Information Systems and Engineering Economics : Syllabus Unit IV Money and Economic Value 08 Hours Engineering Economic Decisions, Time Value of Money, Understanding Money Management, Case Studies- Economic decisions done in Multi-national companies.

Engineering Economic Decisions 2

Engineering Economy It is a collection of mathematical techniques that simplify economic comparisons. It provides a rational and systematic approach for evaluating different economic decision e.g. purchase of a new manufacturing equipment evaluating different manufacturing methods in terms of economic value to the company replacing existing manufacturing equipment or method 3

Alternatives Alternatives are always present for any economic decision Identifying appropriate alternatives is as important as - if not more important than - evaluating the alternatives Alternatives evaluation variables: initial cost; Interest rate (rate of return) anticipated life of equipment (economic life) annual maintenance/operating cost or benefit resale or retrieve value 4

Engineering Economic Decisions Manufacturing Profit Planning Investment Marketing 5

Why Is Engineering Economics Important? Engineers , Architects are DESIGN things and perform PROJECTS Therefore, engineers and Architects must be concerned with the economic aspects of designs that they recommend, and projects that they perform 6

What Kinds of Questions Can Engineering Economics Answer? Engineering economics is needed for many kinds of decision making Example: Buying a car Alternatives: ● $18,000 now, or $600 per month for 3 years Which is better? 7

What Kinds of Questions Can Engineering Economics Answer? It will help you make good decisions: In your professional life And in your personal life! Knowledge of engineering economics will have a significant impact on you personally! 8

What Kinds of Questions Can Engineering Economics Answer? ENGINEERING ECONOMICS INVOLVES: FORMULATING ESTIMATING EVALUATING ECONOMIC OUTCOMES WHEN CHOICES OR ALTERNATIVES ARE AVAILABLE 9

How Does It Do This? MATHEMATICAL RELATIONSHIPS TO COMPARE THE CASH FLOWS OF THE DIFFERENT ALTERNATIVES 10

A Simple Illustrative Example: Car to Finance – Saturn or Honda? Recognize the decision problem Collect all needed (relevant) information Identify the set of feasible decision alternatives Select the decision criterion to use Select the best possible and implementable decision alternative Need to lease a car Gather technical and financial data Choose between Saturn and Honda Want minimum total cash outlay, safety, good performance Choice between Saturn and Honda (or others) Select a car (i.e., Honda, Saturn or another brand) 11

What Makes Engineering Economic Decisions Difficult What Makes Engineering Economic Decisions Difficult? Predicting the Future Estimating the required investments Estimating product manufacturing costs Forecasting the demand for a brand new product Estimating a “good” selling price Estimating product life and the profitability of continuing production 12

Key Factors in Selecting Good Engineering Economic Decisions Objectives, available resources, time and uncertainty are the key defining aspects of all engineering economic decisions 13

Large-Scale Engineering Projects These typically require a large sum of investment can be very risky take a long time to see the financial outcomes lead to revenue and cost streams that are difficult to predict 14

Role of Engineers in Business Create & Design Engineering Projects Evaluate Expected Profitability Timing of Cash Flows Degree of Financial Risk Evaluate Impact on Financial Statements Firm’s Market Value Stock Price Analyze Production Methods Engineering Safety Environmental Impacts Market Assessment 15

Accounting Vs. Accounting Evaluating past performance Evaluating and predicting future events Accounting Engineering Economy Past Future Present 16

Two Factors in Engineering Economic Decisions The factors of time and uncertainty are the defining aspects of any engineering economic decisions 17

A Large-Scale Engineering Project Requires a large sum of investment Takes a long time to see the financial outcomes Difficult to predict the revenue and cost streams 18

Types of Strategic Engineering Economic Decisions in Manufacturing Sector Service Improvement Equipment and Process Selection Equipment Replacement New Product and Product Expansion Cost Reduction 19

Types of Strategic Engineering Economic Decisions in Service Sector Commercial Transportation Logistics and Distribution Healthcare Industry Electronic Markets and Auctions Financial Engineering Retails Hospitality and Entertainment Customer Service and Maintenance 20

Service Improvement How many more jeans would Levi need to sell to justify the cost of additional robotic tailors? 21

Example - Healthcare Delivery Which plan is more economically viable? Traditional Plan: Patients visit each service provider. New Plan: Each service provider visits patients : patient : service provider 22

Equipment Replacement Problem Now is the time to replace the old machine? If not, when is the right time to replace the old equipment? 23

New Product and Product Expansion Shall we build or acquire a new facility to meet the increased demand? Is it worth spending money to market a new product? 24

