Economic Decision Making Ulrich and Eppinger Chapter 15 Deiter & Schmidt Chapter 18 .

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Economic Decision Making

Ulrich and Eppinger Chapter 15

Deiter & Schmidt Chapter 18http://highered.mcgraw-hill.com/sites/dl/free/0072837039/595507/Chapter18Corr06_09.pdf

Adapted from Dr. Stamper

PlanningPlanning

Product Development Process

ConceptDevelopment

ConceptDevelopment

System-LevelDesign

System-LevelDesign

DetailDesign

DetailDesign

Testing andRefinement

Testing andRefinement

ProductionRamp-Up

ProductionRamp-Up

Concept Development Process

Perform Economic Analysis

Benchmark Competitive Products

Build and Test Models and Prototypes

IdentifyCustomerNeeds

EstablishTargetSpecifications

GenerateProductConcepts

SelectProductConcept(s)

Set FinalSpecifications

PlanDownstreamDevelopment

MissionStatement Test

ProductConcept(s)

DevelopmentPlan

Overview• Monday: (Dieter, Chap 18 and Ulrich, Chap 15 Appendix)

• Time Value of Money, Cash Flow Diagrams, Net Present Value, Depreciation

• Thursday• Economic Analysis Process for Product Development (Ulrich Chap

15)• Profitability

• Monday• More analysis

• Wednesday:• Lab exercises

Objectives

• Learn some of the language of the business community

• Provide techniques to evaluate the financial attractiveness of various alternatives that are presented to engineers

• Apply the economic evaluation techniques to personal and professional decisions

Time Value of Money

Proposition:• The value of money changes over time: generally

$1 in the future is worth less than $1 now

Evidence:• Organizations are willing to borrow money in the

present and then return more than what they borrowed at some point in the future (renting money).

Example 1: Simple Interest Future Value

• Assume:– Invest $100 now (P=$100)– At 8% annual interest rate (i=8%=0.08)– A single 1 year period (n=1)

• Find: Future Value (F)– F = (1+i)P = (1+0.08)100= $108

Example 2: Simple Interest Present Value

• Assume:– Desire a future payout of $100 (F=$100)– At 8% annual interest/discount rate (i=8%=0.08)– After a single 1 year period (n=1)

• Find: Present value to give F=$100– Same equation: F = (1+i)P, but solve for P– P=F/(1+i) = $100/(1+0.08)= $92.59

Example 3: Compound Interest Future Value

• Assume:– Invest $100 now (P=$100)– At 8% annual interest rate (i=8%=0.08)– For a 3 year period (n=3)

• Find: Future Value (F)– Fafter 1 year = (1+i)P = (1+0.08)100= $108

– Fafter 2 years = (1+i)(1+i)P = (1+0.08)(1+0.08)100= $116.64

– Fafter 3 years = (1+i)(1+i)(1+i)P = $125.97

Example 4: Compound Interest Present Value

• Assume:– Desire a future payout of $100 (F=$100)– At 8% annual interest rate (i=8%=0.08)– After a 3 year period (n=3)

• Find: Present value to give F=$100– Same equation: F = (1+i)(1+i)(1+i)P, but solve for P– P=$100/[(1+0.08)(1+0.08)(1+0.08)]= $79.38

General Equations for Compound Interest• Future Value:

• Present Value:

• Where:– F is future value– P is present value– i is interest rate (or discount rate)– n is number of periods

How Do We Compare Alternatives?(Economic Decision Making)

• We need some form of “equivalence”• Present Value and Future Value can provide

that equivalence

Cash Flow Diagrams & Net Present Value

Page 867 Dieter and Schmidthttp://highered.mcgraw-hill.com/sites/dl/free/0072837039/595507/Chapter18Corr06_09.pdf

Note the cash flow diagram. • Incomes point into the line• Expenses point away from the

line• Time starts in year 0 (start of year

1)• All other flows are at the end of

the year

Net Present Value of the Costs of Machine APresent Value of Year 0 Costs:

– $25,000

Present Value of Year 1 Costs:– (2000-500)/(1+0.10)^1= $1363.63

Present Value of Year 2 Costs:– (2000-500)/(1+0.10)^2= $1239.67

Present Value of Year 3 Costs:– (2000-500)/(1+0.10)^3= $1126.97

Present Value of Year 4 Costs:– (2000-500)/(1+0.10)^4= $1024.52

Present Value of Year 5 Costs:– (2000-500-3000)/(1+0.10)^5= -$931.38

Net Present Value of the Costs:

25,000+1363.63+1239.67+1126.97+1024.52 -931.38$ 28,823

Does it make sense that the PV of year 0 is the same as year 0?

