Engineering Economics (CE 655) – Chapterwise Question Bank
Tribhuvan University | Institute of Engineering | III/II BE | Compiled from 2065–2081 Exam Papers
Chapters
- Chapter 1: Introduction (Principles, Role of Engineers, Cash Flow)
- Chapter 2: Interest and Time Value of Money
- Chapter 3: Basic Methodologies of Engineering Economic Analysis
- Chapter 4: Comparative Analysis of Alternatives
- Chapter 5: Replacement Analysis
- Chapter 6: Risk Analysis
- Chapter 7: Depreciation and Corporate Income Taxes
- Chapter 8: Inflation and Its Impact on Project Cash Flows
Chapter 1: Introduction
Principles of Engineering Economy · Role of Engineers in Decision Making · Cash Flow Diagram1.1 Principles of Engineering Economy
- 1Explain the principles of engineering economics.
- 2Why do engineers need knowledge of economics in decision making process? Explain briefly about the principles of engineering economics.
- 3Define economics. Briefly explain the scope of engineering economics with appropriate example.
- 4"The study of economic is important for engineers." Justify the statement with a suitable example.
- 5"Use the consistent view point" and "Make uncertainty explicit." Explain these two principles of engineering economics.
- 6Explain the role of engineers in Engineering Economic Decisions.
- 7Explain the principles of engineering economy in detail with appropriate examples.
- 8Why does an engineer must have the knowledge of economics during decision making process? List out principles of engineering economics.
- 9Define Engineering Economics. Why is engineering economics considered as an important aspect for making decisions for engineers? Explain.
- 10Define opportunity cost. Why is engineering economics considered as important aspect for making decisions for engineers? Explain.
- 11State and explain principles of engineering economics.
- 12Define term Engineering Economy. Explain principles of engineering economy.
- 13Define Engineering Economics. Write down the principles of Engineering Economic Analysis.
- 14Define Engineering Economics. Write down the principles of Engineering Economic Analysis.
- 15Scarcity is an emerging issue in engineering field. How does the study of economics help engineers in decision making process? Discuss.
- 16Explain the roles of engineers in making economic decision with appropriate examples.
- 17Define engineering economy. Enlist the principles of engineering economy.
- 18What are the principles of engineering economics? How does it help to decision making process?
- 19"Engineers play the important role in making the economic decision." Do you agree with this statement? Discuss.
- 20Define economic system. Write advantages of socialistic economy. Explain overhead cost and opportunity cost.
- 21Explain the terms, socialistic economy and cash flow diagram.
- 22Define economic system. Discuss briefly on the characteristics of capitalistic economy.
Chapter 2: Interest and Time Value of Money
Simple & Compound Interest · Nominal vs Effective Rate · Equivalence · Interest Formulas · Gradient Series2.1–2.3 Nominal vs Effective Interest Rate & Compounding
- 1Differentiate the nominal and effective interest with the support of 10% bank's interest which compounds daily.
- 2What are the factors to be considered in calculating the time value of money? Distinguish between nominal and effective interest with suitable example. Bank 'A' offers 6.25% interest that compounds daily and Bank 'B' offers 6.4% interest that compounds yearly — which bank do you prefer and why?
- 3Differentiate between simple and compound interest. What is the relation between nominal and effective interest rate? How long will it take for Rs 25,000 to triple itself, if the interest rate is 8% per year?
- 4What is the time value of money? Explain and differentiate nominal and effective interest rate. How long will it take for Rs 25,000 to triple itself, if the interest rate is 8% per year?
- 5What is the time value of money? Suppose that you make the monthly deposits of Rs. 5,000 each into a bank account that pays an interest rate of 8% compounded weekly for 5 years. After 5 years, interest rate changes to 6% per year. How much money will you have accumulated in this bank account at the end of 8 years?
- 6Explain the continuous compounding. What is future equivalent of a continuous funds flow amounting $10,000 per year, N = 12 years, M = ∞, 20% compounding continuously.
- 7A bank is starting its nominal interest rate of 9% p.a. and compounding quarterly. Calculate the effective interest rate (i) a year (ii) a quarter (iii) a month (iv) half year.
- 8Define time value of money, nominal and effective interest rate. Calculate future sum at the end of 5th year when monthly deposit is Rs 6,000 for 3 years that earns 7% interest per year.
- 9What is nominal and effective interest rate? Evaluate FW at the end of 10 years with 12% interest rate compounded monthly of a cash flow of Rs. 40,000 at the beginning of each year for 5 years.
