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									 Economic Decision Making

Decisions using Time Value of Money
              Models


                   Valerie Tardiff and Paul Jensen
               Operations Research Models and Methods
 8/25/04         Copyright 2004 - All rights reserved
Engineering Economic
Analysis
   Recall the seven steps of
    Engineering Economic Analysis




                                    2
Why Do We Make
Investments?
   Buy stock in return for dividends and a
    higher stock price in the future.
   Invest in a machine in return for reduced
    operating costs or greater sales in the
    future.
   Buy a bond in return for interest payments
    and the return of principal in the future.

We give up money now in the hopes of
 receiving more money in the future.
How much return is enough?                       3
Why Do We Borrow Money?
 We borrow money to buy a $100,000
  house in return for paying $300,000 over
  the next 30 years.
 We borrow $10,000 to buy a car and pay
  an extra $2,000 in interest.
 We buy a new set of clothes and put the
  bill on our credit card.
We spend money now in return for paying
  more money in the future.
How much are we willing to pay?
                                             4
Example
   Your brother borrows $200 from you.
   He will pay back in monthly payments of
    $14.44 for the next 16 months.
   Is this an acceptable investment for you,
    the lender (or investor)?
   Is this an acceptable loan for your brother,
    the borrower?



                                                   5
        For the Lender
Cash Flow for investor
14.44
                                        ……
  0
        0   1   2   3   4   5   6   7   8   ……   16




-200


                                                  6
   Is this an acceptable
 What is the total summed over all cash
   investment?
    flows?
    -200 + $14.44 (16) = $31.04
   Is this enough?


   Say the lender requires a minimum return
    of 1% per month. Does this investment
    obtain the required return?


                                               7
   Choose
    
        Selection Procedure (MARR) (for
        a Minimum Acceptable Rate of Return
        the lender) or
   Evaluate the Equivalent Value of the Cash Flow
    using the MARR:
       Compute the Net Present Worth (NPW),
       compute the Future Worth (FW) , or
       compute the Net Annual Worth (NAW).
    If the NPW, FW or NAW is ≥ 0, accept the
    investment or the loan.
       Otherwise, accept the do-nothing alternative. In other
        words, reject the investment or the loan.


                                                                 8
   The NPW as a function of
 Recall our Example,
   MARR
       NPW = –200 + 14.44 (P/A, MARR, 16)
                    Net Present Worth
 35
 30
 25
 20
 15
 10
   5
                          MARR (Percent)
   0
  -5   0      0.5           1              1.5   2    2.5
-10
-15


  When MARR is 1%, NPW is > 0, so accept the investment.


                                                           9
Minimum Acceptable Rate of
Return (MARR)

   In theory, the minimum acceptable
    rate of return (MARR) is the interest
    rate that could be received if the
    funds were invested elsewhere.
   It helps to think of it as the hurdle
    rate.


                                            10
MARR (cont’d)
   In practice, it depends on much more than
    the bank interest rate. It is determined by
    top-management and depends on
       the amount of money available for investment,
        the source and cost of these funds,
       the number of good projects available for
        investment and their purpose,
       the amount of risk associated with investment
        opportunities and the cost of administering
        investments, and
       the type of organization.
   How an organization determines its MARR             11
     Net Present Worth method, also
    The NetPresent Worth
    called the Present Worth (PW), method
    compares all cash flows only after they
    have been measured at a common point in
    time determined to be the present date.
   To find the NPW as a function of i%, we
    transform cash flow amounts to their
    equivalent in the present and add them.
   NPW = CF0 + CF1 (P/F, i,1) + CF2 (P/F, i,2)
    + …. + CFN (P/F, i, N)

                                              12
    Cash Flow for the Borrower
 200

                     For the borrower



   0
         0   1   2    3   4   5   6   7   8…… 16
                                          ……
-14.44




                                                   13
Is this an acceptable loan?

   What is the total summed over all cash
    flows?
    200 - $14.44 (16) = - $31.04
   Is this reasonable?


