Now we will get into the heart of the fundamental conversion factors that will help us to analyze difficult problems later in the class.

Present-Worth Factor (Single Payment)

Now let's find a Present (P) value knowing a Future (F) value. This would pose the question :
How much money need I put away, now (P) at i for N periods to have F?

P = F(P/F, i , N)

P(1 + i)^N = F => P = F(1/(1 + i)^N) = F(1 / (F/P , i ,N))

Hence (P/F, i , N) = (1 / (F/P , i ,N))

For our previous example, if you needed $1610 in 5 yrs, what amount would you need to put away now at 10% compounded annually?

P = 1610(P/F, 10,5)
   =  1610(0.6209)= $1000
    
Note that : (F/P, i , N)(P/F, i , N) = 1
     
Try with    i = 8%          N = 10 yrs       
             
(F/P, i , N) (P/F, i , N) = (2.159)(0.4632) = 1.000

Multiple Cash Flows

What happens if we have additional payments between Time 0 and the end of the time line?     

Example: How much money will I have in 5 yrs if I deposit $500 now & $600 after 2 yrs if  i = 6%/year?

       F = F₅₀₀ + F₆₀₀
          = P₀(F/P, i,N) + P₂(F/P, i,N) 
          = 500(F/P, 6,5) + 600(F/P,6,3) 
          = 500(1.338) + 600(1.191)
          = $669 + $714.60 = $1383.60 

Note that even though the $600 deposit occurs 2 years in the future, it is treated as a P when standing at end of year 2.

We could have used a Two-Step process, to find the the value of P ₀ at N=2, add it to the $600 and find the equivalence of that sum at N = 5. 
                 
 F₂ = 500(F/P, 6,2) + 600 = 1162     
 F₅ = 1162(F/P, 6,3)   = $1383.94(Rounding) 

In the real world 

• Banks want payments on loans more than just at the end of the loan.
• Your paycheck comes more frequently than once a year.
• You usually don't have large sums of money sitting around for investments towards a future  purchase.
• Parts of paychecks are regularly invested.
• Loans are paid of  on a monthly basis.
• Annuities pay holder on a periodic basis.

Series Compound Amount Factor (Uniform Series)

Let's look at a series of identical payments over several years.
 
How much money do we get in N periods at i % with a payment of A per period?
                 F = A(F/A, i, N) 

• Payments are considered to occur at the end of each year.
• Last payment yields no interest.
• A₁= A₂ = A₃ = A _N-1 = A_N
• F is not equal to (A* N)    -- Due to interest on each A for the time it is deposited 

Look at A = $100 / yr  for N = 5 yrs

Year                Amount at end year
  1                   100
  2                   100 + 100(1 + 0.1)
  3                   100 + 100(1.1) + 100(1.1)²
  4                     ...................
  5                   100 + 100(1.1) + 100(1.1)² + 100(1.1)³ + 100(1.1)⁴
       
This would result in the following :
        F = A((1 + i)^N-1) / i = A(F/A, i, N)                      

In our example above the total at the end of 5 yrs = $610 and with the uniform series factor, we obtain  F = 100(F/A, 10,5) = 100(6.10510) = $610

Sinking Fund Factor (Uniform Series)

What payment (A) is required to be paid periodically at (i%) to obtain an amount (F) after  (N) periods?

A = (F/A, i, N)                     
Since F = A((1 + i)^N-1) / i ,  then  A = F        1            =    F( i /(1+i)^N-1)
                                                              ( (1+i)^N-1)/i

Finding the required annual deposit for the final $610.50 from above, yields :    A=$610.50(A/F,10,5)=                          $610.50(0.16380) = $100 / yr

Capital Recovery Factor (Uniform Series)

What payment (A) is required to be paid for (N) periods at (i%) to be equivalent to the amount (P) received now?
                     A = P(A/P,i ,N) = P(i+(1 + i)^N) / ((1 + i)^N-1)

Example :
A $1000 loan at 10% annually for 5 years requires what payment?

A = 1000(A/P,10,5) = 10(0.26380) = $263.80/year 

Total payment = (5) (263.80) = $1319

As opposed to $1610 one time payment at the end of  the loan. 

Why? Each payment lowers some of the original P on which interest is levied.

