University:
Michigan State University
Course:
EC 301 | Intermediate Microeconomics
Academic year:

2020
Views:

122

Pages:

11
Author:

Arielle Armstrong

Chapter 3: Cost Estimation Techniques Integrated cost estimation approach Components 1. Work breakdown structure (WBS): Successive levels of the work elements and their interrelationships. 2. Cost and revenue structure (classification): Delineation of cost and revenue categories and elements for different WBS levels. 3. Estimating techniques (models): Selected mathematical models to estimate future costs and revenues. Example: Develop the first 3 WBS levels for the construction of a small commercial building. Estimating techniques (models) The goal is to develop cash flow projections, not exact future data (which is almost impossible). Cost and revenue estimates can be classified to:  Order-of-magnitude estimates  Semi-detailed (or budget) estimates  Definitive (detailed) estimates Selected models Indexes (ratio technique) - An index is a dimensionless number used to estimate present and future costs from historical data. 𝐼𝐼𝑛𝑛̅ 𝐶𝐶𝑛𝑛 = 𝐶𝐶𝑘𝑘 × 𝐼𝐼𝑘𝑘̅ Where: k: reference year. n: year to be estimated at. Cn and Ck: cost or price in years n and k, respectively. Īn and Īk are the index values for the years n and k, respectively. Example A company wants to install a new boiler. The price of the boiler in the year 2000 was $525,000 when the index was 468. What is the price of the boiler in 2014 given that the index value is 542 in the year 2014? 𝐶𝐶2014 𝐶𝐶2014 𝐼𝐼2014 = 𝐶𝐶2000 × 𝐼𝐼2000 542 = 525,000 × = $608,013 468 Selected models Unit technique - Widely used and understood. - Good for preliminary estimates. Factor technique - Extension of the unit technique. - Good for preliminary estimates. 𝐶𝐶𝐶𝐶𝐶𝐶𝐶𝐶 = � 𝐶𝐶𝑑𝑑 + � 𝐶𝐶𝑚𝑚 𝑈𝑈𝑚𝑚 𝑑𝑑 𝑚𝑚 Power sizing technique 𝐶𝐶𝐴𝐴 𝑆𝑆𝐴𝐴 = 𝐶𝐶𝐵𝐵 𝑆𝑆𝐵𝐵 𝑋𝑋 Example: The purchase price of a commercial boiler (capacity S) was $181,000 eight years ago. Another boiler of the same basic design, except with capacity 1.42 S, is currently being considered for purchase. If the cost index was 162 for this type of equipment when the capacity S boiler was purchased and is 221 now, and the applicable cost capacity factor is 0.8, what is your estimate of the purchase price for the new boiler? Let 𝐶𝐶𝐴𝐴 = cost of new boiler (𝑆𝑆𝐴𝐴 = 1.42 𝑆𝑆) today and 𝐶𝐶𝐵𝐵 = cost of old boiler today (𝑆𝑆𝐵𝐵 221 = $246,920. 𝐶𝐶𝐵𝐵 = $181,000 × 162 1.42𝑆𝑆 𝐶𝐶𝐴𝐴 = $246,920 × 𝑆𝑆 0.8 = $326,879 = 𝑆𝑆). Learning curve 𝑍𝑍𝑢𝑢 = 𝐾𝐾(𝑢𝑢𝑛𝑛 ) Example: The time required to assemble the first car is 100 hours and the learning rate is 80%. What is the time required to assemble the 10th car? 𝑆𝑆 = 0.8 𝐾𝐾 = 100 hours 𝑢𝑢 = 10 cars 𝑍𝑍10 = 100 × log 0.8 10 log 2 = 47.65 hours *** This is not the total time to produce 10 units … it’s the time to produce the 10th unit *** Example You have been asked to estimate the cost of 100 prefabricated structures, each structure provides 1,000 sq.ft of floor space, with 8-ft ceilings. In 2003, you produced 70 similar structures consisting of the same materials and having the same ceiling height, but each provided only 800 sq.ft of floor space. The material cost for each structure was $25,000 in 2003, and the cost capacity factor is 0.65. The cost index values for 2003 and 2014 are 200 and 289, respectively. The estimated manufacturing cost for the first 1,000 sq.ft structure is $12,000. Assume a learning curve of 88% and use the cost of the 50th structure as your standard time for estimating manufacturing cost. Estimate the total material cost and the total manufacturing cost for the 100 prefabricated structures. Material cost 𝐼𝐼2003 = 200 𝑆𝑆2003 = 800 𝐶𝐶2003 = $25,000 𝐼𝐼2014 = 289 𝑆𝑆2014 = 1000 𝐶𝐶2014 Manufacturing cost 𝑠𝑠 = 0.88 𝑋𝑋 = 0.65 289 1000 = $25,000 × × 200 800 𝐾𝐾 = $12,000 𝑍𝑍50 = $12,000 × 50 log 0.88 log 2 Total cost = ($41,764+$5,832)×100 = $4,759,600. 0.65 = $41,764 = $5,832/unit

Chapter 3 Cost Estimation Techniques

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