- 10What is nominal and effective interest rate? Evaluate FW at the end of 15 years with 10% interest rate compounded monthly of a cash flow of Rs. 50,000 at the beginning of each year for first 10 years.
- 11What is effective and nominal interest rate? Evaluate FW at the end of 10 years with 8% interest rate compounded continuously of a cash flow of Rs. 500 at the beginning of each year for first 5 years.
- 12Differentiate between nominal and effective interest. Calculate both nominal and effective annual interest if you deposit now Rs 1,00,000 and you can draw Rs 1000 per month for ever.
- 13Explain the uncertainty and its sources. Differentiate between nominal interest rate and effective interest rate. If monthly interest rate is 1%, what will be the quarterly interest rate?
2.4–2.5 Interest Formulas: Single / Equal Payment / Gradient Series
- 1Assume you contribute Rs. 5,000 monthly to a retirement account that earns an interest rate of 10%, compounded quarterly. Calculate the total balance after 12 years, considering that interest is applied to the deposits made during each compounding period.
- 2An individual makes 5 annual deposits of Rs. 2,000 in a saving account that pays interest rate of 4% per year. One year after making the last deposit, the interest rate changes to 6% per year. Five years after the last deposit, the accumulated money is withdrawn from the account. How much money is withdrawn?
- 3The two cash flow transactions shown below are said to be equivalent at 10% interest, compounded annually. Find the unknown X value which satisfies the equivalence. [LHS: Rs. 200, Rs. 150, Rs. 100, Rs. 100, Rs. 150, Rs. 200 at years 0–5; RHS: 6 equal payments of X at years 0–5]
- 4Mr. Hari, father of Ram, had deposited Rs. 2,50,000 in a bank 10 years ago at an interest of 10% p.a. compounded quarterly. After knowing this, Ram is planning to deposit Rs. 50,000 at the end of this year and planned to increase the deposit annually by 15% till 5 years' end but with revised interest scheme in the same account. The revised interest scheme is 9.5% interest compounding monthly. What would be the accumulated cash in a bank at the end of 10th year of Ram's first deposit?
- 5If you deposited Rs. 5,00,000 now in a bank account which gives 6% interest per year, how many times would you be able to draw Rs. 10,000 per month with that many?
- 6Airplane ticket price will increase 8% in each of the next four years. The cost of end of the 1st year is Rs 2,000. How much needs to be put away now to cover the passenger travel at the end of year for the next 4 years, taking i = 5%?
- 7Suppose that you make a deposit of Rs 5000 per month in a saving account which gives 12% interest compounded quarterly for the first three years and 9% compounded monthly for the last two years. What amount do you expect at the end of 5 years?
- 8A company is considering investing Rs. 5,50,000 in new equipment. Net cash flow estimate during first year will be Rs. 50,000 and will increase by Rs. 25,000 per year the next year and each year thereafter. The equipment has 10 years service life and salvage value of Rs. 60,000. Assuming MARR = 12%: (i) Determine annual capital cost for the equipment. (ii) Determine the equipment annual savings. (iii) Determine if this is a wise investment.
- 9A couple is planning for their child's education. They wish to deposit Rs. 10,000 now in a bank account that gives 12% per year compounded monthly and increase the amount by Rs. 2,000 each year from the previous year for next 9 years. How much amount they will expect at the end of 10 years?
- 10The two cash flow transactions shown below are said to be equivalent at 10% interest, compounded annually. Find the unknown X value which satisfies the equivalence. [Rs. 200, Rs. 150, Rs. 100, Rs. 100, Rs. 150, Rs. 200 at years 0–5 = 6 equal X payments]
- 11A man is planning to retire in 25 years. He wishes to deposit regular money every month until he retires so that he will receive annual payments of Rs. 4,50,000 after the first year of his retirement for the next 10 years. How much he deposit if the interest rate is 8%, compounded monthly?
- 12A machine needs uniform semi-annual cash flow of $10,000 for fuel for 5 years. If interest rate is 12% compounded quarterly, what is its equivalent present worth?
- 13If you wish to withdraw Rs 2000 at the end of 1st year and expecting to be increased by 15% per annum then after till end of 8 years, what amount need to be deposited in a bank right now which has an interest of 15% p.a.?
- 14If you make equal monthly deposits of Rs. 5000 into the bank for 10 years, saving accounts that pays interest rate of 6% compounded monthly, what would be the amount at the end of 15 years?
