time value of money lecture no.2 professor c. s. park fundamentals of engineering economics...
TRANSCRIPT
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Time Value of Money
Lecture No.2Professor C. S. ParkFundamentals of Engineering EconomicsCopyright © 2005
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Time Value of Money
Interest: The Cost of Money
Economic Equivalence
Interest Formulas – Single Cash Flows
Equal-Payment Series
Dealing with Gradient Series
Composite Cash Flows.
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What Do We Need to Know? To make such comparisons, we
must be able to compare the value of money at different point in time.
To do this, we need to develop a method for reducing a sequence of benefits and costs to a single point in time. Then, we will make our comparisons on that basis.
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Time Value of Money Money has a time value
because it can earn more money over time (earning power).
Money has a time value because its purchasing power changes over time (inflation).
Time value of money is measured in terms of interest rate.
Interest is the cost of money-a cost to the borrower and an earning to the lender
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Delaying Consumption
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N = 0 $100
N = 1 $104(inflation rate = 4%)
N = 0 $100
N = 1 $106(earning rate =6%)
Case 2: Earning power exceeds inflation
N = 0 $100
N = 1 $108(inflation rate = 8%)
N = 0 $100
N = 1 $106(earning rate =6%)
Case 1: Inflation exceeds earning power
Cost of RefrigeratorAccount Value
N = 0 $100
N = 1 $104(inflation rate = 4%)
N = 0 $100
N = 1 $106(earning rate =6%)
Case 2: Earning power exceeds inflation
N = 0 $100
N = 1 $108(inflation rate = 8%)
N = 0 $100
N = 1 $106(earning rate =6%)
Case 1: Inflation exceeds earning power
Cost of RefrigeratorAccount Value
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Elements of Transactions involve Interest1. Initial amount of money in transactions involving debt
or investments is called the principal (P).2. The interest rate (i) measures the cost or price of
money and is expressed as a percentage per period of time.
3. A period of time, called the interest period (n), determines how frequently interest is calculated.
4. A specified length of time marks the duration of the transactions and thereby establishes a certain number of interest periods (N).
5. A plan for receipts or disbursements (An) that yields a particular cash flow pattern over a specified length of time. [monthly equal payment]
6. A future amount of money (F) results from the cumulative effects of the interest rate over a number of interest periods.
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Which Repayment Plan?End of Year
Receipts Payments
Plan 1 Plan 2
Year 0 $20,000.00
$200.00 $200.00
Year 1 5,141.85 0
Year 2 5,141.85 0
Year 3 5,141.85 0
Year 4 5,141.85 0
Year 5 5,141.85 30,772.48The amount of loan = $20,000, origination fee = $200, interest rate = 9% APR (annual percentage rate)
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Cash Flow DiagramRepresent time by a horizontal line marked off with the number of interest periods specified. Cash flow diagrams give a convenient summary of all the important elements of a problem.
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Methods of Calculating Interest Simple interest: the practice of
charging an interest rate only to an initial sum (principal amount).
Compound interest: the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn.
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Simple Interest
P = Principal amount
i = Interest rate N = Number of
interest periods Example:
P = $1,000 i = 8% N = 3 years
End of Year
Beginning
Balance
Interest
earned
Ending Balanc
e
0 $1,000
1 $1,000 $80 $1,080
2 $1,080 $80 $1,160
3 $1,160 $80 $1,240
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Simple Interest Formula
( )
where
= Principal amount
= simple interest rate
= number of interest periods
= total amount accumulated at the end of period
F P iP N
P
i
N
F N
$1,000 (0.08)($1,000)(3)
$1,240
F
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Compound Interest the practice of charging an interest rate to an initial sum and to any previously accumulated interest that has not been withdrawn. P = Principal
amount i = Interest
rate N = Number of
interest periods
Example: P = $1,000 i = 8% N = 3 years
End of
Year
Beginning
Balance
Interest
earned
Ending Balance
0 $1,000
1 $1,000 $80 $1,080
2 $1,080 $86.40 $1,166.40
3 $1,166.40
$93.31 $1,259.71
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Compounding Process
$1,000
$1,080
$1,080
$1,166.40
$1,166.40
$1,259.710
1
2
3
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0
$1,000
$1,259.71
1 2
3
3$1,000(1 0.08)
$1,259.71
F
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Compound Interest Formula
1
22 1
0 :
1: (1 )
2 : (1 ) (1 )
: (1 )N
n P
n F P i
n F F i P i
n N F P i
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Practice Problem
Problem StatementIf you deposit $100 now (n = 0) and $200 two years from now (n = 2) in a savings account that pays 10% interest, how much would you have at the end of year 10?
