Cost Analysis Strategies in Civil Engineering Systems

CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 1 of 11 Problem 1 ( 20 ) After analyzing the graph and considering the different trendlines, it is evident that all of them are fairly accurate in representing the data, except for the Linear model, which dips below the x-axis. Among the models, the exponential model appears to be the most reasonable choice. This is due to the observed relationship where an increase in Preventive Maintenance (PM) Expenditure leads to a decrease in Repair Expenditure, which aligns with logical expectations. When more funds are allocated to preventive measures, the repair costs naturally decrease. Therefore, we have opted for the exponential model to address the problem at hand, with y representing Repair Expenditure and x representing PM Expenditure. The specific equation is y = 157.62e^(-0.027x). If we select x = 55, the corresponding y value, y1, is approximately 35.70. Now, if we increase x by 15%, the new value of x becomes 55(1 + 0.15) = 63.25, resulting in a new y value, y2, of 28.57. To calculate the reduction in repair cost (∆Y) and the increment in preventive maintenance cost (∆X), we perform the following calculations: ∆Y = 35.70 - 28.57 = 7.13 ∆X = 55 - 63.25 = -8.25 The net incremental benefit is determined by adding ∆Y and ∆X: Net Increment Benefit = ∆Y + ∆X = $-1.12. As the net incremental benefit is negative, it indicates that increasing the maintenance expenditure from $55/sf to $63.25/sf would not be advantageous. y = -2.1965x + 160.76 R² = 0.6366

CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 2 of 11 y=157.62e (-0.027x) The derived equation for the Y’ variable is Y' = -4.2557e (-0.027x) = 1, with x equal to $53.64. Any Preventive Maintenance (PM) expenditure exceeding $53.64 would yield incremental benefits lower than the incremental cost. dy/dx=-4.2557e^(-0.027x)

CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 3 of 11 Problem 2 (20) We can express the Total Cost (TC) as the sum of a constant term (k) and a linear term involving the concrete surface area inspected and monitored (V). The equation can be represented as TC = k + f(V), where V represents the square footage of the concrete surface area. Let us assume that the function is Linear. c= k+aV; k is $10 million and a=1.25 $/ft^2. (a) annual variable costs= 1.25V 0 200000 400000 600000 800000 1000000 1200000 1400000 1600000 0 500000 1000000 1500000 2000000 2500000 (b) total annual costs= 10,000,000 + 1.25V 0 200000 400000 600000 800000 1000000 1200000 1400000 1600000 9000000 9500000 10000000 10500000 11000000 11500000 12000000 12500000

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CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 4 of 11 (c) average total costs= 10,000,000/V + 1.25 (d) average marginal costs=− 10,000,000/V 2

CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 5 of 11 Problem 3 (20) (a) To determine the monthly water usage (W) that will maximize net revenue, we need to calculate the total fees and total operating costs as a function of W and then find the value of W that maximizes the difference between them. The total fees charged for water usage can be calculated as follows: Total Fees = Water Fee + Fixed Amount Water Fee = $150/m3 * W Fixed Amount = $100,000 Therefore, Total Fees = $150 * W + $100,000 The total operating costs can be calculated using the given cost function: Total Operating Costs = Fixed Operating Cost + Variable Costs Fixed Operating Cost = $25,000 Variable Costs = 0.0001W^3 - 0.001W^2 - 50W Therefore, Total Operating Costs = $25,000 + (0.0001W^3 - 0.001W^2 - 50W) Net revenue is the difference between total fees and total operating costs: Net Revenue = Total Fees - Total Operating Costs Net Revenue = ($150W + $100,000) - ($25,000 + 0.0001W^3 - 0.001W^2 - 50W) Simplifying, Net Revenue = -0.0001W^3 + 0.001W^2 + 100W + $75,000 To find the monthly water usage (W) that maximizes net revenue, we can take the derivative of the net revenue function with respect to W and set it equal to zero. Then solve for W. d(Net Revenue)/dW = -0.0003W^2 + 0.002W + 100 = 0 Solving the quadratic equation, we get two possible values for W. We can calculate the net revenue for both values and choose the one that maximizes it. Let's solve the equation to find the values of W that satisfy the condition: -0.0003W^2 + 0.002W + 100 = 0 Using the quadratic formula, we have:

CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 6 of 11 W = (-0.002 ± sqrt((0.002)^2 - 4 * (-0.0003) * 100)) / (2 * (-0.0003)) Simplifying the equation: W = (-0.002 ± sqrt(0.000004 + 0.012)) / (-0.0006) W = (-0.002 ± sqrt(0.012004)) / (-0.0006) W = (-0.002 ± 0.109629) / (-0.0006) There are two possible values for W: 1. W1 = (-0.002 + 0.109629) / (-0.0006) ≈ 183.4 m3 (approximately) 2. W2 = (-0.002 - 0.109629) / (-0.0006) ≈ -1851.1 m3 (approximately) Since water usage cannot be negative, the feasible value for W is W1 ≈ 183.4 m3. Therefore, the monthly water usage (W) that will maximize net revenue is approximately 183.4 cubic meters. (b) To plot the total operating costs and total fees, we need to calculate them for different values of W (monthly supply in cubic meters) within the given range of 0 to 2500 m3. The total operating cost (TOC) can be calculated using the fixed operating cost (FOC) and the variable cost (VC) as follows: TOC = FOC + VC Fixed operating cost (FOC) = $25,000 per month Variable cost (VC) = 0.0001W^3 - 0.001W^2 - 50W The total fee (TF) can be calculated as follows: TF = $150W + $100,000 Let's calculate the total operating costs (TOC) and total fees (TF) for different values of W and plot the results using Python:

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CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 7 of 11 (c) To plot the average total operating cost function, we need to calculate the total cost for different values of W and then find the average cost. The total cost (TC) is the sum of the fixed cost (FC) and the variable cost (VC). The fixed cost is given as $25,000 per month, and the variable cost is represented by the cost function VC = 0.0001W^3 - 0.001W^2 - 50W. Therefore, the total cost function (TC) can be written as: TC = FC + VC TC = $25,000 + (0.0001W^3 - 0.001W^2 - 50W) To calculate the average total operating cost, we need to divide the total cost by the quantity of water supplied (W). Average Total Operating Cost (ATC) = TC / W Let's plot the average total operating cost function using this information: (d) To analyze the economy of scale implications of total operating costs, we need to consider the relationship between the fixed costs, variable costs, and the total operating costs.

CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 8 of 11 The fixed operating cost of the water supply treatment plant is given as $25,000 per month, which remains constant regardless of the monthly supply (W) in cubic meters (m3). The variable costs (VC) are represented by the cost function: VC = 0.0001W^3 - 0.001W^2 - 50W To calculate the total operating cost (TC), we add the fixed cost to the variable cost: TC = FC + VC Let's analyze the behavior of the total operating costs at different operating points (W values): 1. Economies of Scale: Economies of scale occur when the total operating costs decrease as the volume of production increases. In this case, as the monthly supply (W) of water increases, the total operating costs decrease. Economies of scale are achieved if the variable costs decrease faster than the fixed costs increase. This happens when the rate of decrease in variable costs offsets the fixed costs. In other words, the decrease in variable costs outweighs the increase in fixed costs, resulting in lower total operating costs. 2. Diseconomies of Scale: Diseconomies of scale occur when the total operating costs increase as the volume of production increases. In this case, as the monthly supply (W) of water increases, the total operating costs increase. If the variable costs increase faster than the fixed costs decrease, diseconomies of scale are observed. This happens when the increase in variable costs surpasses the decrease in fixed costs, leading to higher total operating costs. To determine the operating point at which the economy of scale changes to the diseconomy of scale, we need to find the value of W where the transition occurs. This can be done by analyzing the behavior of the cost function VC = 0.0001W^3 - 0.001W^2 - 50W. By observing the cost function equation, we can see no explicit fixed cost term. However, the $25,000 fixed operating cost can be considered a constant added to the variable costs. To find the operating point where the economy of scale changes to a diseconomy of scale, we need to identify the value of W at which the total operating costs start increasing. This can be done by calculating the total operating costs for different values of W and identifying the point where the costs start to rise instead of decrease. Please note that the given monthly water fee of $150/m3 plus a fixed amount of $100,000 is irrelevant for this analysis as it is an additional charge to cover other expenses and is independent of the supply volume (W). With specific data points to evaluate the cost function and perform the calculations, it is possible to provide a precise answer regarding the operating point where the economy of scale changes to the diseconomy of scale.

CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 9 of 11 Problem 4 (10) (a) Assume city-cost indices are: Boston=115, Springfield = 102. The same treatment plant was constructed in 1990 at Boston if constructed in Springfield with 50 mgd = 150 * 102/115 = 133.04 million dollars. Estimated cost of the new treatment = $133.04 * (30/50) 0.75 = 90.7 million dollars (b) Imagine, the 2025 projected cost index = 235.6, and the historical cost index 1990 was 94.3. Monitoring plants 2025 estimated cost = $150 * 235.6/94.3 = 374.76 million dollars Now estimated cost must be scaled to calculate the difference in size between the proposed plant and the monitoring plants. 2025 cost estimate = $374.76 * (30/50) = 224.9 million dollars. Problem 5 (20) (a) To determine the effective annual interest rate with monthly compounding periods, we need to use the formula: r eff = (1 + r nom /n) n - 1 Where: r eff = effective annual interest rate r nom = nominal annual interest rate n = number of compounding periods per year In this case, the nominal annual interest rate is 8%, and the compounding is done monthly (12 periods per year). Plugging in the values, we get: r eff = (1 + 0.08 /12) 12 - 1 Calculating this, we find: r eff = 8.30% Therefore, the effective annual interest rate with monthly compounding periods is approximately 8.30%. (b) The formula gives the effective annual interest rate with an infinite number of compounding periods: r eff = (1 + r nom /n) n - 1 As the number of compounding periods approaches infinity, the formula becomes:

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CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 10 of 11 r eff = (1 + r nom /∞) ∞ -1 Using the limit property of the compound interest formula, we find: r eff = e rnom - 1\) where e is the base of the natural logarithm (approximately 2.71828). Plugging in the value of r nom = 0.08, we have: r eff = e {0.08} - 1 Calculating this, we find: r eff = 8.33% Therefore, the effective annual interest rate with an infinite number of compounding periods is approximately 8.33%. (c) To calculate the combined nominal interest rate on the loan including the effects of inflation, we need to add the inflation rate to the nominal interest rate. The combined nominal interest rate can be calculated as follows: Combined nominal interest rate = nominal interest rate + inflation rate In this case, the nominal interest rate is 8% and the annual inflation rate is 2%. Plugging in the values, we have: Combined nominal interest rate = 8% + 2% = 10% Therefore, the combined nominal interest rate on the loan, including the effects of inflation, is 10%. (d) To find the effective monthly interest rate assuming monthly payments, we can use the formula: r monthly = (1 + r nom /n) {1/n} - 1 In this case, the nominal interest rate is 8%, and the compounding is done monthly (12 periods per year). Plugging in the values, we get: r monthly = (1 + 0.08/12) {1/12} - 1) Calculating this, we find: r monthly = 0.65% Therefore, the effective monthly interest rate, assuming monthly payments, is approximately 0.65%. Problem 6 (10)

CIVE 3720 Civil Engineering Systems Name: Amanda Mello HW 4: Cost Analysis; Max Score = 100 Page 11 of 11

Assignment 4

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