- 15How much rupees should you deposit now so that you will be able to draw Rs. 5000 at the end of this month which increases by 2 percent per month for 15 years. Bank interest rate is 5% per year.
- 16Find the value of P if i = 10%. Use gradient formula also. [Diagram: payments of 10, 12, 14, 16, 16, 16, 16 at years 0–6]
- 17If you deposit Rs. 10000 in a saving account now which gives 10% nominal interest rate, what will be the amount after 5 years if interest is compounded (i) semi-annually (ii) monthly?
- 18Ramesh, a Civil Engineer, is planning to place 20% of his salary (Rs. 250,000 per year now) in a mutual fund. He expects a 7% salary increase each year for next 15 years. If the mutual fund will return 10% annual return, what will be the sum-amount at the end of 15 years? If salary increases by Rs 25000 per year, what will be the amount?
- 19Find the value of A and G if i = 10%, A = 3G. [Diagram: complex gradient cash flow with payments from year 0–10]
- 20Find the value of G if i = 10%. [Diagram: A = 5000 at years 0–5, gradient cash flows]
- 21Find the value of A if i = 15%. [Diagram: complex cash flow including gradient with 25,000 payment at year 0, payments at years 1–10, and 10,000 at year 15]
- 22Mr. X receives a loan of Rs 120,000 from a bank at an interest rate of 12%. He wishes to repay the loan in monthly installments with Rs. 3000 per month — how many installments are necessary? What annual interest rate is he paying if the bank asks him to pay Rs 5000 per month for 30 times?
Chapter 3: Basic Methodologies of Engineering Economic Analysis
MARR · Payback Period · PW / FW / AW Methods · IRR · ERR · B/C Ratio · Lifecycle Costing · Financial vs Economic Analysis3.1 MARR (Minimum Attractive Rate of Return)
- 1Define MARR. What are the determining factors of MARR in the economy? Calculate CR and make investment decision using AW method for the project when Initial investment = Rs. 1,00,000; Annual revenue = Rs. 20,000; Annual expense = Rs. 5,000; Salvage value = Rs. 25,000; Project life = 10 years; MARR = 10% per year.
- 2What is MARR? Explain the life cycle costing.
- 3Explain the factors affecting determination of MARR.
3.2 Payback Period Method
- 1Write advantages and limitations of payback period method with suitable example.
- 2Assess the feasibility of the project using Discounted PB period, AW, IRR, and ERR methods based on the following information. Assume MARR = 15% and ε = 18%. [EOY 0–5 cash flows: -12, +7, -6, +14, +5, +2 (×000)]
- 3Make investment decision for the following project by using (i) IRR (ii) B/C (iii) Discounted Payback methods: Initial cost = Rs. 4,00,000; Annual Revenue = Rs. 1,60,000 for 1st year, decreasing by Rs. 10,000; Annual Expenses = Rs. 40,000 for 1st year, increasing by Rs. 5,000; Salvage value = Rs. 1,00,000; Life = 8 years; MARR = 9%.
- 4Evaluate the project by using AW formulation at i = 12%. [EOY 0–5 cash flows: -3000, 800, 1000, 1100, 1210, 1464]
3.3 Equivalent Worth Methods (PW, FW, AW)
- 1An Engineering College is considering to purchase a new machine costing Rs. 4,00,000 having salvage value Rs. 1,00,000 at the end of 5th year. The use of machine will increase revenue by Rs. 1,50,000 that needs fuel cost of Rs. 30,000 per year. Find IRR and B/C ratio when MARR = 10%.
- 2Determine both types of B/C ratio using FW and AW formulation: Initial investment = Rs. 250,000; Annual Revenue Rs. 75,000 at end of 1st yr. increasing by Rs. 5,000; Annual O&M = Rs. 15,000; Salvage value = 25,000; MARR = 10% per yr.
- 3Consider the following cash flow of project: Initial investment = Rs 25,000; Net annual revenue = Rs 8,000; Salvation value after 5 years = Rs 5,000. Calculate IRR. Is the investment accepted? Assume MARR = 20%. Show unrecovered project balance in graphical as well as tabular form.
- 4Use IRR to evaluate the following project when MARR is 15% per years. [EOY 0: -60,000; 1: 20,000; 2: 40,000; 3: -40,000; 4: 50,000; 5: 70,000]. Make also unrecovered balance diagram.
- 5Your college is considering to purchase a machine of Rs. 3,00,000 expecting salvage value Rs. 50,000 at the end of 10th year. The use of machine saves Rs. 80,000 per year when it needs Rs. 20,000 operating cost for each year. Find: (i) Both types of B/C ratio using AW formulation (ii) Both types of payback periods.
