

Example : What annual interest rate must you get if you need $7000 in 4 years and have $5000 to  
                invest now?                 
                          P = $5000,       F = $7000,    N = 4 years,     i = ?
            
               7000 = 5000(1 + i)4
                     7000 / 5000 = (1 + i)4
                    (1.4)1/4 = 1 + i
              1.0878 = 1 + i
               0.0878 = i = 8.78% / year

Functional Notation

Standardized notation has been established to avoid writing the equation each time, and to give a logical method by which to find the correct factor to use.

   

Read the factor as saying "Find F given P at i% for N periods".

Example:  What is F for $1000 deposit for 5 years @ 10% / year?
                 F = 1000( F/P,10,5) 
                 ( F/P,10,5) = 1.611   --- See table on Page 198
                 F = 1000(1.611) = $1611

Example : Find i for P = $5000,   F = $7000,   N = 4 years
                 Using  F = P(F/P, i, N)
              7000 = 5000(F/P, i, 4)
                   1.4 = (F/P, i, 4) 
Search the Compound Interest Factor tables (Pages 180 - 208) to find the Interest rate that matches the factor.

          i = 9%      (F/P, 9, 4) = 1.412
          i =  8%      (F/P, 8, 4) =1.360         
          
Since an even i value can't be found to match (F/P,i,4) = 1.4, you must interpolate to find the solution. Use the following as a guide for interpolation.
           
Note that the solution here by interpolation does not match exactly to the one found previously. This is due to linear interpolation of a Non-linear function. You may use tabled values or the equation, which ever fits the situation and allows an easier solution.  

More Frequent Compounding

Compounding can take place at intervals more frequently than yearly, say quarterly, daily, weekly, monthly etc.
One must make adjustments for the frequency of compounding.

Example: Using the same $1000 @ 10%  now compounded quarterly.   

There are 4 periods in 1 year which yields  i = 10% / yr  divided by 4 qtrs/yr = 2.5% / qtr            

Year    Quarter    Beginning Period    Interest    End Period

  1           1               1000                   25            1025
  1           2               1025                   25.62       1050.62
  1           3               1050.62              26.26       1076.88
  1           4               1076.88              26.92       1103.80 which is>1100 with yearly compounding
  
Hence more frequent compounding increases F.

What is the actual yearly interest rate?

Effective annual i = i eff= (1103.83 - 1000)/1000 = 0.1038
                            i eff= 10.38% = 10% compounded quarterly > 10% yearly
                             
One could find the effective interest rate without the aid of P & F values, by using the following equation :
          i eff = (1 + r / m)m -1   
                   where r = nominal rate / yr (Usually quoted with a Compounding Frequency)
                            m= # of compounding periods for nominal period
                              i = r/m =  Interest rate/Period  
   
In future discussions, we will use i most exclusively to set up problems. The key is to determine the Frequency of Compounding. Assume i is a yearly rate with yearly compounding, unless it is stated otherwise.

For the above example:  i = 10% per year compounded quarterly
                  Hence          r =  10% = 0.10
                                    m= 4 (qtrs)
                                   i eff = (1 + 0.10/ 4)4 -1   
                                        = (1.025)4-1
                                        = 1.1038-1 
                                        = 0.1038 = 10.38%

Remember:
Nominal is the annual rate with a Frequency of Compounding given.
Effective is annual rate with compounding already considered .

Example: A $1000 deposit for 5 years at 10% / yr compounded quarterly yields what future value?

                    Quarterly rate = 10% / 4 = 2.5%
                    # of Periods (qrts) in 5 yrs = (4 qrts/yr)(5 yrs) = 20 Periods

Find an F given P using 2.5%/Period for 20 Periods.
                   F = P(F/P,2.5,20)      
                   Not    (F/P,10,5)     or  (F/P,2.5,5)     or   (F/P,10,20)    
                    F = 1000(1.639) = $1639 

Remember to use # of periods that corresponds to frequency of compounding
                    i = r / m    where i = Interest rate / period
                                              r = nominal rate / year
                                              m = # of compounding periods/year



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Last updated: April 26, 2002.
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