Docstoc

Slides - Download Now PowerPoint

Document Sample
Slides - Download Now PowerPoint Powered By Docstoc
					             Topics Today
•   Conversion for Arithmetic Gradient Series
•   Conversion for Geometric Gradient Series
•   Quiz Review
•   Project Review




                                                1
     Series and Arithmetic Series
• A series is the sum of the terms of a
  sequence.
• The sum of an arithmetic progression (an
  arithmetic series, difference between one
  and the previous term is a constant)
   sn  a  (a  d )  (a  2d )  (a  3d )  ...  (a  (n  1)d )

• Can we find a formula so we don’t have to
  add up every arithmetic series we come
  across?

                                                                       2
      Sum of terms of a finite AP
S n  a  (a  d )  (a  2d )  ... [a  (n  2)d ]  [a  (n  1)d ]
S n  a  (a  d )  (a  2d )  ... (a  nd  2d )  (a  nd  d )
S n  (a  nd  d )  (a  nd  2d )  ... (a  2d )  (a  d )  a
2S n  a  (2a  nd )  (2a  nd )  ... (2a  nd )  (2a  nd )  a
There are (n) 2a terms  2a  n  2an;
There are (n - 1) nd terms  nd  (n - 1)  nd (n - 1) ; Therefore,
2Sn  2an  nd (n  1)
2S n  n[2a  (n  1)d ]
     n
S n  [2a  (n  1)d ]
     2
                                                                          3
       Arithmetic Gradient Series
• A series of N receipts or disbursements that increase
  by a constant amount from period to period.
• Cash flows: 0G, 1G, 2G, ..., (N–1)G at the end of
  periods 1, 2, ..., N
• Cash flows for arithmetic gradient with base annuity:
  A', A’+G, A'+2G, ..., A'+(N–1)G at the end of
  periods 1, 2, ..., N where A’ is the amount of the base
  annuity


                                                            4
Arithmetic Gradient to Uniform Series
    • Finds A, given G, i and N
    • The future amount can be “converted” to an
      equivalent annuity. The factor is:
                                 1       N
               ( A / G, i , N )  
                                 i (1  i )N  1
    • The annuity equivalent (not future value!)
      to an arithmetic gradient series is A =
      G(A/G, i, N)


                                                   5
Arithmetic Gradient to Uniform Series
     • The annuity equivalent to an arithmetic
       gradient series is A = G(A/G, i, N)
     • If there is a base cash flow A', the base
       annuity A' must be included to give the
       overall annuity:
           Atotal = A' + G(A/G, i, N)
     • Note that A' is the amount in the first year
       and G is the uniform increment starting in
       year 2.
                                                      6
Arithmetic Gradient Series with
        Base Annuity




                                  7
           Example 3-8
• A lottery prize pays $1000 at the end
  of the first year, $2000 the second,
  $3000 the third, etc., for 20 years. If
  there is only one prize in the lottery,
  10 000 tickets are sold, and you can
  invest your money elsewhere at 15%
  interest, how much is each ticket
  worth, on average?

                                        8
Example 3-8: Answer
• Method 1: First find annuity value of prize
  and then find present value of annuity.
   A' = 1000, G = 1000, i = 0.15, N = 20
   A = A' + G(A/G, i, N) = 1000 + 1000(A/G,
     15%, 20)
     = 1000 + 1000(5.3651) = 6365.10

• Now find present value of annuity:
   P = A (P/A, i, N) where A = 6365.10, i =
     15%, N = 20
   P = 6365.10(P/A, 15, 20)
     = 6365.10(6.2593) = 39 841.07

• Since 10 000 tickets are to be sold, on
  average each ticket is worth (39
  841.07)/10,000 = $3.98.                       9
Arithmetic Gradient Conversion Factor
         (to Uniform Series)

• The arithmetic gradient conversion factor (to
  uniform series) is used when it is necessary
  to convert a gradient series into a uniform
  series of equal payments.