- 6Your college is considering to purchase a vehicle of Rs. 3,00,000 expecting salvage value Rs 50,000 at end of 10th year. The use of vehicle saves Rs. 80,000 per year, when it needs Rs. 20,000 operating cost for each year. Find: (i) Both types of B/C ratio by FW formulation (ii) Both types of payback period.
- 7What is the difference between financial and economic analysis? Determine both types of B/C ratio from the following cash flow: Initial investment = 3,00,000; Annual revenue = 85,000; Annual costs = 15,000; Salvage value = 20% of initial investment; Life = 6 years; MARR = 10%.
- 8An equipment costing Rs. 5,00,000 is estimated to have life of 10 years. Expected annual revenue is Rs. 1,10,000 with annual cost of Rs. 20,000. Determine investment decision from PW, AW, and FW method when salvage value is Rs. 1,00,000 and MARR = 12%.
- 9Equipment costs 2,50,000 and has salvage value of 50,000 at end of 5 years. Annual expenses = 40,000. Revenue = 120,000 per year. MARR = 20% = ε. (i) Evaluate IRR using AW formulation. (ii) Evaluate both types of B/C ratio with FW formulation. (iii) Find ERR.
- 10Evaluate both types of B/C ratio using PW: Initial cost = Rs. 25 lakh; Salvage value = Rs. 5 lakh; Useful life = 10 years; Annual benefits = Rs. 10 lakh; Annual O&M = Rs. 5 lakh; MARR = 8%.
3.4 Rate of Return Methods (IRR and ERR)
- 1If you invest Rs. 10,000 and Rs. 8,000 at the beginning of first and second year respectively and you earn Rs. 6,000 from second year and earning increases 6% each year up to tenth year. Find the ERR if external reinvestment rate is 15%. Is the project a sound one if firm's MARR is 20%? Explain.
- 2Explain with suitable example how inconsistent ranking occurs in IRR method.
- 3Find IRR of the following project with initial investment of Rs. 500,000 and Salvage value of Rs. 100,000. The benefit and annual O&M cost are given below. Also draw the investment balance diagram. [EOY 1–5: Benefit 105,000–145,000 (+10,000/yr); O&M 5,000–25,000 (+5,000/yr)]
- 4Evaluate IRR (FW formulation) using linear interpolation. MARR = 10%. Also draw UIB diagram in table and graph. [EOY 0–5: Cash inflow –, 500, 560, 520, 580, 540; Cash outflow 1000, 100, 200, 200, 300, 400]
- 5Explain the limitations of IRR with suitable examples. Compute ERR for a project with cash flows: [EOY 0–6: -3,00,000; 1,50,000; 2,00,000; -1,00,000; 2,00,000; 1,50,000; -50,000]. Take MARR = 12%, ε = 15%.
- 6Calculate both IRR and ERR. MARR = ε = 12%. [EOY 0: -45,000; 1: -4,000; 2: +9,000; 3: +40,000; 4: +60,000; 5: +10,000]
- 7Calculate the ERR of the following cash flow. MARR = 12%, reinvestment rate = 14%. [EOY 0: -100,000; 1: 25,000; 2: 40,000; 3: -10,000; 4: 50,000; 5: 50,000]
- 8Use IRR method to evaluate the following project when MARR is 15%. Make also unrecovered balance graph. [EOY 0: -60,000; 1: 20,000; 2: 40,000; 3: -40,000; 4: 50,000; 5: 70,000]
- 9Define IRR. Find IRR and ERR of the following project. MARR = ε = 15%. [Year 0: -50; 1: -10; 3: 30; 4: 40; 5: 50]
- 10Compute IRR by using trial and error process of the following project. Determine also investment decision. [Initial investment = 25,000; Annual revenue = 8,000; Salvage value = 5,000; Useful life = 5 years; MARR = 20%]
- 11Use IRR method to evaluate the following project when MARR is 20%. [EOY 0–5 cash flows: -60,000; 20,000; 40,000; 50,000; 50,000; 70,000]
- 12Find the IRR of the following cash flow of a project. If MARR = 20%, comment on the acceptability of the project. Show investment balance diagram. [EOY 0: -20000; 1: +8000; 2: +17000; 3: +19000; 4: +18000; 5: -10000]
- 13Select the best project using IRR method. Useful life of all projects = 15 years. MARR = 10%. [Projects A, B, C with initial investments 7,500,000 / 5,500,000 / 4,000,000; Annual revenues 960,000 / 720,000 / 600,000; Salvage values equal to initial investments]
- 14Use ERR method to evaluate the project with following cash flow. MARR = ε = 10%. [Year 1–6 cash flows: 2,00,000; 2,00,000; -50,000; 4,00,000; 4,00,000 (initial outflow -8,00,000)]
- 15Select the best project using ERR method. MARR = 18%, E = 12%. [Projects A & B with cash flows over 5 years]
- 16Find IRR and show the unrecovered investment balance in the graphical and tabular form. [Investment (First) Cost = Rs. 2,50,000; Revenues/Year = Rs. 1,00,000; Expenses/Year = Rs. 30,000; Salvage Value = Rs. 50,000; Useful life = 5 years]
3.5 Public Sector (B/C Ratio) & Financial vs Economic Analysis
- 1If you planned to invest in a project which has stated following information: First Cost = Rs 2 lakhs; Project's Life = 4 years; Salvage value = Rs 50 thousands; Gross revenue = Rs 1 lakh; O&M = Rs 30 thousands. Draw your decision based on discounted payback period method and modified benefit cost ratio. MARR = 14%.
