Time Value of Money Chapter 2.

Time Value of Money Chapter 2

Take a Lump Sum or Annual Installments
Mr. Robert Harris and his wife Tonya won a lottery worth $276 million on February 25, 2008. The couple weighted a difficult question: whether to take $167 million now or $275 million over 26 years (0r $10.57 million a year). What basis should the couple compare these two options? Fundamentals of Engineering Economics, 2nd edition © 2008

What Do We Need to Know? To make such comparisons (the lottery decision problem), 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. Fundamentals of Engineering Economics, 2nd edition © 2008

Interest Formulas – Single Cash Flows Equal-Payment Series
Time Value of Money The Cost of Money is established and measured by an interest rate, a percentage that is periodically applied and added to an amount of money over a specified length of time. Economic Equivalence Interest Formulas – Single Cash Flows Equal-Payment Series Dealing with Gradient Series Composite Cash Flows.

Money has a time value. Why?
Time Value of Money Money has a time value. Why? because it can earn more money over time (earning power). because its purchasing power changes over time (inflation). Time value of money is measured in terms of interest rate.

Figure 2-1 Gains achieved or losses incurred by delaying consumption

Elements of Transactions involve Interest
Initial amount of money in transactions involving debt or investments is called the principal (P). The interest rate ( i ) measures the cost or price of money and is expressed as a percentage per period of time. A period of time, called the interest period (n), determines how frequently interest is calculated. A specified length of time marks the duration of the transactions and thereby establishes a certain number of interest periods (N). A plan for receipts or disbursements (An) that yields a particular cash flow pattern over a specified length of time. [monthly equal payment] A future amount of money (F) results from the cumulative effects of the interest rate over a number of interest periods.

Assumptions (unless stated otherwise)
Interest rates used are the Market Interest Rate (considers the earning power of money as well as the effect of inflation perceived in the marketplace) Cash flow transactions are given in terms of Actual Dollars (effect of inflation, if any, is reflected in the amount)

EXAMPLE OF INTEREST TRANSACTION
Suppose that you apply for an education loan of $30,000 from a bank at a 9% annual interest rate. In addition you pay a $300 loan origination fee when the loan begins. The bank offers two repayment plans, one with equal payments made at the end of every year for the next five years (installment plan) and the other with a single payment made after the loan period of five years (deferment plan).

Table 2.1 Repayment plan offered by the lender
Which Repayment Plan? Table 2.1 Repayment plan offered by the lender End of Year Receipts Payments Plan 1 Plan 2 Year 0 $30,000.00 $300.00 Year 1 $7,712.77 Year 2 Year 3 Year 4 Year 5 46,158.72 The amount of loan = $30,000, origination fee = $300, interest rate = 9% APR (annual percentage rate) F = P (1+0.09)5 = $30,000 x (1.09)5 = $46,158.72

= $7,712.77 = $7,712.77 Figure 2-2 A cash flow diagram for plan 1 of the loan repayment example

A = P + I n A I P Balance $30,000 1 $7,712.77 $2,700.00 $5,013.00
$30,000 1 $7,712.77 $2,700.00 $5,013.00 $24,987 2 2,249.00 5,464.00 19,523 3 1,757.00 5,956.00 13,657 4 1,222 6,490 7,167 5 645 7,067

End of period convention
Assumption: End of period convention Placing all cash flows at the end of the transactions period

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. Engineering economic analysis uses the compound interest scheme exclusively, as it is most frequently practiced in real world.

Simple Interest (Example 2.1)
P = Principal amount i = Interest rate N = Number of interest periods Example: P = $1,000 i = 8% N = 3 years Find: F End of Year Beginning Balance Interest earned Ending Balance $1,000 1 $80 $1,080 2 $1,160 3 $1,240

Simple Interest Formula

Compound Interest (Example 2.1)
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 Find: F End of Year Beginning Balance Interest earned Ending Balance $1,000 1 $80 $1,080 2 $86.40 $1,166.40 3 $93.31 $1,259.71

Compound Interest Formula

Example: i = 8%, Given: P = $1,000, N = 3 years Find: F $1,259.71
3 $1,000

Practice Problem: Warren Buffett’s Berkshire Hathaway (BRK.A)
Went public in 1965: $18 per share Worth today (January 9, 2009): $94,750 Annual compound growth: 21.50% Current market value: $ Billion If his company continues to grow at the current pace, what will be his company’s total market value when he reaches 100? (78 years old as of 2009) Fundamentals of Engineering Economics, 2nd edition © 2008