Series Present Worth Factor

What amount must be paid now (P) in order that at (i%), equal amounts (A) are received for (N) periods?
                  P = A(P/A, i, N) = A((1 + i)^N-1) / (i(1 +i)^N)
                                                   
Example: If we want $500/yr for maintenance of equipment for the next 5 years, what amount (P) should be deposited now at 10% annual compounding?

    P = 500( P/A, i, N) = 500(3.79079) = $1895

Arithmetic Gradient Conversion Factor

Sometimes the payment amount increases or decreases over time.

In order to analyze problems with this situation, a conversion of the increasing series must be made into a uniform payment prior to other analysis to P or F values.

First find A_G = G(A/G,i,N)
Then find A= A' + A_G  where A = Annualized payment
                                             A' = First payment
                                             G = Increase/Decrease in Annual payment.
Finally use A with  P/A to find P.

Or use : (P/G,i,N) = (A/G,i,N)(P/A,i,N) -->Tables without using A'.  

Note that: (A/G,i,N) = [(1+i)ⁿ -ni -1] / i [(1+i)ⁿ -1]
  
Example:  A beginning payment of $150, increasing by $50 per period at  i = 10% for 5 years is equivalent to what present & future amount? (Two alternatives are shown to obtain P) 

Alternate (1)
1st step :   A = A' + A_G
                   = 150 + G(A/G,i,N)
                   = 150 + 50(A/G,10,5)   
                   = 150 + 50(1.810)
                   =150 + 90.50 
                   = $240.50

2nd step :  P = A(P/A,i,N)
                   = 240.50(P/A,10,5) 
                   = 240.50(3.791)
                   = $911.74
     or

Alternate(2)      
     P = A'(P/A,i,N) + G(P/G,i,N)
        = 150(P/A,10,5) + 50(P/G,10,5)
        = 150(3.791) + 50(6.862)
        = 568.65 + 343.10 = $911.75   

Finally  F = P(F/P,i,N) = 911.75(F/P,10,5) = $1468.88

Remember : A' is the base payment from year1 to year N.
                   G is the increase (decrease) in payment starting in year 2 to year N.
                   A_G is the uniform equivalence for the increasing /decreasing cash flow for years 1 to N.

Geometric Gradient Factor

If the interest in the payment is geometric (rising at constant %) each period, the following conversion to a present value is required :
         P = A₁[ 1-((1+g)ⁿ (1+i)^-n)] /(i - g)     where i is not equal to g and
                     
         where : A₁ = First payment in series
                     g = % increase in payment/period 
                     i = interest rate/period
                     P = Present worth

If    i = g, then  P =A₁ + [n / (i + 1)]

Example: A 401-K account makes 8% / yr  with a 10% withholding of annual salary deposited. Current pay is $40,000 annually with an expected 3% raise each year. How much money will you have in 20 years?

A₁ = 40,000(0.10) = $4000,        g = 3%,             i = 8%
P = 4000[1-((1+0.03)²⁰ (1+0.08)^-20)] / [ 0.08-0.03]  =  $49,000.16

F = 49,000.16(F/P,8,20) = 49,000.16(4.6610)
    = $228,389.75

Manipulating Cash Flows & Equivalencies

Tailoring : Adding additional cash flows to create a uniform series which can be more easily handled with standard factors.

P = 200(P/A,5,6)-100(P/F,5,2)+50(P/F,5,4)-300(P/A,5,5)
P = 200(5.076)-100(0.9070)+50(0.8227)-300(0.7835)
P = $730.59
           
When using the brute force method (with P/F Factors), six separate factors would have been required. With tailoring we used only four factors. It's not too big of a deal here, but it could be a problem on  a 20-30  period cash flow.

Shifting : Use multiple equivalence transactions to gain final solution.
    
Step 1 :
                   
Find an equivalent present value at the end of year 7. We will call it P₇. Remember the timeline can be shifted to use the (P/A) factor.
P₇ = A(P/A,7,5) = 500(4.100)
    = $2050
Step 2 :
Let's now call P₇ a future value F₇
                 
P = 2050(P/F,7,7) = 2050(0.6227)
P = $1276.53

Remember P, F and A values can be anywhere on the timeline. However :
1) P is always before F and exactly one year before any A's.
2) F is always after P and on the same year as the last of a series of A's.

Final Fundamentals Review