   Say the borrower wants a cost of
    borrowing no more than 2%. Does this
    loan cost more than the maximum
    acceptable rate for borrowing?
   Compute the NPW using the MARB. If it is
                                               14
      The NPW as a function of
     Recall our example,
 
      MARB - 14.44 (P/A, MARB, 16)
     NPW = +200
                    Net Present Worth
15
10
  5
                     MARB (Percent)
  0
 -5 0      0.5        1           1.5    2       2.5
-10
-15
-20
-25
-30
-35


  When MARB is 2%, NPW is > 0, so accept the loan.

                                                       15
Maximum Acceptable Rate
of Borrowing (MARB)

   The maximum acceptable rate of
    borrowing (MARB) is the interest rate
    that would be paid if the funds were
    borrowed elsewhere.
   The interest rate charged by a bank
    is often the MARB used.


                                            16
Future Worth
   The Future Worth (FW) method compares
    all cash flows only after they have been
    measured at a common point in time
    determined to be the end of the analysis
    period.
   To find the FW as a function of i%, we
    transform cash flow amounts to their
    equivalent in the future and add them.
   FW = CF0 (F/P,i,N) + CF1 (F/P,i,N-1)
    + CF2(F/P,i,N-2) + … + CFN-1 (F/P,i,1) +
    CFN
                                               17
      Net Annual Worth
    The Net Annual Worth (NAW) method, also called
    the Annual Worth (AW), method compares all
    cash flows only after they have been transformed
    into time-equivalent annuities.
   To find the NAW as a function of i%, we transform
    the cash flows over time as uniform payments
    over the length of the project period.
   NAW = NPW (A/P, i, N) = FW(A/F, i, N)




                                                  18
Equivalence
   NPW, FW, and NAW all yield the same
    decision since they are all the
    representations of the same cash flows at
    different points of time.
   NAW = NPW (A/P, i, N)
   FW = NPW (F/P, i, N)
               where i = MARR for the investor
                     i = MARB for the borrower

                                                 19
      Capital Recovery Cost
    In most engineering situations, the capital
    cost has two components
       I, the investment expended at time 0, and
       S, the salvage value received at time N.
                                        S
             0

                                      N

                 I

                                                    20
   Capital Recovery Cost
   (cont’d)
 Capital Recovery (CR) Cost is the annual

    equivalent of the capital cost.
   CR (i) = I (A/P, i, N) - S (A/F, i, N)
   Other Formulas
       CR (i) = (I -S) (A/P, i, N) + i S and
       CR (i) = (I -S) (A/F, i, N) + i I




                                                21
   Your company is planning to manufacture a
      Example
    new product. The product requires a machine
    that the company does not now own. The cost
    of the machine is $10,000, its life is 4 years,
    and its salvage value (at the end of the 4th
    year) is $1,000. With this machine, the new
    profit from each product is $12. What annual
    production makes the investment worthwhile?
    The MARR is 10%.


                                                      22
Example (cont’d)
   Find the capital recovery costs
       I = $10,000
       S = $1,000
       MARR = 10%
       Useful life N = 4
   CR = I (A/P, MARR, N) - S (A/F, MARR, N)
       = $10,000 (A/P,10%,4) - $1,000 (A/F,
    10%, 4)
       = $10,000 (0.3155) - $1,000 (0.2155)
       = $3,155-$215.50 = $2939.5
                                               23
Example (cont’d)
   Using the shortcuts,we get
   CR = (I-S) (A/P,10%,4) + (10%) S
        = $9,000 (0.3155) + (0.1)$1,000
        = $2,839.5 + $100 = $2939.5
   CR = (I-S) (A/F,10%,4) + (10%) I
        = $9,000 (0.2155) + (0.1)$10,000
        = $1,939.5 + $1,000 = $2939.5
                                           24
Example (cont’d)
   Capital recovery costs must equal or
    exceed revenues to be profitable.
   CR = $2939.5 = $12x where x is the
    number of products produced per year.
   Solving for x, we get x >= 250 units.




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