Does it make sense that the PV of each year is decreasing with time?

Why is the PV of Year 5 negative?

Interest

Number of periods Payments Made Each Period

Future Value

Using Excel for Year 3:

Present Value of Year 3 Costs:(2000-500)/(1+0.10)^3= $1126.97

Why is the value red ?

Interest

Number of periods

Payments (Costs) for Each Period

Additional Future Value

Using Excel to find the presentValue for the 5 years of $1500 costs each year:

Present Value of the 5 years:(2000-500)/(1+0.10)^1= $1363.63(2000-500)/(1+0.10)^2= $1239.67(2000-500)/(1+0.10)^3= $1126.97(2000-500)/(1+0.10)^4= $1024.52(2000-500)/(1+0.10)^5= $ 931.38

$ 5686

0 if Payments (Costs)made at end of period

Alternatively we can use the NPV (Net Present Value) function in Excel to capture values of each year for this cash flow diagram.

Why do we have to account for year 0 separately?

Net Present Value of the Costs of Machine BPresent Value of Year 0 Costs:

• $15,000

Present Value of Year 1 Costs:• (4000)/(1+0.10)^1= $3636.36

Present Value of Year 2 Costs:• (4000)/(1+0.10)^2= $3305.79

Present Value of Year 3 Costs:• (4000+3500)/(1+0.10)^3= $5634.86

Present Value of Year 4 Costs:• (4000)/(1+0.10)^4= $2732.05

Present Value of Year 3 Costs:• (4000)/(1+0.10)^5= $2483.69

Net Present Value of the Costs:

15,000+3636.36+3305.79+5634.86+2732.05 +2483.69$ 32,793

Net Present Value Comparison

• NPV Costmachine A = $28,823

• NPV Costmachine B = $32,793

• Costmachine A unadjusted = $29,500

• Costmachine B unadjusted = $38,500

In-Class Exercise: 1For Example 18.3 of Dieter and Schmidt we showed in how the

Present Value (PV) and Net Present Value (NPV) functions in Excel could be used to calculate the Present Value of the costs of Machine A. Create an Excel spreadsheet that shows the annual costs and calculates the Present Value of the costs of Machine B in example 18.3.

Do two separate calculations, the first which uses the PV function, and the second which uses the NPV function.

Raise your hand when you have finished so that you can check your answer with your instructor.

Economic Metrics to Evaluate Projects

• Return on Investment (ROI)• Payback period

Return on Investment (ROI)

• Often given as a ratio of some desired economic outcome to the investment for that outcome.

• Typical numerators:– Annual profit before taxes– Annual profit after taxes– Annual cash flow before taxes– Annual cash flow after taxes

• Typical denominator: capital investment

ROI example:

• ROI = benefit/ cost = (gains-cost)/cost• Buying 100 shares of Arcelor Mittal stock at $18 per

share would cost $1800.• If you later sold those shares for $2000, your gains

minus cost would be $200.• The resulting ROI (ratio of benefit to investment) is

$200/$1800 or 11.1%• Note that time value of money is not considered.• What is your ROI for attending Rose-Hulman?• How would you use that information?

Payback Period

• Typical definition: Ratio of the investment to the annual benefit… giving an estimate of the time to recover the investment

• If benefits are not uniform over time… it is the time at which the cumulative sum of the benefits equal the investment

• Typically does not take into account the time-value of money

Payback Period Example

• Suppose you buy a Mini-Donut maker for $8000 and set it up for your neighborhood’s biannual garage sale. After expenses for dough and grease, you make $500 per year.

• What is the Payback Period?• Looks like 16 years before you have recouped

the initial cost. Once again, we have ignored the time value of money.

What is the Payback Period and 10 year ROI for your Rose Education?

• Payback: Assume $50,000 annual cost for Tuition, Room and Board, etc. and opportunity cost of $16,000 for the lost job at McDonalds.