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Solution
0 1 2 3 4 5 6 7 8 9 10
$100$200
F
10
8
$100(1 0.10) $100(2.59) $259
$200(1 0.10) $200(2.14) $429
$259 $429 $688F
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Practice problem
Problem StatementConsider the following sequence of deposits and withdrawals over a period of 4 years. If you earn 10% interest, what would be the balance at the end of 4 years?
$1,000 $1,500
$1,210
0 1
2 3
4?
$1,000
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$1,000$1,500
$1,210
0 1
2
3
4
?
$1,000
$1,100
$2,100 $2,310
-$1,210
$1,100
$1,210
+ $1,500
$2,710
$2,981
$1,000
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SolutionEnd of Period
Beginningbalance
Deposit made
Withdraw Endingbalance
n = 0 0 $1,000 0 $1,000
n = 1 $1,000(1 + 0.10) =$1,100
$1,000 0 $2,100
n = 2 $2,100(1 + 0.10) =$2,310
0 $1,210 $1,100
n = 3 $1,100(1 + 0.10) =$1,210
$1,500 0 $2,710
n = 4 $2,710(1 + 0.10) =$2,981
0 0 $2,981
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Economic Equivalence
What do we mean by “economic equivalence?”
Why do we need to establish an economic equivalence?
How do we establish an economic equivalence?
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Economic Equivalence (EE) Economic equivalence exists between cash
flows that have the same economic effect and could therefore be traded for one another.
EE refers to the fact that a cash flow-whether a single payment or a series of payments-can be converted to an equivalent cash flow at any point in time.
Even though the amounts and timing of the cash flows may differ, the appropriate interest rate makes them equal.
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Economic Equivalence (EE)
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If you deposit P dollars today for N periods at i, you will have F dollars at the end of period N.
N
F
P
0
NiPF )1(
Equivalence from Personal Financing Point of View
P F
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F dollars at the end of period N is equal to a single sum P dollars now, if your earning power is measured in terms of interest rate i.
N
F
P
0
NiFP )1(
Alternate Way of Defining Equivalence
N0
=
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Practice Problem
0
1 2 33 4 5
$2,042
5
F
0
At 8% interest, what is the equivalent worth of $2,042 now 5 years from now?
If you deposit $2,042 today in a savingsaccount that pays 8% interest annually.how much would you have at the end of5 years?
=
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Solution
5$2,042(1 0.08)
$3,000
F
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Example 2.2
$3,000$2,042
50
At what interest rate would these two amounts be equivalent?
i = ?
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Equivalence Between Two Cash Flows
Step 1: Determine the base period, say, year 5.
Step 2: Identify the interest rate to use.
Step 3: Calculate equivalence value.
$3,000$2,042
50
i F
i F
i F
6%, 042 1 0 06 733
8%, 042 1 0 08 000
10%, 042 1 0 10
5
5
5
$2, ( . ) $2,
$2, ( . ) $3,
$2, ( . ) $3,289
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Example - Equivalence
Various dollar amounts that will be economically equivalent to $3,000 in 5 years, given an interest rate of 8%.
0 1 2 3 4 5
P F
$2,042 $3,000$2,205 $2,382 $2,778$2,572
5
$3,000$2,042
(1 0.08)P
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Example 2.3
0 1 2 3 4 5
$100$80
$120$150
$200
$100
0 1 2 3 4 5
V=
Compute the equivalent lump-sum amount at n = 3 at 10% annual interest.
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0 1 2 3 4 5
$100
$80
$120$150
$200
$100
V
Approach
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0 1 2 3 4 5
$100$80
$120$150
$200
$100
V
3 2100(1 0.10) $80(1 0.10) $120(1 0.10) $150
$511.90
1 2$200(1 0.10) $100(1 0.10)
$264.46
3 $511.90 $264.46
$776.36
V
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Practice Problem
How many years would it take an investment to double at 10% annual interest?
P
2P
0
N = ?
