Practice Problems: Time Value of Money

Time Value of Money - PE Exam Question Bank

Question 1: A civil engineer is evaluating the construction of a wastewater treatment plant for a municipality. The initial construction cost is $2,500,000, and annual operating and maintenance costs are $180,000. The facility will require a major equipment replacement costing $400,000 at the end of year 10. Using an interest rate of 6% per year, what is the present worth of the total project costs over a 20-year analysis period?
(a) $4,287,500
(b) $4,623,850
(c) $4,958,270
(d) $5,234,100

Question 2: An electrical engineer is comparing two motor options for a manufacturing facility. Motor A costs $12,000 with annual energy costs of $3,200. Motor B costs $18,500 with annual energy costs of $2,100. Both motors have a 15-year service life with no salvage value. Using a minimum attractive rate of return (MARR) of 8% per year, what is the incremental benefit-cost ratio for the higher initial cost alternative?
(a) 0.94
(b) 1.05
(c) 1.12
(d) 1.28

Question 3: A mechanical engineer must replace an industrial boiler system. The new boiler costs $85,000 installed and will save $18,500 per year in energy costs compared to the existing system. The boiler has an expected life of 12 years with a salvage value of $8,000. If the company requires a rate of return of 10% per year, what is the net present value (NPV) of this investment?
(a) $46,280
(b) $52,950
(c) $43,120
(d) $38,760

Question 4: A transportation engineer is analyzing a toll bridge project. The bridge construction requires an initial investment of $45,000,000 and will generate annual toll revenues of $6,200,000. Annual maintenance costs are $950,000. The bridge has a 40-year design life. At an interest rate of 5% per year, what is the approximate capitalized cost of the bridge considering perpetual service through future replacements?
(a) $64,000,000
(b) $88,500,000
(c) $105,250,000
(d) $126,800,000

Question 5: An environmental engineer is evaluating a groundwater remediation project. The cleanup equipment costs $325,000 and will require $42,000 annually for operation. After 8 years, the equipment can be sold for $65,000. Using an interest rate of 7% per year, what is the equivalent uniform annual cost (EUAC) of this remediation system?
(a) $88,450
(b) $93,280
(c) $96,720
(d) $101,540

Question 6: A structural engineer is designing a parking structure that will generate revenue of $285,000 per year. The structure costs $1,850,000 to build and will require major renovations costing $420,000 at the end of year 15. The structure has a 30-year life with a salvage value of $200,000. At an interest rate of 6% per year, what is the present worth of the net revenue?
(a) $2,138,700
(b) $2,487,300
(c) $2,256,400
(d) $1,943,800

Question 7: A chemical engineer is evaluating a process improvement that requires an investment of $560,000. The improvement will generate cost savings with the following cash flow pattern: $80,000 in year 1, increasing by $15,000 each year thereafter for 10 years. Using an interest rate of 9% per year, what is the present worth of the savings?
(a) $782,340
(b) $845,920
(c) $908,760
(d) $1,024,500

Question 8: An industrial engineer is comparing two automation systems. System X costs $140,000 with annual savings of $38,000 over 8 years. System Y costs $195,000 with annual savings of $48,000 over 8 years. Both have zero salvage value. Using a MARR of 12% per year, what is the incremental rate of return for investing in System Y versus System X?
(a) 8.2%
(b) 10.5%
(c) 12.4%
(d) 14.1%

Question 9: A water resources engineer is designing a reservoir system. The construction cost is $12,500,000. Annual operating costs are $340,000 for the first 10 years and will increase to $480,000 per year thereafter. Using a 50-year project life and 4% interest rate, what is the present worth of all operating costs?
(a) $11,850,000
(b) $9,720,000
(c) $13,420,000
(d) $10,560,000

Question 10: A manufacturing engineer evaluates replacing an existing machine. The defender has a current market value of $45,000, annual operating costs of $28,000, and 5 years of remaining life. The challenger costs $95,000, has annual operating costs of $16,000, and a 10-year life. Both have zero salvage value. At 10% interest, what is the annual cost advantage of the challenger?
(a) $3,240
(b) $4,580
(c) $2,870
(d) $5,120

