1. Economic Estimation Models Engineering Economics Lecture 19 19 January 2010

2. Estimation To calculate approximately A tentative evaluation or rough calculation A judgment based on one's impressions; an opinion

3. Difference between estimation and forecasting Estimation require calculation, forecasting may or may not require calculation Estimation require calculation, forecasting may or may not require calculation Forecasting always deals with future, estimation can be for future or past Forecasting always deals with future, estimation can be for future or past Forecasting require extensive data, estimation require accurate assumptions and valid approximations Forecasting require extensive data, estimation require accurate assumptions and valid approximations Estimation is approximation, forecasting is projection into future Estimation is approximation, forecasting is projection into future

4. Why estimation fail You are not a good estimator You are biased You are ignorant You are fixated and rigid You are to calculative You are sick You are over pushy - don’t try to hit the bull

5. Why estimation succeed Calm and balance Unbiased Know the facts and figures High morale Healthy Don’t aim high

6. Engineering Economic Analysis Seven Steps 1. Recognition and formulation of the problem. 2. Development of the feasible alternatives. 3. Selection of a criteria for determining best alternative. 4. Analysis and comparison of the alternatives. 5. Selection of the preferred alternative. 6. Implementation of the preferred alternative. 7. Performance monitoring and post-evaluation.

7. Developing Estimation – why necessary Because engineering economy studies deal with outcomes that extend into the future, estimating for feasible alternatives is a critical step in the analysis procedure.

8. Estimating Techniques Indexes Unit Technique Factor Technique Power-Sizing Technique Learning Curve

9. Indexes An index is a dimensionless number that indicates how a cost or a price has changed with time with respect to the base year. C n = cost or selling price of an item in year n C k = cost or price of the item at an earlier point in time (say year k ) I n = index value in year n I k = index value in year k C n = C k (I n /I k )

10. Example 1 Company XYZ installed an exchange in 1989 for $350,000 when the index had value of 312. This same company must install another exchange of the same size in 1996. The index in 1996 is 468. what will be the cost in 1996? C n = cost of an item in year n = ? C k = cost of item at an earlier point in time (year k ) = $ 350,000 I n = index value in year n = 468 I k = index value in year k = 312 C n = C k (I n /I k ) Approximate cost of new boiler = C 1996 = $350,000 (468/312) = $525,000

11. Unit Technique Involves a “per unit factor” that can be estimated effectively. Examples: – Operating cost per day – Construction cost per square foot – Maintenance cost per hour

12. Example 2 We need a preliminary estimate of the cost of a particular IT laboratory. Use the factor of, say, $550 per IT equipment kit and assume that the laboratory require approximately 2,000 kits. What will be the estimated cost? Estimated cost = $550 x 2,000 = $1100,000

13. Factor Technique The factor technique is an extension of the unit technique We need a refined estimate of the cost of the house. Assume that the house is approximately 2,000 square feet of living space, has one porch and two garages. Use the factor of, say, $50 per square foot of living space, $5,000 per porch and $8,000 per garage. Estimated cost of the house = $50 x 2,000 + $5,000 + ($8,000 x 2)= $121,000

14. Power-Sizing Technique Also sometimes referred to as the exponential model Often used to cost industrial plants and equipment C A = cost for plant A C B = cost for plant B S A = size of plant A S B = size of plant B X = cost-capacity factor C A /C B = (S A /S B ) X C A = C B (S A /S B ) X

15. Example 4 Make a preliminary estimate of the cost of a 600 MW power plant. It is known that a 200 MW plant cost $100 million 20 years ago when the appropriate cost index was 400. That cost index is now 1,200. The power-sizing factor is 0.79. Today’s estimated cost of a 200-MW plant = –$100 million x (1,200/400) = $300 million Today’s estimated cost of a 600-MW plant = –$300 million x (600/200) 0.79 = $714 million

16. Mathematics of the Learning Curve u = the output of desired number Z u = the number of input resources for output number u K = the number of input resources for the first output unit s = the learning curve slope parameter Z u = K u n where n = log s / log 2

17. Example 5 A student team is designing a telecom systems for national competition. The time required for the team to assemble the first system is 100 hours. Their learning rate is 0.8. The time it will take to assemble the 10th car = ? Z u = K u n where n = log s / log 2 Z 10 = 100 (10) log 0.8/log2 = 100 (10) -0.322 = 100 / 2.099 = 47.6 hours

18. Example 6 A rare product is made in batches of 50 units. Within a batch, each unit take less and less time to be produced because of a learning process of 75%. The time needed to assemble the first unit = 2.3123 hrs The time needed to assemble additional units is Z u = 2.3123 ( u ) log 0.75/log2 = 2.3123 ( u ) -0.415 The total time taken for all 50 units = Z 1 + Z 2 + Z 3 +...+ Z 50 = 36.48 hours

19. Compensatory Quiz