Estimated Market Value in 2031
Assume that the company’s stock will continue to appreciate at an annual rate of 21.50% for the next 22 years. Fundamentals of Engineering Economics, 2nd edition © 2008

With EXCEL In 1626 the Indians sold Manhattan Island to Peter Minuit of the Dutch West Company for $24. If they saved just $1 from the proceeds in a bank account that paid 8% interest, how much would their descendents have now? As of Year 2009, the total US population would be close to 300 millions. If the total sum would be distributed equally among the population, how much would each person receive? Fundamentals of Engineering Economics, 2nd edition © 2008

Practice Problem Problem Statement
If 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?

Solution F $100 $200

? Practice problem Problem Statement
Consider 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? ? $1,210 3 1 4 2 $1,500 $1,000 $1,000

? $1,210 1 3 2 4 $1,000 $1,000 $1,500 $1,100 $1,000 $1,210 $2,981 $2,100 $2,310 + $1,500 -$1,210 $1,100 $2,710

Solution: Tabular Approach
End of Period Beginning balance Deposit made Withdraw Ending n = 0 $1,000 n = 1 $1,000( ) =$1,100 $2,100 n = 2 $2,100( ) =$2,310 $1,210 $1,100 n = 3 $1,100( ) =$1,210 $1,500 $2,710 n = 4 $2,710( ) =$2,981 $2,981 Fundamentals of Engineering Economics, 2nd edition © 2008

Economic Equivalence

At Issue – How Do You Compare Two Different Money Transactions?
Consider the following two cash receipts. Which option would you prefer? $1,200 $1,000 Option 1 Option 2

Which Option Would You Prefer?

What do we mean by “economic equivalence?”
Why do we need to establish an economic equivalence? How do we establish an economic equivalence?

Definition - Economic Equivalence
Economic equivalence exists between cash flows that have the same economic effect and could therefore be traded for one another. Even though the amounts and timing of the cash flows may differ, an appropriate interest rate (discount rate) makes them equal.

Equivalence Calculation: A Simple example
Figure 2-4 Using compound interest to establish economic equivalence

= Practice Problem F At 8% interest, what is the equivalent worth
of $2,042 now 5 years from now? $2,042 If you deposit $2,042 today in a savings account that pays 8% interest annually, how much would you have at the end of 5 years? 1 2 3 4 5 3 F = 5

Solution

would these two payments be equivalent?
Example 2.2 Equivalence At what interest rate would these two payments be equivalent? i = ? $2,007 $1,500 5

i = (F / P) 1/N – 1 = (2007 / 1500 )1/5 – 1 i = 0.05999 or 6%
EXAMPLE Practice Problem Given: F = $2,007, N = 5 years, P = $1,500, Find: i $2,007 = $1,500 (1+i )5 i = (F / P) 1/N – 1 = (2007 / 1500 )1/5 – i = or 6% If i is less than 6%, you prefer the promise of $2,007 in five years to $1,500 today. If i is greater than 6%, you would prefer $1,500 now. At a lower interest rate, P must be higher in order to be equivalent to the future amount. For example, at i = 4%, P = $1,650.

Figure 2-5 b Equivalence calculation at varying interest rate

Example 2.3 Equivalence Calculation
1 2 3 4 5 $100 $80 $120 $150 $200 = 1 2 3 4 5 Compute the equivalent lump-sum amount at n = 3 at 10% annual interest.

Equivalent Worth Calculation at n = 3.

Figure 2-6 Equivalent worth calculation at n = 3
EXAMPLE Equivalence Calculation The cash flows given in figure 2-6, FIND: V3 (equivalent worth at n = 3) and i = 10%. Step 1 $100(1+0.1)3+ $80(1+0.1)2+$120(1+0.1)1+$150 = $511.90 Step 2 $200(1+0.1)-1+ $100(1+0.1)-2 = $264.46 Step 3 V3= $ $ = $776.36 Figure 2-6 Equivalent worth calculation at n = 3

Compound Amount Factor
Interest Formulas for Single Cash Flows Compound Amount Factor Figure 2-7 Compounding process: Find F, given P, i, and N