Cost Reduction Should a company buy equipment to perform an operation now done manually? Should spend money now in order to save more money later? 25

The Four Fundamental Principles of Engineering Economics 1: An instant dollar is worth more than a distant dollar… 2: Only the relative (pair-wise) difference among the considered alternatives counts… 3: Marginal revenue must exceed marginal cost, in order to carry out a profitable increase of operations 4: Additional risk is not taken without an expected additional return of suitable magnitude 26

Principle 1 An instant dollar is importance more than a distant dollar… Today 6 months later 27

Principle 2 Only the cost (resource) difference among alternatives counts Option Monthly Fuel Cost Monthly Maintenance Cash paid at signing (cash outlay ) Monthly payment Salvage Value at end of year 3 Buy $960 $550 $6,500 $350 $9,000 Lease $2,400 The data shown in the green fields are irrelevant items for decision making, since their financial impact is identical in both cases 28

Principle 3 Marginal (unit) revenue has to exceed marginal cost, in order to increase production Manufacturing cost 1 unit Marginal revenue Sales revenue 1 unit 29

Principle 4 Additional risk is not taken without a suitable expected additional return Investment Class Potential Risk Expected Return Savings account (cash) Lowest 1.5% Bond (debt) Moderate 4.8% Stock (equity) Highest 11.5% A simple illustrative example. Note that all investments imply some risk: portfolio management is a key issue in finance 30

Summary The term engineering economic decision refers to any investment or other decision related to an engineering project The five main types of engineering economic decisions are (1) service improvement, (2) equipment and process selection, (3) equipment replacement, (4) new product and product expansion, and (5) cost reduction The factors of time, resource limitations and uncertainty are key defining aspects of any investment project Notice that all listed decision types can be seen and modeled as a constrained decision (optimization) problem 31

Time Value of Money

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.

Time value of money is measured in terms of interest rate. 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

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.

Which Repayment Plan? End of Year Receipts Payments Plan 1 Plan 2 $20,000.00 $200.00 Year 1 5,141.85 Year 2 Year 3 Year 4 Year 5 30,772.48 The amount of loan = $20,000, origination fee = $200, interest rate = 9% APR (annual percentage rate)

Definitions Project : an investment opportunity generating cash flows over time c Cash Flow: the movement of money (in or out) of a project c Interest: used to move money through time for comparisons. The rent for loaned money (cost of using money) 39

Cash Flow Diagram Represent 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.

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.

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 Balance $1,000 1 $80 $1,080 2 $1,160 3 $1,240

Simple Interest Formula

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 $1,000 1 $80 $1,080 2 $86.40 $1,166.40 3 $93.31 $1,259.71

Compounding Process $1,080 $1,166.40 $1,259.71 1 $1,000 2 3 $1,080 $1,259.71 1 $1,000 2 3 $1,080 $1,166.40

$1,259.71 1 2 3 $1,000

Compound Interest Formula

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 0 1 2 3 4 5 6 7 8 9 10 $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 1 4 2 3 $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 n = 0 n = 1 n = 2 n = 3 n = 4 End of Period Beginning balance Deposit made Withdraw Ending n = 0 $1,000 n = 1 $1,000(1 + 0.10) =$1,100 $2,100 n = 2 $2,100(1 + 0.10) =$2,310 $1,210 $1,100 n = 3 $1,100(1 + 0.10) =$1,210 $1,500 $2,710 n = 4 $2,710(1 + 0.10) =$2,981 $2,981

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

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.

Economic Equivalence (EE)

Equivalence from Personal Financing Point of View If you deposit P dollars today for N periods at i, you will have F dollars at the end of period N.

Alternate Way of Defining Equivalence P 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 N

= F At 8% interest, what is the equivalent worth Practice Problem 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 4 5 3 F = 5

Solution

would these two amounts be equivalent? Example 2.2 At what interest rate would these two amounts be equivalent? i = ? $3,000 $2,042 5

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 5

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

Example 2.3 V 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.

Approach 1 2 3 4 5 $100 $80 $120 $150 $200 V

1 2 3 4 5 $100 $80 $120 $150 $200 V

Practice Problem 2P How many years would it take an investment to double at 10% annual interest? N = ? P

Solution P 2P N = ?

Practice Problem Given: i = 10%, $500 $1,000 0 1 2 3 0 1 2 3 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 0 1 2 3 0 1 2 3 A B C

Solution $500 $1,000 0 1 2 3 0 1 2 3 A B C For A: For B: To Find C:

Practice Problem $500 $1,000 0 1 2 3 0 1 2 3 $502 $502 $502 A B At what interest rate would you be indifferent between the two cash flows?