• Assume annual salary after graduation of $60,000. (Note that the delta due to Rose is $60,000-$16,000 or $44,000)

• Evaluate ROI as a percentage.

Rose Payback

• Total cost over 4 years is $66,000*4=$264K• Total annual benefit is $44,000• It will take 6 years to pay back the cost of

education at Rose.• How is this information helpful for decision

making?

10 Year Rose ROI

• Total cost is $264K• Total 10 year benefit is $44,000*6=$264K• ROI is $264/$264=1• You could view this as a 100% ROI

Homework Problem #7

• Honda Civic– Hybrid vs. Conventional

Homework #5

• Publishers Clearinghouse v. Megamillions• Sketch cash flow diagram for PC• Determine PV

Depreciation and Taxes

• Since the capital used to produce goods, services, and energy declines in value over time, tax law currently allows the owners of capital equipment to reduce their taxes each year to reflect that declining value.

Types of Expenditures

• Capital– Funds used to purchase facilities and equipment

that are useful for more than 1 year– These purchases are “capitalized”

• Expense– Funds used to purchase consumables (e.g. labor,

material, utilities)– These purchases are “expensed”

The categorization of expenditures has important tax implications

Depreciation of Capital Assets• Accounting systems assume that capital

equipment (not land) loses value over time• The loss of value of capital equipment is called

depreciation• Depreciation is important in the economic

analysis of engineering projects because depreciation can be used to reduce the taxes that are paid on corporate income

Taxes and Depreciation

• The amount of tax a company pays is calculated by multiplying the corporate tax rate (approximately 35% for many companies) by the company’s taxable income

• Where:– income = revenues – costs– taxable income = revenues – costs - depreciation

Example Cash Flow with Tax and Depreciation

From Dieter and Schmidt

Calculating Depreciation

• Step 1: determine the period over which the capital asset should be depreciated.

• Step 2: determine how the depreciating value should be distributed over the selected period

Determining the Period of Depreciation

• See your business office for accounting rules• Examples:

– Computers, trucks: 5 years– Office furniture, railroad track, Ag buildings: 7 years– Durable goods manufacturing equipment: 10 years– Sewage treatment plant: 15 years

What do you expect the time frame to be for a wind turbine?

Determining the Distribution• Straight line depreciation• Declining balance depreciation• Sum–of–years-digits depreciation

Straight-Line Depreciation

Initial Cost

Salvage Value Periods

Declining Balance Depreciation

Initial Cost Salvage Value

Total Number of Periods

Period for which depreciationIs being calculated

Depreciation in the jth year

Sum-of-Years-Digits Depreciation

Initial Cost

Salvage ValueTotal Number of Periods

Period for which depreciationIs being calculated

Repaying a Loan

• Generally you will make a down payment and annual payments.

• The down payment occurs in year 0.• The amount of the loan is the cost of the

purchase minus the down payment• The payment of the loan is easily found using

Excel

Using the PMT Function to find Payments on a Loan

Principal

Number of Periods30 years*12 months

Monthly Interest rateAnnual rate/12

Machine ComparisonYou are concerned with the purchase of a heat-treating furnace for gas carburizing

of steel parts. Furnace A will cost $325,000 and will last 10 years; furnace B will cost $400,000

and will also last 10 years. However, furnace B will provide closer control on case depth, which means that the

heat treater can shoot for the low side of the specification range on case depth. This will mean that the production rate for furnace B will be 2740 lb/hr compared

with 2300 lb/hr for furnace A. Total yearly production is required to be 15,400,000 lb. The cycle time for furnace

A is 16.5 hr and that for furnace B is 13.8 hr. The hourly operating cost is $64.50 per hr.

Assume that money is worth 10% and the tax rate is 50%. Also use straight line depreciation.

How might you compare the two alternatives?