Interest Rate Factors (15 %) N
Single Payment Equal Payment Series Gradient Series N Compound Amount Factor (F/P, i, N) Present Worth Factor (P/F, i, N) Compound Amount Factor (F/A, i, N) Sinking Fund Factor (A/F, i, N) Present Worth Factor (P/A, i, N) Capital Recovery Factor (A/P, i, N) Gradient Uniform Series (A/G, i, N) Gradient Present Worth (P/G, i, N) 1 1.1500 0.8696 1.0000 0.0000 2 1.3225 0.7561 2.1500 0.4651 1.6257 0.6151 3 1.5209 0.6575 3.4725 0.2880 2.2832 0.4380 0.9071 2.0712 4 1.7490 0.5718 4.9934 0.2003 2.8550 0.3503 1.3263 3.7864 5 2.0114 0.4972 6.7424 0.1483 3.3522 0.2983 1.7228 5.7751

Figure 2-8 Cash flow diagram
Example 2.4 If you had $1,000 now and invested it at 7% interest compounded annually, how much would it be worth in 8 years? Given: P = $1,000, i = 7 %, and N = 8 years; Find: F F = $1,000 (1+0.07)8 = $1, or using this F = P (F/P, i, N) factor notation together with table value F = $1,000 (1.7182) = $1,718.19 Figure 2-8 Cash flow diagram

Present-Worth Factor

Example 2.5 Given: F = $1000, i = 6%, and N = 8 years Find: P P = $1,000 (1+0.06)-8 = $1,000 (0.6274) = $627.40 Figure 2-10 Cash flow diagram

Example 2.6 Given: P = $10, F = $20 and N = 5 years, Find: i
F = $20 = $10 (1+ i )5 $2 = (1+ i )5 = (F / P, i, 5) or look at N = 5 row under the (F/P, i, 5) column until you can locate the value of 2. or 1+ i = = i = – 1 = or = 14.87%

Example 2.7 Given: P = $3,000, F = $6,000 and i = 12%, Find: N F = P (1+ i )N = P (F/P, i, N) = which is in this case $6,000 = $3,000 ( )N = expressed as $3,000 (F/P, 12%, N) or 2 = (1.12) N = (F/P, 12%, N) Log 2 = N log 1.12 solve for N gives N = log 2 / log 1.12 = / = 6.116 ≈ 6.12 years Figure 2-12 Cash flow diagram

Practice Problem $1,000 $500 A Given: i = 10%, 0 1 2 3 Find: C
A B C Given: i = 10%, Find: C that makes the two cash flow streams to be indifferent

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 A B C

Solution For A: For B: To Find C: $500 $1,000 A B C

Practice Problem $500 $1,000 $502 $502 $502 A B At what interest rate would you be indifferent between the two cash flows?

Approach $1,000 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 A $ $ $502 B

Establish Equivalence at n = 3
Find the solution by trial and error, say i = 8%

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?

Solution F = $10,000; N = 6 years; i = 7%; Find P
6 P

Example 2.8 Tuition Prepayment
Option (TPO) Figure 2-13 Decomposition of uneven cash flow series

At What Interest Rate Are These Two Transactions Equivalent?
$1,200 $1,000 Option 1 Option 2

Select the base period at n = 3.
$1,200 $1,000  Option 1 Base period Option 2

Lottery Problem – What interest rate are the two payoff options equivalent?
$10.57M

Solution – about 4.5%

Example 2.7 – Find N, Given P, F, and i
$6,000 How many years would it take an investment to double at 12% annual interest? Exact Solution: N = ? $3,000

Rule of 72 Approximating how long it will take for a sum of money to double

Using either F/P or P/F factors
Practice Problems Using either F/P or P/F factors

Problem 1 Today you deposited $1,000 into a savings account paying 9% interest. How much should you have in 10 years?

Problem 2 A loaf of bread today costs $1.50. If grocery prices are going up at the rate of 7% per year, how much will a loaf of bread cost in 10 years?

Problem 3 The average price of new homes is $180,000. If new home prices are increasing at a rate of 6% per year, how much a new home cost in 10 years?

Problem 4 You are interested in buying a piece of real estate property that could be worth $200,000 in 5 years. Assuming that your money is worth 9%, how much would you be willing to settle for?

Problem 5 If 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?