• Example: What would be the equal annual
  series, A, that would have the same net
  present value at 20% interest per year to a
  five year gradient series that started at $1000
                                                  10
  and increased $150 every year thereafter?
Arithmetic Gradient Conversion Factor
         (to Uniform Series)
1       2        3       4       5       1           2           3           4           5


                                             $1000

                                                         $1150
    A        A       A       A       A
                                                                     $1300
                                                                                 $1450

                       (1  i ) n  (1  ni)                                                 $1600
            A  Ag  G
                         i[(1  i ) n  1]
                              (1  0.20 ) 5  (1  5 * 0.20 )
               $1,000  $150
                                 0.20[(1  0.20 ) 5  1]
               $1,246
                                                                                             11
Arithmetic Gradient Conversion Factor
             (to Present Value)

• This factor converts a series of cash
  amounts increasing by a gradient value,
  G, each period to an equivalent present
  value at i interest per period.
• Example: A machine will require $1000
  in maintenance the first year of its 5
  year operating life, and the cost will
  increase by $150 each year. What is
  the present worth of this series of
  maintenance costs if the firm’s minimum
  attractive rate of return is 20%?       12
Arithmetic Gradient Conversion Factor
                      (to Present Value)
                                                                  $1600
                                                      $1450
                                          $1300
                              $1150
                  $1000
              1           2           3           4           5




         P
       (1  i ) n  1    1  (1  ni)(1  i )  n
   PA                G
        i (1  i ) n
                                   i2
               (1  0.20 ) 5  1         1  (1  5 * 0.20 )(1  0.20 ) 5
      $1,000                      $150
              0.20 (1  0.20 )  5
                                                     (0.20 ) 2
      $3,727
                                                                             13
 Geometric Gradient Series
• A series of cash flows that increase or decrease
  by a constant proportion each period
• Cash flows: A, A(1+g), A(1+g)2, …, A(1+g)N–1
  at the end of periods 1, 2, 3, ..., N
• g is the growth rate, positive or negative
  percentage change
• Can model inflation and deflation using
  geometric series
                                                14
           Geometric Series
• The sum of the consecutive terms of a
  geometric sequence or progression is
  called a geometric series.
• For example:
     Sn  a  ak  ak 2  ak 3  ....  ak n  2  ak n 1
 Is a finite geometric series with quotient
 k.
• What is the sum of the n terms of a finite
  geometric series
                                                             15
  Sum of terms of a finite GP
        Sn  a  ak  ak 2  .... ak n  2  ak n 1
      kSn  ak  ak 2  .... ak n  2  ak n 1  ak n
Sn  kSn  a  0  0  ..... 0  0  ak n
Sn (1  k )  a (1  k n )
               (1  k n )
        Sn  a
                (1  k )

• Where a is the first term of the geometric progression,
  k is the geometric ratio, and n is the number of terms
  in the progression.
                                                        16
 Geometric Gradient to
    Present Worth
• The present worth of a geometric series is:
          A       A(1  g )     A(1  g )N 1
     P                    
        (1  i ) (1  i )2
                                  (1  i )N
• Where A is the base amount and g is the
  growth rate.
• Before we may get the factor, we need what
  is called a growth adjusted interest rate:
             1 i              1      1 g
      i          1 so that       
             1 g             1 i  1 i
                                                17
       Geometric Gradient to Present Worth
              Factor: (P/A, g, i, N)
                                    (1  i  )N  1 1 
               (P / A, g, i , N )                    
                                              N 1 g
                                     i (1  i )       
                                (P/A,i ,N)
                              
                                  ( 1  g)
Four cases:
(1) i > g > 0:         i° is positive  use tables or formula
(2) g < 0:     i° is positive  use tables or formula
(3) g > i > 0: i° is negative  Must use formula
(4) g = i > 0: i° = 0                            A 
                                            P  N     
                                                 1 g 

                                                                18
 Compound Interest Factors
 Discrete Cash Flow, Discrete Compounding

To Find   Given       Name of Factor          Factor
                  Compound Amount
  F        P      Factor (single payment)     (1  i) n
                  Present Worth Factor
  P        F      (single payment)           (1  i )  n

                  Compound Amount           (1  i ) n  1
  F        A      Factor (uniform series)         i
                                                  i
  A        F      Sinking Fund Factor       (1  i ) n  1


                                                             19
Compound Interest Factors
Discrete Cash Flow, Discrete Compounding

To Find   Given      Name of Factor                 Factor
                                                  i (1  i ) n
  A        P      Capital Recovery Factor        (1  i ) n  1
                                                  (1  i) n  1
                  Present Worth Factor
  P        A      (uniform series)                 i (1  i) n
                  Arithmetic Gradient
                                             (1  i ) n  (1  ni )
                  Conversion Factor (to
  A        G      uniform series)              i[(1  i ) n  1]
                  Arithmetic Gradient
                  Conversion Factor (to     1  (1  ni)(1  i )  n
  P        G      present value)                      i2
                                                                       20
 Compound Interest Factors
Discrete Cash Flow, Continuous Compounding

To Find   Given       Name of Factor          Factor
                  Compound Amount
  F        P      Factor (single payment)     e rn
                  Present Worth Factor
  P        F      (single payment)           e  rn

