1. Cost Concepts and Design Economics
Lecture 2
Cost Analysis
NE 364 Engineering Economy

2. Cost Estimating
A term used to describe the process by which the present and future cost consequences of engineering designs are forecast.
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3. Cost estimating used to
Provide information used in setting a selling price for quoting, bidding, or evaluating contracts
Determine whether a proposed product can be made and distributed at a profit (for simplicity, price = cost + profit)
Evaluate how much capital can be justified for process changes or other improvements
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4. Cost Classifications Fixed and Variable Costs
Direct and Indirect Costs
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5. Cost Classifications (cont.)
Fixed costs are those unaffected by changes in activity level over a feasible range of operations for the capacity or capability available.
(e.g. insurance and taxes on facilities, general management and administrative salaries, license fees, and interest costs on borrowed capital).
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6. Cost Classifications (cont.)
Variable costs are those associated with an operation that vary in total with the quantity of output or other measures of activity level.
Example of variable costs include : costs of material and labor used in a product or service, because they vary in total with the number of output units -- even though costs per unit remain the same.
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7. Cost Classifications (cont.)
Direct costs can be reasonably measured and allocated to a specific output or work activity -- labor and material directly allocated with a product, service or construction activity.
Indirect costs are difficult to allocate to a specific output or activity -- costs of common tools, general supplies, equipment maintenance and overhead costs.
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8. Cost Terminology
Investment Cost or capital investment is the capital (money) required for most activities of the acquisition phase;
Working Capital refers to the funds required for current assets needed for start-up and subsequent support of operation activities;
Operation and Maintenance Cost includes many of the recurring annual expense items associated with the operation phase of the life cycle;
Disposal Cost includes non-recurring costs of shutting down the operation;
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9. Total Cost Calculation
Total Cost (TC) = Fixed Costs (CF) + Variable Costs (CV) TC = CF + CV TC = CF + cv * D
cost
Quantity (D) CF Cv CT
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10. Profit = Total Revenue (TR) – Total Cost (TC)
Profit Calculation
Profit = Total Revenue (TR) – Total Cost (TC)
Scenario1:
Constant Price
Scenario2:
Variable Price
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11. Scenario 1: Constant Price
Profit = Total Revenue (TR) – Total Cost (TC)
TR = p * D , where p is constant
Profit = p * D – TC
Break-even point occurs when Profit = 0
TR = TC
p * D’ =CF + cv * D’
D’ = CF/ (p – cv)
NE 364 Engineering Economy

12. Breakeven Chart (Scenario 1)TR
cost
Quantity (D) CF Cv CT
BEP (D’)
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13. Scenario 2: Variable Price
Profit = Total Revenue (TR) – Total Cost (TC)
TR = p * D , where p is variable
p = a – b*D
Demand (D)
Price (p)
p = a - b D
a is the intercept at the price axis
-b is the slope
p is the selling price per unit
D is the demand
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14. The Total Revenue FunctionTR
Total Revenue = p x D
= (a – bD) x D
=aD – bD2
QUANTITY ( OUTPUT )
TR = Max
D=a/2b
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15. Breakeven Chart (Scenario 2)
Cost / Revenue
Quantity ( Output )
Demand CT
D’1
D’2 D*
Profit
Total Revenue
Maximum Profit
(not necessarily at max revenue)
D’1 and D’2 are breakeven points
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16. Finding BEP: Scenario 2
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17. Scenario 2: Profit Maximization D*
Occurs where total revenue exceeds total cost by the greatest amount;
Occurs at d(profit)/dD=0  
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18. Solved Examples
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19. Example 1: Finding BEP: Scenario 1
A engineering consulting firm measures its output in a standard service hour unit, which is a function of the personnel grade levels in the professional staff. The variable cost is $62 per standard service hour. The charge-out rate (i.e., selling price) is $85.56 per hour. The maximum output of the firm is hours per year, and its fixed cost is $2,024,000 per year. For this firm:
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20. Example 1 (cont.)
what is the breakeven point in standard service hours and in % of total capacity
What is the % reduction in the breakeven point
if fixed cost s are reduced by 10%;
if variable cost per hour is reduced by 10%;
if both costs are reduced by 10%; and
if the selling price per unit is increased by 10%?
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21. Example 1: Solution At BEP: Sensitivity Analysis
TR=TC
p D’ = CF + cvD’
D’ = CF / (p – cv)
D’=$2,024,000/($85.56 – $62)=85,908 hours per year
D’=85,908/160,000=0.537 (or 53.7% of capacity)
Sensitivity Analysis
BEP at 10 % reduction in CF:
D’= 0.9($2,024,000)/($85.56 – $62)=77,318 hours per year
(85,908 – 77,318)/85,908=0.10 (or a reduction of 10% in D’)
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22. Example 1: Solution (cont.)
BEP at 10% reduction in cv:
D’= ($2,024,000)/($85.56 – 0.9($62))=68,011 hours per year
(85,908 – 68,011)/85,908=0.208 (or a reduction of 20.8% in D’)
10 % reduction in both costs
D’=$(0.9)(2,024,000)/($85.56 – (0.9)$62)= 61,210 hours per year
(85,908 – 61,210)/85,908=0.287 (a reduction of 28.7% in D’)
10% increase in p
D’=$2,024,000/((1.1)$85.56 – $62)=63,021 hours per year
(85,908-63,021)/85,908=0.266 (26.6% reduction in D’)
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23. Example 1: Solution Summary
Change in Factor value (s)
Decrease in BEP
10% reduction in CF
10.0%
10% reduction in cv
20.8%
10% reduction in CF and in cv
28.7%
10% increase in p
26.6%
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24. Example 2: Finding BEP: Scenario 2
A company has established that the relationship between the sales price for one of its products and the quantity sold per month is approximately D= p units. (D is the demand or quantity sold per month, and p is the price per dollars.) The fixed cost is $800 per month, and the variable cost is $30 per unit produced.
What number of units, D*, should be produced per month and be sold to maximum net profit?
What is the maximum profit per month related to the product?
Determine D'1 and D'2.
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25. Example 2: Solution Given: D = 780 - 10p (units/month)
Fixed Cost (CF) = $800/month
Variable Cost per Unit (cv) = $30/unit
p= (780 – D )/10
p = 78 – 0.1 * D
D* = (78 – 30)/0.2 = 240 units/month
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26. Example 2: Solution (cont.)
D* = 240,
Profit = 48D - 0.1D
Profit = 48(240) - 0.1(240) = $4,960
Maximum Profit = $4,960/month
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27. Example 2: Solution (cont.)
𝐷′ 1,2 = −(78−30)± (78−30) 2 −4(−0.1)(−800) 2(−0.1)
D1' = or 18 units/month
D2' = or 462 units/month
Profitable range of demand:
18 units/month To 462 units/month
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28. See you next week!
NE 364 Engineering Economy