CE Systems HW8

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3720

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Civil Engineering

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Apr 3, 2024

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CIVE 3720 Civil Engineering Systems Name: Karen Vinal March 22, 2024 HW 8: Economic Decision Analysis (Max Score 100) Page 1 of 8 Problem 1: (10) Given: As a city engineer, you seek the capitalized cost of perpetual service from a water storage tank. Due to the highly corrosive nature of your coastal environment, the tank (which costs $40,000) is maintained at an annual cost of $1,000 and is replaced every 10 years. Whenever its service life ends, the tank is intended to be sold to a steel mill as metal scrap for $3000. Use 10% interest rate. Find: (a) Find the capitalized cost of the storage tank investment. (b) Give two reasons why the formula you used in (a) is most appropriate for questions involving civil systems of this type. Solution: (a) Present Value of Initial Cost- P = $40,000 Present Value of Annual Cost- P = A(P|A,10%,10) P=A ( (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 ) P=$1,000( (1+0.1) 10 −1 0.1(1+0.1) 10 ) P = $6,145 Present Value of Salvage Value- P = F(P|F,10%,10) P = F(1+i) -n P = $3,000 (1+0.1) -10 P = $1,157 Total Capitalized Cost- CC = IC + AC – S CC = $40,000 + $6,145 -$1,157 CC = $44,988

CIVE 3720 Civil Engineering Systems Name: Karen Vinal March 22, 2024 HW 8: Economic Decision Analysis (Max Score 100) Page 2 of 8 (b) The formula for capitalized cost considers all relevant costs associated with the system investment. This provides a fully comprehensive view of the total cost of ownership over time. This formula also accounts for the time value of money by discounting future cash flows back to their present value using the interest rate. This ensures that all costs that are accrued during the system ’ s life are appropriately weighted in the total cost of ownership. Problem 2: (10) Given: The county municipal council owns a toll bridge that costs the county $250,000 every year to operate and $130,000 a year to maintain. The facility brings in revenue of $500,000 per year. There is a proposal to sign a 10-year lease of the bridge to a private operator who will give the city $1 million upfront with the understanding that the private operator will be responsible for the operation and maintenance of the bridge throughout the lease period and will always keep its level of service above a certain threshold. Assume a 4% interest rate. Find: Should the City accept the proposal from the private operator? Solution: Benefit = Rev – OC – MC Present Value of Revenue: P = A(P|A,4%,10) P = A ( (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 ) P=$500,000 ( (1+0.04) 10 −1 0.04(1+0.04) 10 ) P = $4,055,448 Present Value of Operating Costs: P = A(P|A,4%,10) P = A ( (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 ) P = $250,000( (1+0.04) 10 −1 0.04(1+0.04) 10 ) P = $2,027,724

CIVE 3720 Civil Engineering Systems Name: Karen Vinal March 22, 2024 HW 8: Economic Decision Analysis (Max Score 100) Page 3 of 8 Present Value of Maintenance Costs: P = A(P|A,4%,10) P = A ( (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 ) P = $130,000 ( (1+0.04) 10 −1 0.04(1+0.04) 10 ) P = $1,054,416 Benefit = $4,055,448 - $2,027,724 - $1,054,416 Benefit = $973,307 The city should accept the deal, since the $1,000,000 from the private operator is greater than the proposed benefit of $973,307. Problem 3: (10) Given: You are the city engineer and wish to determine the life-cycle costs for the construction and maintenance of a road system assuming a 25-year service life. The initial cost of the project is $2M. Annual maintenance costs are estimated to be $50,000 during the first year and are expected to rise by $10,000 each year thereafter throughout the life of the project. Find: Determine the present worth life-cycle cost of the road system assuming a 5% annual interest rate. Solution: Present Worth of Annual Maintenance Costs P = A (P|A,5%,25) P = A ( (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 ) P = $50,000 ( (1+0.05) 25 −1 0.05(1+0.05) 25 ) P = $704,697 Present Worth of Arithmetic Gradient

