Introduction to Engineering Economic

1. Introduction to Engineering Economic Anastasia Lidya M.

2. Short Syllabus • This course studies the basic concept of time value of money and methods for alternative and investment evaluation. The study covers definition and scope of engineering economics, cash flow, concept of time value of money, present equivalent value, annual value, internal rate of return, payback method, profitability index method, sensitivity analysis, depreciation, inflation and deflation, replacement analysis, tax analysis, public investment cost-benefit analysis. The course provides ability in conducting analysis and decision making for alternative selection or investment evaluation using economic criteria.

3. Grading • Requirements: • Min 80% Present in Class • Score/Grade : • Project : 25% • Quiz : 25% • Mid test (UTS) : 25% • Final Test (UAS) : 25% • No additional assignment to increase your score • Project should be submitted on time

4. Outline

5. The Role of Engineering Economic Analysis The problems most suitable for solution by EEA have these qualities: • The problem is sufficiently important • The problem can’t be worked in one’s head ( a careful analysis requires that we organize the problem and all the various consequences) • The problem has economic aspects there are sufficiently important to be a significant component of the analysis leading to a decision

6. Examples of Engineering Economic Analysis • Engineering economic analysis focuses on costs, revenues, and benefits that occur at different times. • For example, when a civil engineer designs a road, a dam, or a building, the construction costs occur in the near future; • the benefits to users begin only when construction is finished, but then the benefits continue for a long time.

7. Engineering economic analysis is used to answer many different questions • Which engineering projects are worthwhile? Has the mining or petroleum engineer shown that the mineral or oil deposit is worth developing? • Which engineering projects should have a higher priority? Has the industrial engineer shown which factory improvement projects should be funded with the available dollars?

8. Engineering economic analysis is used to answer many different questions • How should the engineering project be designed? Has the mechanical or electrical engineer chosen the most economical motor size? Has the civil or mechanical engineer chosen the best thickness for insulation? Has the aeronautical engineer made the best trade-offs between : 1) lighter materials that are expensive to buy but cheaper to fly and 2) heavier materials that are cheap to buy and more expensive to fly?

9. Engineering economic analysis is used to answer many different questions • How to achieve long-term financial goals: How much should you save each month to buy a house, retire, or fund a trip around the world? • How to compare different ways to finance purchases: Is it better to finance your car purchase by using the dealer's low interest rate loan or by taking the rebate and borrowing money from your bank or credit union?

10. Rational Decision-Making Process • Recognize the decision problem • Collect all needed (relevant) information • Identify the set of feasible decision alternatives • Define the key objectives and constraints • Select the best possible and implementable decision alternative

11. Recognize the decision problem Collect all needed (relevant) information Identify the set of feasible decision alternatives Define the key objectives and constraints Select the best possible and implementable decision alternative Need to lease a car Gather technical and financial data Select cars to consider Wanted: small cash outlay, safety, good performance, aesthetics,… Choice between Saturn and Honda (or others) Select a car (i.e., Honda, Saturn or another brand) A Simple Illustrative Example: Car to Lease – Saturn or Honda?

12. Engineering Economic Decisions Needed e.g. in the following (connected) areas: Profit! Then continue at the next stage… Manufacturing Design Financial planning Investment and loan Marketing

13. 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

14. Example 1: Healthcare Service Improvement • 1 Traditional Plan: Patients visit the service providers • 2 New Strategy: Service providers visit the patients Which one of the two plans is more economical? The answer typically depends on the type of patients and the services offered. Examples? 1 2 service providers patients

15. Example 2:Equipment and Process Selection • How do you choose between using alternative materials for an auto body panel? • The choice of material will dictate the manufacturing process and the associated manufacturing costs

16. Example 3:Equipment Replacement Problem • Key question: When is the right time to replace an old machine or equipment?

17. Example 4:New Product and Product Expansion • Shall we build or acquire a new facility to meet the increased (increasing forecasted) demand? • Is it worth spending money to market a new product?

18. Example 5: MACH 3 Project • R&D investment: $750 million(!) • Product promotion through advertising: $300 million(!) • Priced to sell at 35% higher than the preceding Sensor Excel model (i.e., about $1.50 extra per razor) • Question 1: Would consumers pay $1.50 extra for a shave with greater smoothness and less irritation? • Question 2: What happens if the blade consumption drops more than 10% – due to the longer blade life of the new razor?...

19. Example 6: Cost Reduction • Should a company buy new equipment to perform an operation that is now done manually? • Should we spend money now, in order to save more money later? • The answer obviously depends on a number of factors.