Interest Formulas – Equal Payment Series

Figure 2-14 Equal payment series: Find equivalent P or F

Equal Payment Series – Compound Amount Factor
1 2 N A A A F N 1 2 1 2 N A A A

Equal Payment Series Compound Amount Factor: Find F, Given A, i, and N
Figure 2-15 Cash flow diagram of the relationship between A and F

Equal Payment Series Compound Amount Factor (Future Value of an annuity)
N A Example 2.9: Given: A = $5,000, N = 5 years, and i = 6% Find: F Solution: F = $5,000(F/A, 6%, 5) = $28, or use interest factor table and get the value F = $5,000 (5.6371) = $28,185.46

Validation

Example 2.10 Handling Time Shifts in a Uniform Series
First deposit occurs at n = 0 i = 6% $5,000 $5, $5, $5,000 $5,000

Solution: =FV(6%,5,5000,0,1) Annuity Due Excel Solution
Beginning period =FV(6%,5,5000,0,1)

Finding an Annuity Value
N A = ? Example: Given: F = $5,000, N = 5 years, and i = 7% Find: A Solution: A = $5,000(A/F, 7%, 5) = $869.50

Sinking fund 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.

Sinking Fund Factor is 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. F N A 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

i = 7% Given: F = $100,000 N = 8 years Find: A from table (0.0975)
Solution: A = $100,000(A/F, 7%, 8) = $9,746.78

Capital Recovery Factor (Annuity Factor)
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. A series of equal payments occurring at equal periods of time. 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.

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 which indicates a series of payments of a fixed amount for a specified number of periods. P N A = ? Example 2.12: You borrowed $21, to finance the educational expenses for your senior year of college. The loan will be paid off over five years. The loan carries an interest rate of 6% per year and is to be repaid in equal annual installments over the next five years. Assume that the money was borrowed at the beginning of your senior year and that the first installment will be due a year later. Compute the amount of the annual installments.

Example 2.12 Paying Off Educational Loan
A A A A A Example 2.12: Paying Off Education Loan Given: P = $21,061.82, N = 5 years, and i = 6% Find: A (0.2374) from table Solution: A = $21,061.82(A/P,6%,5) = $5,000

i = 6% i = 6% Example 2.13 Deferred (delayed) Loan Repayment Plan
Grace period A A A A A P ’ = $21,061.82(F / P, 6%, 1) = $22,325.53 i = 6% A’ A’ A’ A’ A’

Two-Step Procedure

Using F/A, A/F, P/A, or A/P Factor
Practice Problems Using F/A, A/F, P/A, or A/P Factor

Problem 1 You are paying into a mutual fund that earns 6%. If the payments are $10,000 per year, how much will be in the funds in 15 years?

Problem 2 Your company is required to pay into a sinking fund each year in order to replace an industrial equipment costing $100,000 at the end of 10 years from now. The sinking fund is expected to earn 6% interest. How much must your company deposit each year to meet these needs?

Problem 3 You borrow $10,000 agreeing to pay the balance in 10 equal annual installments at an interest rate of 8%. What should the payments be?

Problem 4 You are interested in replacing an industrial equipment to save labor cost. Labor savings are expected to be $30,000 per year over 10 years. The equipment costs $180,000. Should you buy this equipment if your interest rate is 12%?

Present Value of Perpetuities
P = A/i Example Perpetual steam of $1000. if the interest rate today is 10%, what's is the worth of the perpetuity? P= 1000/0.1 = $10,000

Interest Formulas (Gradient Series)

Two Types of Gradient Series
A1+7G Linear Gradient G – Constant $ amount Geometric Gradient g – Constant % A1+2G A1+G A1 A1(1+g) A1

Linear Gradient Series
Engineers frequently meet situations involving periodic payments that increase or decrease by a constant amount (G) from period to period. Figure 2-25 Cash flow diagram of a strict gradient series

Gradient Series as a Composite Series of a Uniform Series of N Payments of A1 and the Gradient Series of Increments of Constant Amount G.

Present-Worth Factor: Linear Gradient Find P, Given G, N, and i
Figure 2-27 Cash flow diagram of a strict gradient series

Figure 2-28 Cash flow diagram representing a graduated payment plan
Example Creating a Graduated Loan Repayment with a Linear Gradient Series; Given P, A1, N and i Find: G Figure 2-28 Cash flow diagram representing a graduated payment plan SOLUTION: Since the loan payment series consist of two parts – (1) a $1,500 equal payment series and (2) A strict gradient series (unknown, yet to be determined) – we can calculate the present value of each series and equate them with $10,000. $10,000 = $1,500 (P/A, 10%, 5) + G (P/G, 10%, 5) get the table values from appendix B $10,000 = $5, G $10,000 - $5, = G G = $628.67

Present value calculation for a gradient series
$2,000 $1,750 $1,500 $1,250 $1,000 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? P =?