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Solution
2 (1 0.10)
2 1.1
log 2 log1.1
log 2
log1.1
7.27 years
N
N
F P P
N
N
P
2P
0
N = ?
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Given: i = 10%,
Find: C that makes the two cash flow streams to be indifferent
Practice Problem
$500
$1,000
0 1 2 3
0 1 2 3
A
B
C C
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Approach
Step 1: Select the base period to use, say n = 2.
Step 2: Find the equivalent lump sum value at n = 2 for both A and B.
Step 3: Equate both equivalent values and solve for unknown C.
$500
$1,000
0 1 2 3
0 1 2 3
A
B
C C
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Solution
For A:
For B:
To Find C:
2 12 $500(1 0.10) $1,000(1 0.10)
$1,514.09
V
$500
$1,000
0 1 2 3
0 1 2 3
A
B
C C2 (1 0.10)
2.1
V C C
C
2.1 $1,514.09
$721
C
C
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At what interest rate would you be indifferent between the two cash flows?
$500
$1,000
0 1 2 3
0 1 2 3
$502 $502 $502
A
B
Practice Problem
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Approach
Step 1: Select the base period to compute the equivalent value (say, n = 3)
Step 2: Find the net worth of each at n = 3.
$500
$1,000
0 1 2 3
0 1 2 3
$502 $502 $502
A
B
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Establish Equivalence at n = 3
33
23
Option A : $500(1 ) $1,000
Option B : $502(1 ) $502(1 ) $502
F i
F i i
33
23
Option A : $500(1.08) $1,000
$1,630
Option B : $502(1.08) $502(1.08) $502
$1,630
F
F
Find the solution by trial and error, say i = 8%
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Single Cash Flow Formula
Single payment compound amount factor (growth factor)
Given:
Find: F
P
F
N
0
F P i
F P F P i N
N
( )
( / , , )
1
i
N
P
10%
8
000
years
$2,
F
F P
$2, ( . )
$2, ( / , )
$4, .
000 1 010
000 10%,8
28718
8
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Practice Problem
If you had $2,000 now and invested it at 10%, how much would it be worth in 8 years?
$2,000
F = ?
8
0i = 10%
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Solution
8
Given:
$2,000
10%
8 years
Find:
$2,000(1 0.10)
$2,000( / ,10%,8)
$4,287.18
EXCEL command:
=FV(10%,8,0,2000,0)
=$4,287.20
P
i
N
F
F
F P
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Single Cash Flow Formula
Single payment present worth factor (discount factor)
Given:
Find:P
F
N
0
P F i
P F P F i N
N
( )
( / , , )
1
i
N
F
12%
5
000
years
$1,
P
P F
$1, ( . )
$1, ( / , )
$567.40
000 1 0 12
000 12%,5
5
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Practice Problem
You want to set aside a lump sum amount today in a savings account that earns 7% annual interest to meet a future expense in the amount of $10,000 to be incurred in 6 years. How much do you need to deposit today?
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Solution
0
6
$10,000
P 6$10,000(1 0.07)
$10,000( / ,7%,6)
$6,663
P
P F
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Multiple Payments …..(Example 2.8)
How much do you need to deposit today (P) to withdraw $25,000 at n =1, $3,000 at n = 2, and $5,000 at n =4, if your account earns 10% annual interest?
0
1 2 3 4
$25,000
$3,000 $5,000
P
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0
1 2 3 4
$25,000
$3,000 $5,000
P
0
1 2 3 4
$25,000
P1
0
1 2 3 4
$3,000
P2
0
1 2 3 4
$5,000
P4
+ +
1 $25,000( / ,10%,1)
$22,727
P P F
2 $3,000( / ,10%,2)
$2,479
P P F
4 $5,000( / ,10%,4)
$3,415
P P F
1 2 3 $28,622P P P P
Uneven Payment Series
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Check
0 1 2 3 4Beginning Balance
0 28,622 6,484.20
4,132.62
4,545.88
Interest Earned (10%)
0 2,862 648.42 413.26 454.59
Payment
+28,622 -25,000 -3,000 0 -5,000
Ending Balance
$28,622 6,484.20
4,132.62
4,545.88
0.47
Rounding errorIt should be “0.”