20. 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

21. Principle 1An instant dollar is worth more than a distant dollar… Today 6 months later

22. Principle 2Only the cost (resource) difference among alternatives counts The data shown in the green fields are irrelevant items for decision making, since their financial impact is identical in both cases

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

24. Principle 4Additional risk is not taken without a suitable expected additional return A simple illustrative example. Note that all investments imply some risk: portfolio management is a key issue in finance

25. EXAMPLE

26. Solution

27. solution • The shipping department would reduce its cost from $793.50 to $688.50 by using the outside printer. In that case, how much would the printing department's costs decline? We will examine each of the cost components: 1. Direct Labor. If the printing department had been working overtime, then the overtime could be reduced or eliminated. But, assuming no overtime, how much would the saving be? It seems unlikely that a printer could be fired or even put on less than a 40-hour work week. Thus, although there might be a $228 saving, it is much more likely that there will be no reduction in direct labor.

28. Solution 2. Materials and Supplies. There would be a $294 saving in materials and Supplies 3. Allocated Overhead Costs: There will be no reduction in the printing department’s monthly $5000 overhead, for there will be no reduction in department floor space.

29. ENGINEERING DECISION MAKING FOR CURRENT COSTS (1) Example 1: • A concrete aggregate mix is required to contain at least 31% sand by volume for proper batching. One source of material, which has 25% sand and 75% coarse aggregate, sells for $3 per cubic meter (m3). Another source, which has 40% sand and 60% coarse aggregate, sells for $4.40/m3. • Determine the least cost per cubic meter of blended aggregates.

30. ENGINEERING DECISION MAKING FOR CURRENT COSTS (2)

31. ENGINEERING DECISION MAKING FOR CURRENT COSTS (3) Example 2: • A machine part is manufactured at a unit cost of 40 cent for material and 15 cent for direct labor. An investment of $500,000 in tooling is required. The order calls for 3 million pieces. Half-way through the order, anew method of manufacture can be put into effect that will reduce the unit costs to 34 cent for material and 10 cent for direct labor, but it will require $100,000 for additional tooling. If all tooling costs are to be amortized during the production of the order, and other costs are 250% of direct labor cost, would it profitable to make the change?

32. ENGINEERING DECISION MAKING FOR CURRENT COSTS (4)

33. exercise

34. Computing cash flows The manager has decided to purchase a new $30,000 mixing machine. The machine may be paid for in one of two ways: 1. Pay the full price now minus a 3% discount. 2. Pay $5000 now; at the end of one year, pay $8000; at the end of each of the next four years, pay $6000. • List the alternatives in the form of a table of cash flows

35. Computing cash flows

36. Computing cash flows • A man borrowed $1000 from a bank at 8% interest. He agreed to repay the loan in two end-of-year payments. At the end of the first year, he will repay half of the $1000 principal amount plus the interest that is due. At the end of the second year, he will repay the remaining half of the principal amount plus the interest for the second year. • Compute the borrower's cash flow !

37. Computing cash flows

39. TIME VALUE OF MONEY • Which would you prefer, $100 cash today or the assurance of receiving $100 a year from now?

40. TIME VALUE OF MONEY • You might decide you would prefer the $100 now because that is one way to be certain of receiving it. • If the current interest rate is 9% per year, and you put $100 into the bank for one year, how much will you receive back at the end of the year?

41. TIME VALUE OF MONEY • You will receive your original $100 together with $9 interest, for a total of $109. This example demonstrates the time preference for money: we would rather have $100 today than the assured promise of $100 one year hence; • but we might well consider leaving the $100 in a bank if we knew it would be worth $109 one year hence. This is because there is a time value of money in the form of the willingness of banks, businesses, and people to pay interest for the use of various sums.

42. 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.

43. Time Value of Money • Money has a time value because it can earn more money over time (earning power). • 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

44. EQUIVALENCE

45. 0 1 Equivalence • Different sums of money at different times may be equal in economic value $106 one year from now Interest rate = 6% per year $100 now $100 now is said to be equivalent to $106 one year from now, if the $100 is invested at the interest rate of 6% per year.

46. Repayment Plans

47. Cash Flow Diagram

48. 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.

49. Simple Interest • P = Principal amount • i = Interest rate • N = Number of interest periods • Example: • P = $1,000 • i = 8% • N = 3 years

50. Compound Interest • P = Principal amount • i = Interest rate • N = Number of interest periods • Example: • P = $1,000 • i = 8% • N = 3 years