Question 11: A project manager is analyzing a renewable energy project with the following cash flows: initial investment of $2,200,000 at year 0, annual revenues of $450,000 for years 1-20, and a decommissioning cost of $180,000 at year 20. Using an interest rate of 8% per year, what is the benefit-cost ratio for this project?
(a) 1.78
(b) 1.94
(c) 2.12
(d) 2.35

Question 12: A geotechnical engineer is evaluating a soil stabilization project. The work can be done now for $380,000, or deferred for 3 years when it will cost $485,000. Annual inspection costs of $22,000 will be incurred if the work is deferred. Using an interest rate of 7% per year, what is the present worth of cost savings by doing the work now?
(a) $32,450
(b) $48,720
(c) $56,180
(d) $61,340

Question 13: A consulting engineer invests in office equipment costing $78,000 with a 6-year life and $12,000 salvage value. The equipment generates additional revenue of $24,000 per year. Using a MARR of 15% per year, what is the external rate of return (ERR) if the reinvestment rate is 10%?
(a) 16.8%
(b) 18.3%
(c) 20.1%
(d) 22.5%

Question 14: A power systems engineer is evaluating a transmission line upgrade. The upgrade costs $8,500,000 and reduces power losses valued at $950,000 annually. Maintenance costs decrease by $180,000 per year. The line has a 35-year life. At 5% interest, what is the simple payback period (without considering time value of money)?
(a) 7.5 years
(b) 8.9 years
(c) 9.4 years
(d) 10.2 years

Question 15: A facilities engineer must choose between two HVAC systems. System A costs $225,000 with a 15-year life and annual costs of $34,000. System B costs $285,000 with a 20-year life and annual costs of $26,500. Using an interest rate of 9% per year and equivalent annual worth analysis, which system should be selected and what is the annual worth advantage?
(a) System A, $2,450 advantage
(b) System B, $3,180 advantage
(c) System A, $4,270 advantage
(d) System B, $5,620 advantage

Question 16: A highway engineer is analyzing a pavement rehabilitation project. The project requires $1,850,000 initial investment and will save $285,000 annually in maintenance costs for 12 years. Additional reconstruction of $420,000 is required at year 6. Using a discount rate of 6% per year, what is the net present value?
(a) $412,800
(b) $456,300
(c) $378,950
(d) $523,700

Question 17: A mining engineer evaluates equipment that costs $525,000 and produces net annual revenue following this pattern: $180,000 in year 1, decreasing by $15,000 each year for 8 years. The salvage value is $85,000. At 11% interest, what is the present worth of this investment?
(a) $245,800
(b) $189,400
(c) $312,600
(d) $267,200

Question 18: A petroleum engineer is evaluating an enhanced oil recovery project requiring $4,200,000 investment. Annual revenues are projected at $1,250,000 with annual operating costs of $520,000 for 10 years. Using a MARR of 18% per year, what is the discounted payback period?
(a) 6.8 years
(b) 7.4 years
(c) 8.2 years
(d) 9.1 years

Question 19: An aerospace engineer analyzes a testing facility upgrade costing $1,650,000 with annual maintenance of $95,000. The facility generates testing revenue of $425,000 per year. After 15 years, the facility has a salvage value of $220,000. What interest rate makes the present worth equal to zero (internal rate of return)?
(a) 17.2%
(b) 18.9%
(c) 19.6%
(d) 21.3%

Question 20: A biomedical engineer evaluates two sterilization systems for a hospital. System P costs $165,000, lasts 10 years, has annual costs of $28,000, and salvage value of $18,000. System Q costs $235,000, lasts 15 years, has annual costs of $19,500, and salvage value of $32,000. Using 8% interest and present worth analysis over the least common multiple of lives, what is the present worth difference?
(a) $42,800 favoring System P
(b) $38,650 favoring System Q
(c) $51,200 favoring System Q
(d) $46,320 favoring System P