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. $1,000 $500 A 0 1 2 3 $502 $502 $502 B 0 1 2 3

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

Single Cash Flow Formula Single payment compound amount factor (growth factor) Given: Find: F F N P

Practice Problem If you had $2,000 now and invested it at 10%, how much would it be worth in 8 years? F = ? i = 10% 8 $2,000

Solution

Single Cash Flow Formula Single payment present worth factor (discount factor) Given: Find: F N P

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 $10,000 6 P

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? 1 2 3 4 $25,000 $3,000 $5,000 P

Uneven Payment Series + + P P2 P4 P1 $25,000 $3,000 $5,000 1 2 3 4 1 2 3 4 P $25,000 $3,000 $5,000 + + 1 2 3 4 1 2 3 4 1 2 3 4 P2 P4 P1

Check 1 2 3 4 Beginning Balance 28,622 6,484.20 4,132.62 4,545.88 1 2 3 4 Beginning Balance 28,622 6,484.20 4,132.62 4,545.88 Interest Earned (10%) 2,862 648.42 413.26 454.59 Payment +28,622 -25,000 -3,000 -5,000 Ending Balance $28,622 0.47 Rounding error It should be “0.”

Equal Payment Series: Find equivalent P or F 0 1 2 3 4 5 N-1 N F P

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

Compound Amount Factor A(1+i)N-2 A A A A(1+i)N-1 N 1 2 1 2 N

Given: A = $5,000, N = 5 years, and i = 6% Find: F Equal Payment Series Compound Amount Factor (Future Value of an annuity) F 0 1 2 3 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) = $28,185.46

Validation

Given: F = $5,000, N = 5 years, and i = 7% Find: A Finding an Annuity Value The annuity factor indicates a series of payments of a fixed, or constant, amount for a specified number of periods. F 0 1 2 3 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

Example 2.10 Handling Time Shifts in a Uniform Series First deposit occurs at n = 0 i = 6% 0 1 2 3 4 5 $5,000 $5,000 $5,000 $5,000 $5,000

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. 90

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 0 1 2 3 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 91

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

93

Understanding Money Management Nominal and Effective Interest Rates Equivalence Calculations Changing Interest Rates Debt Management 94

Focus 1. If payments occur more frequently than annual, how do we calculate economic equivalence? If interest period is other than annual, how do we calculate economic equivalence? How are commercial loans structured? How should you manage your debt? 95

Nominal Versus Effective Interest Rates Nominal Interest Rate: Interest rate quoted based on an annual period Effective Interest Rate: Actual interest earned or paid in a year or some other time period 96

18% Compounded Monthly Nominal interest rate Interest period Annual percentage rate (APR) 97

Effective Annual Interest Rate r = nominal interest rate per year ia = effective annual interest rate M = number of interest periods per year 98

18% compounded monthly Question: Suppose that you invest $1 for 1 year at 18% compounded monthly. How much interest would you earn? Solution: = $1.1956 0.1956 or 19.56% : 1.5% 18% 99

= : 1.5% 18% compounded monthly or 1.5% per month for 12 months 19.56 % compounded annually 100

Compounding Semi-annually Compounding Quarterly Nominal and Effective Interest Rates with Different Compounding Periods Effective Rates Nominal Rate Compounding Annually Compounding Semi-annually Compounding Quarterly Compounding Monthly Compounding Daily 4% 4.00% 4.04% 4.06% 4.07% 4.08% 5 5.00 5.06 5.09 5.12 5.13 6 6.00 6.09 6.14 6.17 6.18 7 7.00 7.12 7.19 7.23 7.25 8 8.00 8.16 8.24 8.30 8.33 9 9.00 9.20 9.31 9.38 9.42 10 10.00 10.25 10.38 10.47 10.52 11 11.00 11.30 11.46 11.57 11.62 12 12.00 12.36 12.55 12.68 12.74 101

Effective Annual Interest Rates (9% compounded quarterly) First quarter Base amount + Interest (2.25%) $10,000 + $225 Second quarter = New base amount = $10,225 +$230.06 Third quarter = $10,455.06 +$235.24 Fourth quarter + Interest (2.25 %) = Value after one year = $10,690.30 + $240.53 = $10,930.83 102

Effective Interest Rate per Payment Period (i) C = number of interest periods per payment period K = number of payment periods per year r/K = nominal interest rate per payment period 103

12% compounded monthly Payment Period = Quarter Compounding Period = Month One-year 1st Qtr 2nd Qtr 3rd Qtr 4th Qtr 1% 1% 1% 3.030 % Effective interest rate per quarter Effective annual interest rate 104