                  Compound Amount           e rn  1
  F        A      Factor (uniform series)   er 1
                                            er 1
  A        F      Sinking Fund Factor       e rn  1


                                                       21
 Compound Interest Factors
Discrete Cash Flow, Continuous Compounding

To Find   Given      Name of Factor                Factor
                                                e rn (e r  1)
  A        P      Capital Recovery Factor          e rn  1
                                                    e rn  1
                  Present Worth Factor
  P        A      (uniform series)               e rn (e r  1)
                  Arithmetic Gradient          1     n
                  Conversion Factor (to           rn
  A        G      uniform series)
                                            e 1 e 1
                                             r


                  Arithmetic Gradient
                  Conversion Factor (to     e rn  1  n(e r  1)
  P        G      present value)                e rn (e r  1) 2
                                                                    22
   Compound Interest Factors
Continuous Uniform Cash Flow, Continuous Compounding

  To Find   Given       Name of Factor        Factor
                    Sinking Fund Factor
                                                  r
                    (continuous, uniform
    C        F      payments)                 e rn  1
                    Capital Recovery Factor
                                               re rn
                    (continuous, uniform
    C        P      payments)                 e rn  1
                    Compound Amount
                                              e rn  1
                    Factor (continuous,
    F        C      uniform payments)             r
                    Present Worth Factor
                                              e rn  1
                    (continuous, uniform
    P        C      payments)                  re rn
                                                         23
        Quiz---When and Where
•   Quiz: Tuesday, Sept. 27, 2005
•   11:30 - 12:20 (Quiz: 30 minutes)
•   Tutorial: Wednesday, Sept. 28, 2005
•   ELL 168 Group 1
•   (Students with Last Name A-M)
•   ELL 061 Group 2
•   (Students with Last Name N-Z)



                                          24
     Quiz---Who will be there

•   U, U, U, U, and U!!!!
•   CraigTipping ctipping@uvic.ca
•   Group 1 (Last NameA-M) ELL 168
•   LeYang          yangle@ece.uvic.ca
•   Group 2 (Last Name N-Z) ELL 061


                                         25
    Quiz---Problems, Solutions

• Do not argue with your TA!
• Question? Problems? TAWei
• Solutions will be given on Tutorial
• Bring: Blank Letter Paper, Pen, Formula
  Sheet, Calculator, Student Card
• Write: Name, Student No. and Email


                                            26
    Quiz---Based on Chapter 1.2.3.

•   Important: Wei’s Slides
•   Even More Important: Examples in Slides
•   1 Formula Sheet is a good idea
•   5 Questions for 1800 seconds.
•   Wei used 180 seconds (relax)

                                         27
Quiz---Important Points

•   Simple Interests
•   Compound Interests
•   Future Value
•   Present Value
•   Key: Compound Interest
•   Key: Understand the Question

                                   28
Quiz---Books in Library!!!

                 Engineering Economics in Canada, 3/E


              Niall M. Fraser, University of Waterloo
              Elizabeth M. Jewkes, University of Waterloo
              Irwin Bernhardt, University of Waterloo
              May Tajima, University of Waterloo




 Economics: Canada in the Global Environment
 by Michael Parkin and Robin Bade.


                                                   29
Calculator Talk

•   No programmable
•   No economic function
•   Simple the best
•   Trust your ability
•   Trust your teaching group

                                30
• Questions?

• (Sorry I forget the problems)




                                  31
      Project----Time Table
•   Find your group: Mid-October
•   Select Topic: End of October
•   Survey finished: End of October
•   Project: November (3 Weeks)
•   Project Report Due: Final Quiz


                                      32
Project----Requirements

•   Group: 3-6 Students
•   Topic: Practical, Small
•   Report: On Time, Original
•   Marks: 1 make to 1 report
•   Report: 25 marks out of 100


                                  33
    Project Topic----What to do

•   You Find it
•   Practical
•   Example: Run a Pizza Shop
•   Example: Run a Store for computer renting
•   Example: Survey on the Tuition Increase
•   Example: Why ??? Company failed…..
•   Team Work


                                                34
        Project----Recourse
•   Not your teaching group
•   No spoon feed: Independent work
•   Example: Government Web
•   Example: Library, Database, Google
•   Example: Economics Faculty
•   Example: Newspaper, TV
•   Example: Friends
                                         35
                 Summary
•   Conversion for Arithmetic Gradient Series
•   Conversion for Geometric Gradient Series
•   Quiz: My slides and the examples in slides
•   Project: Good Idea, be open, independent




                                                 36

				
DOCUMENT INFO
Shared By:
Categories:
Stats:
views:103
posted:10/3/2010
language:English
pages:36