Capstone Design. Cost Estimating and Estimating Models

Transcription

1 Capstone Design Engineering Economics II Engineering Economics II (1 of 14) Cost Estimating and Estimating Models Engineering economic analysis involves present and future economic factors It is critical to obtain reliable estimates of future costs, benefits and other economic parameters Estimates can be rough estimates, semi-detailed estimates, or detailed estimates, depending on the needs A characteristic ti of cost estimates t is that t errors in estimating are typically non-symmetric Costs are more likely to be underestimated than overestimated Engineering Economics II (2 of 14) 1

2 Cost Estimating and Estimating Models (continued) Difficulties in developing cost estimates arise from such conditions as: One-of-a-kind estimates Resource availability Estimator expertise Generally the quality of a cost estimate increases as the resources allocated to developing the estimate increase Benefits expected from improving a cost estimate should outweigh the cost of devoting additional resources to the estimate improvement Engineering Economics II (3 of 14) Models for Cost Estimates Several models are available for developing cost (or benefit) estimates The per unit model is a simple but useful model in which a cost estimate is made for a single unit Total cost estimate results from multiplying the estimated cost per unit times the number of units The segmenting model partitions the total estimation task into segments Each segment is estimated, then the segment estimates are combined for the total cost estimate Engineering Economics II (4 of 14) 2

3 Cost Indexes Cost indexes can be used to account for historical changes in costs The widely-reported Consumer Price Index (CPI) is an example Monthly data on changes in the prices paid by urban consumers for a representative basket of goods and services Cost index data are available from a variety of sources Suppose A is a time point in the past and B is the current time Let IV A denote the index value at time A and IV B denote the current index value for the cost estimate of interest To estimate the current cost based on the cost at time A, use the equation: Cost at time B = (Cost at time A) (IV B / IV A ) Engineering Economics II (5 of 14) Power Sizing Model The power sizing model accounts explicitly for economies of scale For example, the cost of constructing a six-story building will typically be less than double the construction cost of a comparable three-story building To estimate the cost of B based on the cost of comparable item A, use: Cost of B = (Cost of A) [ ("Size" of B) / ("Size" of A) ] x where x is the appropriate power sizing exponent An economy of scale is indicated by an exponent < 1.0 An exponent of 1.0 indicates no economy of scale, and an exponent greater than 1.0 indicates a diseconomy of scale "Size" is used here in a general sense to indicate physical size, capacity, or some other appropriate comparison unit Engineering Economics II (6 of 14) 3

4 Learning Curve Cost Estimate Learning curve cost estimating is based on the assumption that as a particular task is repeated, the operator systematically becomes quicker at performing the task Based on the assumption that the time required to complete the task for production unit 2x is a fixed percentage of the time that was required for production unit x, for all positive, integer x The learning curve slope indicates "how fast" learning occurs For example, a learning curve rate of 70% represents much faster learning than a rate of 90% If an operator exhibits learning on a certain task at a rate of 70%, the time required to complete production unit 50, for example, is only 70% of the time required to complete unit 25 Engineering Economics II (7 of 14) Learning Curve Time Let b = learning curve exponent = log (learning curve rate in decimal form) / log 2.0 Then T N = time estimate for unit N (N = 1, 2,...) = (T 1 ) (N) b where T 1 is the time required for unit 1 Engineering Economics II (8 of 14) 4

5 Simple Interest Simple interest is computed only on an original sum Total interest earned (owed) = P i n P = principal sum of money i = interest rate n = period (usually given in years) At the end of the loan period, the total amount due is F = P + P i n = P( 1+ i n) Engineering Economics II (9 of 14) Compound Interest Practical interest is calculated using a compound interest method For a loan, any interest owed but not paid at the end of a specified time period (often one year) is added to the balance due Next year s interest is calculated based on unpaid balance due Compound interest can be thought of a interest on top of interest Engineering Economics II (10 of 14) 5

6 Single payment compound interest formulas Given a present dollar amount P, interest rate i% per year, compounded annually, and a future amount F that occurs n years after the present, then the relationship between these terms is: F = P (1 + i) n Example: If $100 is invested at 6% interest per year, compounded annually, then the future value of this investment after four years is: F = P (1 + i) n = $100 ( ) 4 = $100 (1.06) 4 = $100 (1.2625) = $ Solving the above equation for P yields: P = F (1 + i) -n Engineering Economics II (11 of 14) Single payment compound interest formulas (other periods) If the interest period and compounding period are not stated, then the interest rate is understood to be annual with annual compounding Examples: "12% interest" means that the interest rate is 12% per year, compounded annually "12% interest compounded monthly" means that the interest rate is 12% per year (not 12% per month), compounded monthly Thus the interest rate is 1% (12% / 12 ) per month. "1% interest per month compounded monthly" is unambiguous Engineering Economics II (12 of 14) 6

7 Single payment compound interest formulas (other periods, continued)) When the compounding period is not annual, problems must be solved in terms of the compounding period, not years Example: If $100 is invested at 6% interest, compounded monthly, then the future value of this investment after four years is: F = P (1 + i) n = $100 ( ) 48 = $100 (1.005) 48 = $100 (1.2705) = $ Note that the interest rate used above is (6% / 12) = 0.5% per month = per month, and that the number of periods used is 48 (months), not 4 (years) Engineering Economics II (13 of 14) Single payment compound interest (solving for i or n) The single payment compound interest formula F = P (1 + i) n or single payment interest t table factors can be used to solve for unknown i or n. Example: A $100 investment now in an account that pays compound interest annually will be worth $250 at a point exactly 31 years from now. What annual interest rate does this account pay? Solving the equation: 250 = 100 (1 + i) 31 for i yields an answer of 3% Engineering Economics II (14 of 14) 7