- 2Distinguish between financial and economic analysis.
- 3Differentiate between Financial and Economic analysis.
- 4What do you mean by financial and economic analysis? Briefly explain the concept of lifecycle costing.
- 5Explain any two drawbacks of IRR with example. Differentiate between Economic analysis and financial analysis.
- 6A preliminary estimate of a multipurpose hydropower project produced the following data: Initial investment = Rs. 50 crore; Annual power sales = Rs. 8 crore; Annual irrigation benefit = Rs. 1 crore; Annual recreational benefit = Rs. 2 Crore; Annual operation and maintenance = Rs. 1.5 crore; Life of project = 50 yrs; Salvage value = Rs. 40 crore. Give your suggestion to the government about the implementation of the project. Take MARR = 8%.
Chapter 4: Comparative Analysis of Alternatives
Same/Different Useful Lives · Repeatability · Co-terminated · Capitalized Worth · Mutually Exclusive / Contingent / Independent Projects4.1 Comparing with Same Useful Life (B/C Ratio, IRR, PW, AW, ERR)
- 1Recommend the best project from the following mutually exclusive projects using modified B/C ratio. MARR = 15% per year. [Projects A, B, C: Capital investment 40,000 / 60,000 / 80,000; Annual revenues 3,000 / 3,500 / 4,000; Annual expenses 300 / 400 / 500; Useful life 5 years; Salvage value 400 / 700 / 900]
- 2Use the modified B/C ratio method with AW formulation to select the preferred design from the following mutually exclusive projects. Take MARR = 9% per year and analysis period of 15 years each. [Designs 1, 2, 3 with different investment, salvage, O&M, and benefit values]
- 3You are planning to invest in a project for 7 years. Based on the following information, which option would you prefer and why? Take MARR = 11%. [Project A: Investment Rs. 100,000, Revenue Rs. 25,000, Life 10 yrs; Project B: Rs. 120,000, Rs. 17,000, 7 yrs; Project C: Rs. 150,000, Rs. 18,000, 5 yrs]
- 4Find both types of B/C ratio using FW formulation from the following cash flow of a project. Initial investment = Rs 5,00,000; Revenue = Rs 5,0000 in first year and increases by 15000 each year after that; Expenses = 30000 in first year and increase by 5% each year after that. Salvage value at end of 8 years = 25000. MARR = 8%.
- 5From the following information select the best project. [Projects A & B: periods 4 and 8 years; service period required: (i) 4 years by FW method; (ii) 8 years by IRR method with PW formulation; MARR = 10%]
- 6Three mutually exclusive alternatives are to be compared by rate of return method. MARR is 10%. Salvage value is 20% of first cost. [X: first cost 70,000; Y: 60,000; Z: 100,000; annual income 15000 / 10000 / 18000; life 8 years each]. Which option has the highest IRR and recommend best alternative.
- 7Use IRR method to select best project. MARR = 12%. [A: 1100 investment, 500 annual income, 4 yr life, SV 250; B: 1500, 700, 4 yr, SV 500; C: 2750, 1200, 4 yr, SV 800; D: 2000, 950, 4 yr, SV 1000]. Select best combination if A, B, C are mutually exclusive.
4.2 Comparing with Different Useful Lives (Repeatability / Co-terminated / Capitalized Worth)
- 1Explain the techniques for comparing mutually exclusive alternatives having unequal useful lives.