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CIVE 3720 Civil Engineering Systems Name: Karen Vinal March 22, 2024 HW 8: Economic Decision Analysis (Max Score 100) Page 4 of 8 P = G ( 1−(1+𝑛𝑖)(1+𝑖) −𝑛 𝑖 2 ) P = $10,000 ( 1−(1+25(0.05))(1+0.05) −25 0.05 2 ) P = $1,342,275 Total Present Worth = IC – (Maintenance costs + Gradient Costs) Total Present Worth = $1,000,000 - $704,697 - $1,342,275 Total Present Worth = -$1,046,972 Problem 4: (10) Given: The same project as given in problem 3, however, maintenance costs are estimated to rise at a rate of 2%/year after the first year. Find: Determine the present worth life-cycle cost of the road system assuming a 5% annual interest rate. Present Worth of Annual Maintenance Costs P = A (P|A,5%,25) P = A ( (1+𝑖) 𝑛 −1 𝑖(1+𝑖) 𝑛 ) P = $50,000 ( (1+0.05) 25 −1 0.05(1+0.05) 25 ) P = $704,697 Present Worth of Arithmetic Gradient P = G ( 1−(1+𝑛𝑖)(1+𝑖) −𝑛 𝑖 2 ) P = $10,000 ( 1−(1+25(0.05))(1+0.05) −25 0.05 2 ) P = $1,342,275 Total Present Worth = IC – (Maintenance costs + Gradient Costs) Total Present Worth = $2,000,000 - $704,697 - $1,342,275 Total Present Worth = -$46,972

CIVE 3720 Civil Engineering Systems Name: Karen Vinal March 22, 2024 HW 8: Economic Decision Analysis (Max Score 100) Page 5 of 8 Problem 5: (20) Given: You own a construction business and wish to purchase a new excavator. You have narrowed your search to two choices. The estimated costs and revenues (benefits) for each alternative are shown below. MARR=10% Find: Assuming a 10-year analysis period, which Excavator should you purchase? Method: Use NPV, IRR, B-C, and DPBP Analysis Methods Solution: NPV: Alt 1- (-IC) -OM(P|A,10%,10) +R(P|F,10%,10) +S (P|F,10%,10 -$250 - $150 ( (1+0.1) 10 −1 0.1(1+0.1) 10 ) + $200 (1+0.1) -10 + $30(1+0.1) -10 $1083 Alt 2- (-IC) -OM(P|A,10%,10) +R(P|F,10%,10) +S (P|F,10%,10) -$350 - $175 ( (1+0.1) 10 −1 0.1(1+0.1) 10 ) + $250 (1+0.1) -10 + $50(1+0.1) -10 $1310 Using NPV, Alternative 2 should be selected. IRR: Using Excel ’ s Goal Seek function, it can be determined that the Internal Rate of Return for both Alternative 1 and two are greater than 10%, meaning both alternatives are economically feasible. Using incremental IRR analysis through Excel ’ s IRR function, it can be determined that

CIVE 3720 Civil Engineering Systems Name: Karen Vinal March 22, 2024 HW 8: Economic Decision Analysis (Max Score 100) Page 6 of 8 the highest incremental rate of return is 22.2% when moving from Alternative 1 to Alternative 2. This determines that alternative 2 should be selected. B-C: PWB = R(P|A,10%,10)+S(P|F,10%,10) ALT 1: $1240.48 ALT 2: $1632.53 PWC = IC + OM(P|A,10%,10) ALT 1: $1,172 ALT 2: 1425 B/C ALT 1: 1.06 ALT 2: 1.15 2-1>1 1.546 ---->> Select Alternative Two DPBP: Using Excel ’ s Goal Seek Function, Alternative 1 and 2 have the same discounted payback period, meaning there would be next to no difference between selecting the two alternatives. Problem 6: (20) Conduct a sensitivity analysis of Initial Purchase Price, O&M, Revenue, and MARR variables on the NPV of Alternative 1 in problem 5 above. The analysis should consider that each variable may have +/- 20% error. Plot the results and rank the variables in order of highest to lowest sensitivity.

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CIVE 3720 Civil Engineering Systems Name: Karen Vinal March 22, 2024 HW 8: Economic Decision Analysis (Max Score 100) Page 7 of 8 Highest Sensitivity: Revenue, Operation and Maintenace, Salvage Value, then Initial Cost Problem 7: (20) Given: A probability distribution of annual net cash flow (Revenue – O&M) for Alternative 1 of problem 5 is provided in the following table. p(A) A ($K) 0.2 -$50 0.3 -$20 0.4 $75 0.1 $85 Find: 1. Probability that you will have at least $10K net positive cash flow in any given year. P(A>$10k). Assume normal distribution of annual cash flow. 2. Expected Value of the Present Worth for a 10-year analysis at MARR = 10%. Is this a wise investment? Solution: 1. From Excel, P(A>$10K)=0.488 -10000 -5000 0 5000 10000 15000 -25 -20 -15 -10 -5 0 5 10 15 20 25 IC OM P® P(S)

CIVE 3720 Civil Engineering Systems Name: Karen Vinal March 22, 2024 HW 8: Economic Decision Analysis (Max Score 100) Page 8 of 8 2. P = A(P|A,10%,10) P = $22.5K ( (1+0.1) 10 −1 0.1(1+0.1) 10 ) P = $138.25K This is not a wise investment, because there is a less than 50% chance of having a profit greater than $10,000, therefore there would not be enough funds to cover this purchase.

CE Systems HW8

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