Production RateYearly Required Operating Yearly Depreciation

Production (lb) hours Cost ($/hr) Oper Cost ($) $Furnace A 2300lb/hr 15400000 6696 64.5 431870 32500 Furnace B 2740lb/hr 5620 362518 40000

B-A -69351 7500 Interest Rate 0.1

Year 0 1 2 3 4 5 6 7 8 9 10

Initial

Cost

Furnace A 325,000

Furnace B 400,000

Net Difference

B-A -75,000 73101 73101 73101 73101 73101 73101 73101 73101 73101 73101

PV $75,000 (66,456) (60,414) (54,922) (49,929) (45,390) (41,264) (37,513) (34,102) (31,002) (28,184)

Sum ($374,176)

B cost $75,000 more than A

B saves $69,351 in operating costs

B saves $3,750 in taxes

Let’s compare with NPV

First organize the info

Next, draw a Cash Flow Diagram

Check the NPV

Chapter 15: Product Development Economics

Product Design and DevelopmentFourth Edition

by Karl T. Ulrich and Steven D. Eppinger

Economic Analysis for Product Development

(Ulrich and Eppinger)

1. Build a base-case financial model2. Perform a sensitivity analysis3. Use sensitivity analysis to understand project

trade-offs4. Consider the influence of qualitative factors

on project success

Step 1: Build a Base-Case Model

Step 1: Build a Base-Case Model

Annual interest divided by number of periods per year

Number of periods

Payments Made Each Period

Future Value

Using Excel for Q4 of Year 1:

Present Value of Year 3 Costs:(-2250)/(1+0.10/4)^3= -$2089

Homework Problem #2a

2. a. Use Excel to find the NPV for a drug eluting Cardiac Stent project: • Years 1-4 development: $70M/year • Years 4-8 FDA testing, IP costs, manufacturing ramp up: $ 110 • M/year • Year 10 until expiration of patent

– Volume: 600,000 units / year – Revenue: $2500 / unit – Costs: $1200 / unit

• Patent issues at start of year 8 and is enforceable for 17 years • Cost of money is 5%

Step 2: Perform Sensitivity Analysis(e.g. 20% decrease in development costs)

Step 2: Perform Sensitivity Analysis(e.g. 25% increase in development time)

Step 2: Perform Sensitivity Analysis

Ulrich & Eppinger, “Product Design and Development”

Step 3: Use Sensitivity Analysis to Understand Project Trade-offs

Step 3: Use Sensitivity Analysis to Understand Project Trade-offs

(estimate Trade-off Rules from sensitivity analyses)

Ulrich & Eppinger, “Product Design and Development”

Homework #2b. a. Use Excel to find the NPV for a drug eluting Cardiac Stent project:

• Years 1-4 development: $70M/year • Years 4-8 FDA testing, IP costs, manufacturing ramp up: $ 110 • M/year • Year 10 until expiration of patent

– Volume: 600,000 units / year – Revenue: $2500 / unit – Costs: $1200 / unit

• Patent issues at start of year 8 and is enforceable for 17 years • Cost of money is 5%

b. Find the NPV if the FDA testing takes twice as long as planned (still at $110M/year)

A Question:

What are some situations when you might not pursue an option that presents the best NPV?

Step 4: Consider the Influence of Qualitative Factors

Ulrich & Eppinger, “Product Design and Development”

• Interactions between the Project and the Firm (e.g. strategic fit, risk/liability exposure)

• Interactions between the Project and the Market (e.g. competitors, customers, suppliers)

• Interactions between the Project and the Macro Environment (e.g. economic shifts, government regulations, social trends)

Modeling Uncertain Cash Flows

Dealing With Risk

Probability that the Patent is allowed

NPV= Pa*PVa + Pb*PVb = 0.6($6.5 million) + 0.4($1.5 million) = $4.5 million

Determining NPV with probabilities.

NPV with market testing is $2,650,000

HW Problem 2c

2. a. Use Excel to find the NPV for a drug eluting Cardiac Stent project: • Years 1-4 development: $70M/year • Years 4-8 FDA testing, IP costs, manufacturing ramp up: $ 110 • M/year • Year 10 until expiration of patent

– Volume: 600,000 units / year – Revenue: $2500 / unit – Costs: $1200 / unit

• Patent issues at start of year 8 and is enforceable for 17 years • Cost of money is 5%

b. Find the NPV if the FDA testing takes twice as long as planned (still at $110M/year) c. For the original case, determine the NPV if there is a 5% probability that there is no FDA approval, a 10%

probability of 1.5B intellectual property settlement in year 14, and a 85% probability of business as predicted.

Economics Laboratory

Apply the tools of economic decision making to a large capital project and

a personal project