- 2Define mutually exclusive, independent and contingent projects. What are the conditions to apply capitalized worth method? Why do we need incremental analysis?
- 3How much should you deposit now in an account which gives 8% interest per year if you wish to draw Rs. 5,000 per month + Rs 100,000 each year + Rs 300,000 in every 4 year for infinite period?
- 4Consider the following three sets of mutually exclusive alternatives. Which project would you select based on BCR when MARR = 15%? [Projects A, B, C with cash flows over 3 years]
- 5Use repeatability assumption to select the best project from the following three projects. [Projects A, B, C: Initial Investment 100,000 / 200,000 / 300,000; Annual Expenditure 25,000 / 20,000 / 15,000; Life 3/5/6 years; Salvage Value 40,000 / 50,000 / 60,000; MARR 14%]
- 6Illustrate with example, why incremental analysis is needed and how it can be performed.
- 7Based on following information select the best alternative using ERR method. [Alternative X: Investment Rs. 10,000, Revenue Rs. 5,000; Alternative Y: Rs. 15,000, Rs. 8,000; Life = 5 years; MARR = 10%, Reinvestment rate = 12%; Salvage value = 12% of Investment; O&M = Rs. 1,500]
- 8Recommend the best project from the following projects using repeatability assumption. Assume MARR = 10% per year. [Projects A, B, C: Investment 500,000 / 700,000 / 900,000; Annual Revenue 175,000 / 250,000 / 325,000; Annual cost 25,000 / 40,000 / 60,000; Salvage Value 50,000 / 70,000 / 90,000; Useful life 6/8/10 years]
- 9Select the best alternative using ERR method. [Alternative X: Investment Rs. 10,000, Revenue Rs. 5,000; Alternative Y: Rs. 12,000, Rs. 7,000; Alternative Z: Rs. 15,000, Rs. 8,000; Life = 5 years; MARR = 10%, Reinvestment rate = 12%]
- 10Recommend the best project from the following two projects taking study period as 6 years. Assume MARR = 10% per year. [Project A: Rs. 350,000 investment, Rs. 130,000 revenue, Rs. 15,000 cost, SV 35,000, 5 yr life; Project B: Rs. 500,000, Rs. 175,000, Rs. 25,000, SV 50,000, 8 yr life]
- 11Compare following two projects by IRR method when i = 10% per year. [Project A: Initial Cost 5,00,000; Annual revenue 2,00,000; Annual cost 50,000; Salvage value 80,000; Life 7 yrs; Project B: 7,00,000; 3,00,000; 1,00,000; 1,50,000; 7 yrs]
- 12Select the best project by using repeatability assumption when MARR = 13%. [Project X: Initial cost 4,00,000; Revenue 1,75,000; O&M 50,000; Life 4 yrs; SV 1,00,000; Project Y: 7,00,000; 2,50,000; 70,000; 6 yrs; 1,50,000]
- 13Consider the following two mutually exclusive alternatives; recommend the best alternatives using repeatability assumptions. MARR = 15%. [Project X: Initial Cost 100,000; Annual Cost 25,000; Salvage Value 40,000; Life 6 years; Project B: 150,000; 12,000; 50,000; 10 years]
- 15Compare the following two mutually exclusive projects by using (i) Co-terminated (ii) Repeatability assumption. MARR = 8%. [Project A: Initial cost 1,50,000; Annual revenue 90,000; Operating cost 20,000; Life 4 yrs; SV 80,000; Project B: 2,00,000; 1,00,000; 20,000; 6 yrs; 1,20,000]
- 16Use Repeatability assumption to select the best project from the following three projects. [Projects A, B, C: Initial Investment 100000/200000/250000; Annual Expenditure 25000/20000/15000; Life 3/5/7 years; SV 40000/50000/60000; MARR 14%]
- 17State and explain about the cases of mutually exclusive, contingent and independent projects with example. Compare the following projects by using repeatability assumption when MARR is 12%. [Project A: 2,00,000 investment; 25,000 revenue; 7,000 cost; 6 yr life; SV 10,000; Project B: 3,00,000; 30,000; 9,000; 8 yr; SV 20,000]
4.3 Mutually Exclusive, Contingent & Independent Projects in Combination
- 1Five proposed projects being considered by an engineer in an integrated system. Project A1 and A2 are mutually exclusive and independent on B set. Project B1 and B2 are mutually exclusive and contingent on acceptance of A2. Project C is contingent on B1. Assume MARR = 8% per year, useful life 4 years. Determine best combination if capital is: (i) Unlimited; (ii) Limited to 48,000. [A1: 50,000 inv., 20,000 benefit; A2: 30,000, 12,000; B1: 14,000, 4,000; B2: 15,000, 5,000; C: 10,000, 6,000]
- 2Explain Capitalized Worth. Four projects being considered: Recommend which investment alternative should be selected? MARR = 10%. [Projects A, B, C, D: B & C are mutually exclusive; D contingent on C; A contingent on B. Various initial investments and revenues over 5–8 year lives]
- 3Nepal government is planning to invest in three irrigation projects. Compare mutually exclusive projects to invest. [Koshi, Gandaki, Karnali: initial cost 20,000/22,000/24,000 billion; annual benefit 4,000/4,500/5,000; annual cost 1,000/1,200/1,400; MARR = 10%, life 20 years]
- 4Define independent and contingent projects. Find Present worth from annual cash flow series of Rs. 5,000 forever when i = 8% per year.