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Equal Payment Series: Find equivalent P or F
A
0 1 2 3 4 5 N-1 N
F
P
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Equal Payment Series – Compound Amount Factor
0 1 2 N
0 1 2 NA A A
F
0 1 2N
A A A
F
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Compound Amount Factor
0 1 2 N 0 1 2 N
A A A
F
A(1+i)N-1
A(1+i)N-2
1 2(1 ) (1 )N NF A i A i A
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Equal Payment Series Compound Amount
Factor (Future Value of an annuity)
F Ai
iA F A i N
N
( )
( / , , )
1 1
Example 2.9: Given: A = $5,000, N = 5 years, and i = 6% Find: F Solution: F = $5,000(F/A, 6%, 5) = $28,185.46
0 1 2 3N
F
A
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Validation
F =?
0 1 2 3 4 5
$5,000 $5,000 $5,000 $5,000 $5,000
i = 6%
4
3
2
1
0
$5,000(1 0.06) $6,312.38
$5,000(1 0.06) $5,955.08
$5,000(1 0.06) $5,618.00
$5,000(1 0.06) $5,300.00
$5,000(1 0.06) $5,000.00
$28.185.46
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Finding an Annuity Value
Example: Given: F = $5,000, N = 5 years, and i =
7% Find: A Solution: A = $5,000(A/F, 7%, 5) =
$869.50
0 1 2 3N
F
A = ?
A Fi
i
F A F i N
N
( )
( / , , )
1 1
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Example 2.10 Handling Time Shifts in a Uniform Series
F = ?
0 1 2 3 4 5
$5,000 $5,000 $5,000 $5,000 $5,000
i = 6%
First deposit occurs at n = 0
5 $5,000( / ,6%,5)(1.06)
$29,876.59
F F A
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59
Sinking fund
(1) A fund accumulated by periodic deposits and reserved exclusively for a specific purpose, such as retirement of a debt.
(2) A fund created by making periodic deposits (usually equal) at compound interest in order to accumulate a given sum at a given future time for some specific purpose.
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60
Sinking Fund Factoris an interest-bearing account into which a fixed sum is deposited each interest period; The term within the colored area is called sinking-fund factor.
0 1 2 3N
F
A
A Fi
i
F A F i N
N
( )
( / , , )
1 1
Example 2.11 Example 2.11 – College Savings Plan:
Given: F = $100,000, N = 8 years, and i = 7%
Find: A Solution:
A = $100,000(A/F, 7%, 8) = $9,746.78
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OR
Given: F =
$100,000 i = 7% N = 8 years
0
1 2 3 4 5 6 7 8
$100,000
i = 8%
A = ?
Current age: 10 years old
Find: A Solution: A = $100,000(A/F, 7%, 8)
= $9,746.78
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62
Capital Recovery Factor (Annuity Factor)
Annuity: (1) An amount of money payable to a recipient at regular intervals for a prescribed period of time out of a fund reserved for that purpose. (2) A series of equal payments occurring at equal periods of time. (3) Amount paid annually, including reimbursement of borrowed capital and payment of interest.
Annuity factor: The function of interest rate and time that determines the amount of periodic annuity that may be paid out of a given fund.
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Capital Recovery Factor is the colored area which is designated (A/P, i, N). In finance, this A/P factor is referred to as the annuity factor.
Example 2.12: Paying Off Education Loan
Given: P = $21,061.82, N = 5 years, and i = 6%
Find: A Solution: A = $21,061.82(A/P,6%,5) =
$5,000
1 2 3N
P
A = ?
0
A Pi i
i
P A P i N
N
N
( )
( )
( / , , )
1
1 1
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P =$21,061.82
0 1 2 3 4 5 6
A A A A A
i = 6%
0 1 2 3 4 5 6
A’ A’ A’ A’ A’
i = 6%
P’ = $21,061.82(F / P, 6%, 1)
Grace period
Example 2.13 Deferred (delayed) Loan Repayment Plan
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Two-Step Procedure
' $21,061.82( / ,6%,1)
$22,325.53
$22,325.53( / ,6%,5)
$5,300
P F P
A A P
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Present Worth of Annuity Series The colored area is referred to as the equal-payment-series present-worth factor
Example 2.14: Powerball Lottery
Given: A = $7.92M, N = 25 years, and i = 8%
Find: P Solution: P = $7.92M(P/A, 8%, 25) =
$84.54M
1 2 3N
P = ?