Method 1: Using the (P/F, i, N) Factor
$2,000 $1,750 $1,500 $1,250 $1,000 $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 P =?

Method 2: Using the Gradient Factor

Example 2.17 – Equivalent Cash Value

Equivalent Present Value of Annual Payment Option at 4.5%

Excel Spreadsheet Presentation of the Linear Gradient Series

Geometric Gradient Series Many 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.

Example 2.18 Suppose your retirement benefit during your first year of retirement are $50,000 Assume that this amount is just enough to meet your cost of living during the first year. However, your cost of living is expected to increase at an annual rate of 5% due to inflation Suppose you do not expect to receive any cost of living adjustment in your retirement pension. Then some of your cost of living has to come from your savings retirement pension If your saving account earns 7% interest a year, how much you should set a side in order to meet this future increase in the cost of living over 25 years

Example Required Cost-of-living Adjustment Calculation Find P, Given A1, g, i, N Figure 2-31 Cash flow diagram

Given: g = 5%, i = 7%, N = 25 years, A1 = $50,000
Find the equivalent amount of total benefit paid over 25 years. P = $50,000 (P/A, 7%, 25) = $582,679 Find the equivalent amount of total cost of living with inflation. P = $50,000 (P/A, 5%, 7%, 25) The Required additional savings to meet the future increase in cost of living will be ∆P = $940,167 - $582,679 = $357,488

An Excel Worksheet to Illustrate the Process of Calculating the Savings Required for Retirement

Using (P/G, i, N) or (P/A1,g,i,N)
Practice Problems Using (P/G, i, N) or (P/A1,g,i,N)

Problem 1 $2,000 $1,750 $1,500 $1,250 $1,000 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? P =?

Solution:

Problem 2 Your starting salary as a mechanical engineer is expected to be $55,000. A total of 10% of your salary each year will be placed in the mutual fund of your choice. You can also count on a 5% salary increase for the next 30 years of employment. If the mutual fund will average 9% annual return over the course of your career, what can you expect at retirement?

COMPOSITE CASH FLOWS Compute the equivalent present worth for mixed payment series at 15%.
METHOD 1

METHOD 2 GROUP THE CASH FLOW COMPONENTS

SOLUTION OF END OF CHAPTER PROBLEMS
22, 38, 39,

Composite Cash Flows – Brute Force Approach

Composite Cash Flows – Grouping Approach

Personal Savings Point of View
Situation 1: If you make four annual deposits of $100 in your savings account which earns 10% annual interest, what equal annual amount (A) can be withdrawn over 4 subsequent years?

Economic Equivalence Point of View
Situation 2: What value of A would make the two cash flow transactions equivalent if i = 10%?

Method 1: Select the Base Period at n = 0

Method 2: Select the Base Period at n = 4

Equivalence Calculation with Changing Interest Rates
F = ? Find the balance at the end of year 5. 6% 4% 4% 6% 5% 2 4 5 1 3 $400 $300 $500

Solution

Cash Flows with a Missing Payment
$100 i = 10% Missing payment

Strategy – Modify the Cash Flow Series to Have a Standard Pattern
Add this cash flow to offset the change $100 $100 Pretend that we have the 10th payment i = 10%

Equivalence Between Inflows and Outflows
P = ? $100 i = 10% $100

Unconventional Regularity in Cash Flow Pattern
$10,000 i = 10% 1 C C C C C C C Payment is made every other year

Approach: Modify the Interest Rate
Idea: Since the cash flows occur every other year, let's find out the equivalent compound interest rate that covers the two-year period. How: If interest is compounded 10% annually, the equivalent interest rate for two-year period is 21%. (1+0.10)(1+0.10) = 1.21

Solution $10,000 i = 21% 1 C C C C C C C

HOMEWORK PROBLEMS DUE DATE IS Monday March 14th class time.
5, 8, 14, 20 , 31, 36 , 43 (part a only), 47, 49, 62

Time Value of Money Chapter 2.

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