- 5Define mutually exclusive, independent and contingent projects. How much should you deposit at present that earns 12% interest per year when you can draw Rs 10,000 per month for (i) 50 years (ii) Forever?
- 6Choose the best project among these alternatives using IRR, if MARR = 15% and study period is 10 years. Salvage value is 20%. [Projects A, B, C, D: First Cost Rs. 900/1500/2500/4000; Annual Revenue Rs. 150/276/400/925]
- 7Define mutually exclusive, contingent and independent projects with suitable example.
- 8Select the best combination of the project where A is independent and B is contingent on C. [Projects A, B, C with investment 40,000/70,000/50,000; revenue 15,000/20,000/20,000; annual cost 2,500/3,500/0; life 8 years each; investment limited to Rs. 120,000; MARR = 10%]
- 9Why is incremental analysis essential during comparative analysis of alternatives using BCR, IRR and ERR? Illustrate with help of example. Compare the following projects using IRR method MARR = 14%. [4 projects A, B, C, D with investment 20,00,000–40,00,000 over 8 years]
Chapter 5: Replacement Analysis
Defender vs Challenger · Economic Service Life · Infinite & Finite Planning Horizon · Cash Flow vs Opportunity Cost Approach5.1 Fundamentals: Concepts, Terminology & Approaches
- 1What do you mean by replacement analysis? List out major causes of replacement. Distinguish between cash flow approach and opportunity cost approach in replacement.
- 2Define defender, challenger and explain cash flow approach and opportunity cost approach of comparing. Why is replacement of existing equipment necessary?
- 3Explain the techniques for comparing mutually exclusive alternatives having unequal useful lives. [Also covers replacement approaches]
- 4What is replacement analysis? What factors should be considered in replacement analysis? Explain the cash flow approach and opportunity cost approach.
- 5What do you mean by replacement analysis and economic service life? What are the procedures for replacement when planning horizon is infinite and finite?
- 6Define replacement. Explain the main reasons for replacement. Find economic service life from the given data.
- 7What are the procedures for replacement analysis when planning horizon is infinite? Find economic service life from the following information: Initial cost = Rs 50,000; Operation cost = Rs 10,000 for 1st year, increases by 15% thereafter; Salvage value = Decline each successive year by 20%; Useful life = 8 years; MARR = 15%.
- 8Why is replacement analysis necessary in economic analysis? How do you make replacement analysis when planning horizon is finite?
- 9Explain about the Sunk Cost, Economic life and reasons for replacement of an asset.
- 10Write down the reasons of replacement of an existing asset.
- 11Define defender and challenger and explain economic service life. Company X is going to purchase a router having initial cost Rs. 18,000 having salvage value of Rs. 12,000 at end of first year and decreases by 20% each year. Annual O&M cost is Rs. 5000 in first year and increases by Rs. 2000 each year. Its useful life is 6 years. Calculate economic service life of the router.
5.2 Economic Service Life of Defender and Challenger
- 1Determine the choice between defender and challenger when current market value of defender is Rs. 5,000 and a challenger can be purchased for Rs. 7,500. Both have 3 years service life & MARR is 12%. Operating costs: Defender year 1/2/3 = 1,700/2,000/2,500; Challenger = 500/1,100/1,300. Use both cash flow and opportunity cost approach.
- 2Find economic service life of the following asset: Initial investment = Rs. 50,000; Operating cost = Rs. 10,000 for first year and increases by 15% thereafter; Salvage value decreases each year by 20% from previous year's value; Useful life = 7 years; MARR = 15%.