Effective Interest Rate per Payment Period with Continuous Compounding where CK = number of compounding periods per year continuous compounding => 105

Interest Period = Quarterly Case 0: 8% compounded quarterly Payment Period = Quarter Interest Period = Quarterly 1st Q 2nd Q 3rd Q 4th Q 1 interest period Given r = 8%, K = 4 payments per year C = 1 interest periods per quarter M = 4 interest periods per year 106

Interest Period = Monthly Case 1: 8% compounded monthly Payment Period = Quarter Interest Period = Monthly 1st Q 2nd Q 3rd Q 4th Q 3 interest periods Given r = 8%, K = 4 payments per year C = 3 interest periods per quarter M = 12 interest periods per year 107

Interest Period = Weekly Case 2: 8% compounded weekly Payment Period = Quarter Interest Period = Weekly 1st Q 2nd Q 3rd Q 4th Q 13 interest periods Given r = 8%, K = 4 payments per year C = 13 interest periods per quarter M = 52 interest periods per year 108

Interest Period = Continuously Case 3: 8% compounded continuously Payment Period = Quarter Interest Period = Continuously 1st Q 2nd Q 3rd Q 4th Q  interest periods Given r = 8%, K = 4 payments per year 109

Summary: Effective interest rate per quarter Case 0 Case 1 Case 2 Case 3 8% compounded quarterly 8% compounded monthly 8% compounded weekly 8% compounded continuously Payments occur quarterly 2.000% per quarter 2.013% per quarter 2.0186% per quarter 2.0201% per quarter 110

Equivalence Analysis using Effective Interest Rate Step 1: Identify the payment period (e.g., annual, quarter, month, week, etc) Step 2: Identify the interest period (e.g., annually, quarterly, monthly, etc) Step 3: Find the effective interest rate that covers the payment period. 111

Principle: Find the effective interest rate that covers the payment period Case 1: compounding period = payment period (Example 5.5) Case 2: compounding period < payment period (Examples 5.7 and 5.8) Case 3: compounding period > payment period (Example 5.9) 112

Case I: When Payment Periods and Compounding periods coincide Step 1: Identify the number of compounding periods (M) per year Step 2: Compute the effective interest rate per payment period (i) i = r/M Step 3: Determine the total number of payment periods (N) N = M (number of years) Step 4: Use the appropriate interest formula using i and N above 113

Example 5.5: Calculating Auto Loan Payments Given: Invoice Price = $21,599 Sales tax at 4% = $21,599 (0.04) = $863.96 Dealer’s freight = $21,599 (0.01) = $215.99 Total purchase price = $22,678.95 Down payment = $2,678.95 Dealer’s interest rate = 8.5% APR Length of financing = 48 months Find: the monthly payment 114

Example 5.5: Payment Period = Interest Period $20,000 48 1 2 3 4 A Given: P = $20,000, r = 8.5% per year K = 12 payments per year N = 48 payment periods Find A Step 1: M = 12 Step 2: i = r/M = 8.5%/12 = 0.7083% per month Step 3: N = (12)(4) = 48 months =PMT(0.085/12,48,20000)(put into excel sheet) Or Step 4: A = $20,000(A/P, 0.7083%,48) = $492.97 115

Dollars Up in Smoke What three levels of smokers who bought cigarettes every day for 50 years at $1.75 a pack would have if they had instead banked that money each week: Level of smoker Would have had 1 pack a day 2 packs a day 3 packs a day $169,325 $339,650 $507,976 Note: Assumes constant price per pack, the money banked weekly and an annual interest rate of 5.5% Source: USA Today, Feb. 20, 1997 116

Sample Calculation: One Pack per Day Step 1: Determine the effective interest rate per payment period. Payment period = weekly “5.5% interest compounded weekly” i = 5.5%/52 = 0.10577% per week Step 2: Compute the equivalence value. Weekly deposit amount A = $1.75 x 7 = $12.25 per week Total number of deposit periods N = (52 weeks/yr.)(50 years) = 2600 weeks F = $12.25 (F/A, 0.10577%, 2600)= $169,325 117

Case II: When Payment Periods Differ from Compounding Periods Step 1: Identify the following parameters M = No. of compounding periods K = No. of payment C = No. of interest periods per payment period Step 2: Compute the effective interest rate per payment period For discrete compounding For continuous compounding Step 3: Find the total no. of payment periods N = K (no. of years) Step 4: Use i and N in the appropriate equivalence formula 118

