Reducing alternative cash flows to a present value in order to decide the best course of action, based on a current "Money-time frame" basis.

Requirements for P-W comparisons
    Alternatives must co-terminate. That is, they must have the same time span to ensure equivalent outcomes.

Assumptions for PW comparisons:

1.   Cash flows are known. ( Estimating future cash flows introduces risk)
2.   Cash flows are in constant dollar values. ( No inflation is assumed to occur)
3..   Interest rate is known. ( Set by organization)
4.   Comparisons are made with before tax cash flows. ( Taxes make it complex)
5.   Don't include intangible factors. ( It looks neat, great for PR, etc)
6.   Funds are available to implement alternatives.
7.   Technological stability exists. (Don't account for fact that technological inventions  get cheaper 
      as time goes on.  286 Computer is free now and 1980 VCR- $2500 is $200 now)

 Present - Worth Equivalence (PW)

Determines the present worth equivalence of a set of future transactions.

Example:  Buy a machine @ year 0 for $100,000. Additional start up costs (training and set up) after1 yr for $50,000. No maintenance costs. Years 2 through 5 pays $42,000/yr. Find PW if    i = 10%
       
PW= PW(Benefits) - PW(Costs)

PW = -$100,000 - $50000 (P/F,10,1) + (P/F,10,1) $42000 (P/A,10,4)
        = -$100,000-$50,000(0.90909)+(0.90909)$42,000(3.16987)
PW = -$24,423

A negative Present-worth equivalence indicates a negative decision toward making the  investment alternative. The worth is less than zero.

PW>0  is better than investing @ i %
PW<0  is worse than investing @ i %

If comparing alternates, take the one with the bigger P-W (Less - or More +)

Net P- W Equivalence

Used to combine receipts and disbursements in future.

Net PW= PW(Benefits) - PW(Costs) which is equivalent to PW(Benefits-Costs)

Method 1 :

PW(Benefits)= $25,000(P/A,8,5) + $30,000(P/F,8,5)
                     = $25,000(3.99271) + $30,000(0.68058)
                     = $120,235.15

PW(Costs)   = $75,000 + $10,000(P/A,8,5) + $15,000(P/F,8,5)
                    = $75,000 + $10,000(3.99271) + $15,000(0.68058)
                    = $125,135.80

Net PW       = $120,235.15 - $125,135.80 = -$4900.65

Method 2 :
        
Net PW = -$75,000 + $15,000(P/A,8,4) + $30,000(P/F,8,5)
              = -$75,000 + $15,000(3.31213) + $30,000(0.68058)
               = -$75,000 + $70,099.35 = -$4900.65

In either case PW<0. Do not purchase the equipment. Also notice that PW is the same using either method. Receipts and disbursements are additive at any time t.

Unequal Life Projects

To calculate the Present-Worth for cash flows of unequal durations is incorrect!!
This assumes that the shorter lived alternative gives benefits at no cost even after that alternative has exhausted its service life.

Life Span Equalization Methods

1.  Common multiple method (Repeated replacement) : The least common multiple (LCM)  of      both options is determined and used as the study period for both options.

2.  Study period method:
     A.  Solving for an estimated salvage using t = shorter life.
     B.  Solving for an estimated costs using t = longer life.

Example : A machine to provide a service can be purchased. Two options are available.

Method 1 :  (Repeated Replacement)

• The least common multiple of 5 & 3 is 15
• Assume repeated replacement on each alternative for 15 years.
• This assumes no technological advances (Ha!!)

PW(A1) = -15,000 - 15,000(P/F,7,5) - 15,000(P/F,7,10) - 7,000(P/A,7,15)
              = -15,000(1 + 0.71299 + 0.50835) - 7000(9.10791)
              = -$97,075

PW(A2) = -20,000(1 + 0.81630 + 0.66634 + 0.54393 + 0.44401) - 2,000(9.10791)
              = -$87,627

A2 is best under the assumptions of repeat replacement.

A result is determined. However does the assumption make us feel uneasy about the results? Let's look at the other way to analyze the situation.

Method 2(a) : Study period (t = shorter life)
                     Set study period = 3 years (which corresponds to option A2).

PW₃(A2) = -20,000 - 2,000(P/A,7,3)
                = -20,000 - 2,000(2.62432)
                = -$25,249

PW₃(A1) = -15,000 -7,000(P/A,7,3) + PW(Salvage)
                = -$33,370 + PW(Salvage)

It would be better to solve for a salvage value to see if it sounds reasonable than to guess the amount of worth of machine with 3/5 life used. 

            A2                      A1
     -$25,249    = -33,370 + S(P/F,7,3)
     -$25,249    = -33,370 + S(0.81630)
                S     = $9,948 of A1

Thus, one would have to be able to find a buyer for the machine, at 2/3 the original price with only 2/5 life remaining in the machine, in order for A1 to be attractive. Thus buy A2.

Is time period too short? Will the service provided by the machine still be needed after 3 years?

Let's look at the longer study period method.

Method 2(b) : Study period (t = longer life)
                      Set period = 5 years (which corresponds to option A1). 

PW₅(A1) = -15,000 - 7,000(P/A,7,5)
                = -15,000 - 7,000(4.10020)
                = -$43,701

PW₅(A2) = -20,000 - 2,000(P/A,7,3) - PW(Cost/Penalty)

Now, determine the Cost/Penalty required in  years 4 and 5 on A2 to provide service that the A1 machine does anyway.

-43,701 = -20,000 - 2000(P/A,7,3) - C/P(P/A,7,2)(P/F,7,3)  (where C/P is required annual cost)
-43,701 = -20,000 - 2000(2.62432) - C/P(1.80802)(0.81630)
-43,701 = -20,000 - 5249 - C/P(1.4759)
-18,452 = -C/P(1.4759)
       P/C = $12,502/year

One would have to find the supplier of machine A1's duties for less than $12,500 to make A2 the more attractive option.

With these results you probably could again choose A2, since the first cost of A2 = $20,000. Rental of such a machine should surely be less than $12,500/year.

Three study periods = Three identical answers
Which method is the best?
It depends on the company's view of:  risk, outlook, i %, technology,actual service life really needed.

I prefer to use method 2a, since it is fairly easy to estimate and compare future salvage values. Method 1 gives consistent results without estimation.

Assets with Infinite Lives - Capitalized Cost

What if the Service life/Study period is infinity?
        

Capital cost = P = A(P/A, i, n) = A[( 1(1+ i)^N-1) /  i(1+ i)^N]

As N--> infinity      (P/A, i, N) =  [((1+ i)^infinity - 1)/ i(1+i)^infinity] 

But  (1+ i)^infinity = infinity       

So (P/A, i, N) = (infinity-1)/(i) (infinity)    

But inf-1 = infinity
Thus infinity / (i) (infinity) = 1/i  and  P = A/i

Check this equation against the compound interest tables in your book. Note that the values for P/A as N increases start to approach 1/i .

 Example: After winning the lottery and getting $2,000,000 after taxes, how much could you spend on a house this year and still have $100,000 a year for life if i = 10%?

Capitalized cost   (PW_infinity ) = P + A(1/i)
                        2,000,000 = P + 100,000(1/0.10)
                                     P = $1,000,000 worth of house

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