A
0
P Ai
i i
A P A i N
N
N
( )
( )
( / , , )
1 1
1
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0 1 2 3 4 5 6 7 8 9 10 11 12
44
Option 2: Deferred Savings Plan
$2,000
Example 2.15 Early Savings Plan – 8% interest
0 1 2 3 4 5 6 7 8 9 10
44
Option 1: Early Savings Plan
$2,000
?
?
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Option 1 – Early Savings Plan
10
44
$2,000( / ,8%,10)
$28,973
$28,973( / ,8%,34)
$396,645
F F A
F F P
0 1 2 3 4 5 6 7 8 9 10
44
Option 1: Early Savings Plan
$2,000
?
6531Age
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Option 2: Deferred Savings Plan
44 $2,000( / ,8%,34)
$317,233
F F A
0 11 12
44
Option 2: Deferred Savings Plan
$2,000
?
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0 1 2 3 4 5 6 7 8 9 10
44
0 1 2 3 4 5 6 7 8 9 10 11 12
44
Option 1: Early Savings Plan
Option 2: Deferred Savings Plan
$2,000
$2,000
$396,644
$317,253
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71
Linear Gradient SeriesEngineers frequently meet situations involving periodic payments that increase or decrease by a constant amount (G) from period to period.
P Gi i iN
i i
G P G i N
N
N
( )
( )
( / , , )
1 1
12
P
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Gradient Series as a Composite Series
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$1,000$1,250 $1,500
$1,750$2,000
1 2 3 4 50
P =?
How much do you have to deposit now in a savings account that earns a 12% annual interest, if you want to withdraw the annual series as shown in the figure?
Example – Present value calculation for a gradient series
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Method 1:
$1,000$1,250 $1,500
$1,750$2,000
1 2 3 4 50
P =?
$1,000(P/F, 12%, 1) = $892.86$1,250(P/F, 12%, 2) = $996.49$1,500(P/F, 12%, 3) = $1,067.67$1,750(P/F, 12%, 4) = $1,112.16$2,000(P/F, 12%, 5) = $1,134.85
$5,204.03
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Method 2:
P P A1 000 12%,5
604 80
$1, ( / , )
$3, .
P P G2 12%,5
599 20
$250( / , )
$1, .
P
$3, . $1, .
$5,204
604 08 599 20
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0 1 2 3 4 5 6 7 25 26
$3.44 million
0 1 2 3 4 5 6 7 25 26
Cash Option
$175,000$189,000
$357,000
$196,000G = $7,000
Annual Payment Option
Example 2.16 Supper Lottery
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Equivalent Present Value of Annual Payment Option at 4.5%
[$175,000 $189,000( / , 4.5%,25)
$7,000( / , 4.5%,25)]( / , 4.5%,1)
$3,818,363
P P A
P G P F
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Geometric Gradient SeriesMany engineering economic problems, particularly those relating to construction costs, involve cash flows that increase over time, not by a constant amount, but rather by a constant percentage (geometric), called compound growth.
PA
g i
i gi g
NA i i g
N N
1
1
1 1 1
1
( ) ( )
/ ( ),
, if
if
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Present Worth Factor
PA
g i
i gi g
NA i i g
N N
1
1
1 1 1
1
( ) ( )
/ ( ),
, if
if
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Alternate Way of Calculating P
1
Let '1
( / , ', )(1 )
i gg
g
AP P A g N
g
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Example 2.17: Find P, Given A1, g, i, N Given:
g = 5%
i = 7%
N = 25 years
A1 = $50,000
Find: P
25 251 (1 0.05) (1 0.07)$50,000
0.07 0.05
$940,696
P
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Required Additional Savings
$50,000( / ,7%,25)
$582,679
$940,696 $582,679
$358,017
P P A
P
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$50
$100 $100 $100
$150 $150 $150 $150
$200
Group 1 $50( / ,15%,1)
$43.48
P P F
Group 2 $100( / ,15%,3)( / ,15%,1)
$198.54
P P A P F
Group 3 $150( / ,15%,4)( / ,15%,4)
$244.85
P P A P F
Group 4 $200( / ,15%,9)
$56.85
P P F
$43.48 $198.54 $244.85 $56.85
$543.72
P
0
1 2 3 4 5 6 7 8 9
Composite Cash Flows
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