Discrete Case: Quarterly deposits with Monthly compounding F = ? Year 1 Year 2 Year 3 0 1 2 3 4 5 6 7 8 9 10 11 12 Quarters A = $1,000 Step 1: M = 12 compounding periods/year K = 4 payment periods/year C = 3 interest periods per quarter Step 2: Step 3: N = 4(3) = 12 Step 4: F = $1,000 (F/A, 3.030%, 12) = $14,216.24 119

Continuous Case: Quarterly deposits with Continuous compounding F = ? Year 1 Year 2 Year 3 0 1 2 3 4 5 6 7 8 9 10 11 12 Quarters A = $1,000 Step 1: K = 4 payment periods/year C =  interest periods per quarter Step 2: Step 3: N = 4(3) = 12 Step 4: F = $1,000 (F/A, 3.045%, 12) = $14,228.37 120

Credit Card Debt Annual fees Annual percentage rate Grace period Minimum payment Finance charge 121

Methods of Calculating Interests on your Credit Card Description Interest You Owe Adjusted Balance The bank subtracts the amount of your payment from the beginning balance and charges you interest on the remainder. This method costs you the least. Your beginning balance is $3,000. With the $1,000 payment, your new balance will be $2,000. You pay 1.5% on this new balance, which will be $30. Average Daily Balance The bank charges you interest on the average of the amount you owe each day during the period. So the larger the payment you make, the lower the interest you pay. Your beginning balance is $3,000. With your $1,000 payment at the 15th day, your balance will be reduced to $2,000. Therefore, your average balance will be (1.5%)($3,000+$2,000)/2=$37.50. Previous Balance The bank does not subtract any payments you make from your previous balance. You pay interest on the total amount you owe at the beginning of the period. This method costs you the most. Regardless of your payment size, the bank will charge 1.5% on your beginning balance $3,000: (1.5%)($3,000)=$45. 122

Commercial Loans Amortized Loans Effective interest rate specified Paid off in installments over time Examples: Auto-loans, home mortgage loans, most business loans Add-on Loans Simple interest rate specified to pre-calculate the total interest Examples: financing furniture and appliances 123

Amortized Loan - Auto Loan $20,000 48 1 2 25 24 Given: APR = 8.5%, N = 48 months, and P = $20,000 Find: A A = $20,000(A/P,8.5%/12,48) =$492.97 124

Suppose you want to pay off the remaining loan in lump sum right after making the 25th payment. How much would this lump be? $20,000 48 1 2 25 24 $492.97 $492.97 25 payments that were already made 23 payments that are still outstanding P = $492.97 (P/A, 0.7083%, 23) = $10,428.96 125

Add-on Loans Given: You borrow $5,000 for 2 years at an add-on rate of 12% with equal payments due at the end of each month. Add-on Interest: (0.12)($5,000)(2) = $1,200 Principal + add-on interest $5,000 + $1,200 = $6,200 Monthly Installments A = $6,200/24 = $258.33 Find: the effective interest rate for this add-on loan 126

By trial and error method, we find i = 1.7975% per month $5,000 i = ? 1 2 24 A = $258.33 $5,000 = $258.33 (P/A, i, 24) (P/A, i, 24) = 19.3551 By trial and error method, we find i = 1.7975% per month r = 1.7975% x 12 = 21.57% per year 127

Buying vs. Lease Cost to Buy : $25,886 Cost to Lease : $15,771 Down payment: $2,100 Car Loan at 8.5% (48 payments of $466): $22,368 Sales tax (at 6.75%): $1,418 Cost to Lease : $15,771 Lease (48 payments of $299) : $14,352 Sales tax (at 6.75%): $969 Document fee: $450 Refundable security deposit (not included in total) : $300 128

Buying versus Lease Decision Option 1 Debt Financing Option 2 Lease Financing Price $14,695 Down payment $2,000 APR (%) 3.6% Monthly payment $372.55 $236.45 Length 36 months Fees $495 Cash due at lease end $300 Purchase option at lease end $8.673.10 Cash due at signing $731.45 129

130

Which Option is Better? Debt Financing: Pdebt = $2,000 + $372.55(P/A, 0.5%, 36) - $8,673.10(P/F, 0.5%, 36) = $6,998.47 Lease Financing: Please = $495 + $236.45 + $236.45(P/A, 0.5%, 35) + $300(P/F, 0.5%, 36) = $8,556.90 131

Summary Financial institutions often quote interest rate based on an APR. In all financial analysis, we need to convert the APR into an appropriate effective interest rate based on a payment period. When payment period and interest period differ, calculate an effective interest rate that covers the payment period. 132