- 3One of four ovens at a bakery is being considered for replacement. Its salvage value and maintenance costs are given for several years. A new oven costs Rs 80,000 (includes guarantee of maintenance costs for first two years). MARR = 12%. Should the oven be replaced this year? [Detailed table with defender/challenger SV and maintenance costs for 4 years]
- 4Explain the replacement analysis under infinite planning horizon.
- 5A 5-year old defender has a current market value of Rs 40,000, O&M = Rs 30,000 this year, increasing by Rs 15,000 per year. Future market values decline by Rs 10,000 per year. Challenger costs Rs 60,000 and has O&M of Rs 20,000 per year, increasing by Rs 10,000 each year. SV at end of 5 years = Rs 20,000. Calculate the economic service life of both defender and challenger.
- 6A new machine costs $20,000. Future market values are expected to decrease by $2000 each year over previous year's value. Useful life is 6 years. Operating costs are estimated at $3000 during first year and expected to increase 15% per year thereafter. MARR = 12%. Determine the economic service life of the machine.
- 7Explain the replacement analysis under infinite planning horizon.
- 8Calculate AECs from the following information and determine economic service life: I = 18,000; N = 8 years; O and M = 3,000 for 1st year and increases by 15% thereafter; S = Decline by 20% each successive year over previous price; MARR = 12% per year.
- 9Define economic service life of an asset. From the following information find the economic service life: Initial investment = 50,000; Annual operating cost = 10,000 for 1st year increasing by 15%; Salvage value declining each year by 20% from previous year's value; Useful life = 7 years; MARR = 15%.
5.3 Replacement Strategy (Finite & Infinite Planning Horizon)
- 1The cash flow of two mutually exclusive projects are as follows: [Alt. A: Capital invested Rs. 20,000; Annual expenses Rs. 5,500; Salvage value Rs. 1,000; Useful life 5 yrs; Alt. B: Rs. 38,000; Rs. 4,000; Rs. 4,200; 10 yrs]. Which alternative should be selected if (a) study period is indefinite? (b) study period is 5 years? MARR = 20%.
- 2Consider the annual cost of defender and challenger and determine the best replacement strategy if a firm has a contract to perform a given service for next 8 years. MARR = 15%. [Defender EOY 1–6: 53,800; 52,030; 54,690; 58,440; 62,580; 66,820; Challenger EOY 1–8: 61,840; 57,560; 53,800; 52,500; 61,840; 65,200; 66,000; 67,300]
- 3Suppose that the firm has a contract to perform a given service on which current defender or challenger is considered for 8 years. Determine the best replacement strategy. [Defender EOY 1–5 AEC: -3078; -3070; -3300; -3576; -3860; Challenger EOY 1–6: -5100; -4291; -4094; -4065; -4110; -4189]
- 4The AEC of defender and challenger are given in the table below. What is the best replacement strategy? Use MARR = 10%. Planning horizon = 8 years. [Defender n=1–6: 5400/5200/5500/5700/6200/6600; Challenger n=1–6: 7700/6200/5700/5500/5680/5900]
- 5Explain about the reasons for replacement of asset. The AEC of defender and challenger are given. What is the best replacement strategy? Use MARR = 12%. Planning horizon = 8 years. [Defender n=1–6: 5300/5250/5400/5750/6200/5550; Challenger: 7700/6150/5700/5600/5675/5800]
- 6The new machine costs 10,000, operating cost 2200 in first year, then increases by 20% per year. Market value is 6000 after one year and will decline by 15% each year. N = 5 years. Old machine: market value now 5000, declines 25% each year; overhauling cost 1200; operating costs 2000 in first year, increases by 1500 per year. MARR = 15%. (i) Find economic service life of new machine; (ii) AEC of defender given in table — when should old machine be replaced?
- 7A chemical plant owns a filter press that was bought 3 years ago for Rs. 30,000. Now it has a market value of Rs. 9,000, a life of 5 years, and salvage value of Rs. 2000 at that time. Challenger has a cost of Rs. 36,000, a life of 5 years, and estimated market value Rs. 12,000 after 5 years. 5 years planning period, MARR = 15%. Operating and maintenance costs given. Should the old filter press be replaced now? Use cash flow approach.
- 8A firm has a contract to provide printing service to IOE for next 8 years. It can use its old machine (current defender) or the newly bought machine (challenger). Considering annual equivalent costs of old and new machine — what are their economic service lives and what is the best replacement strategy? [Old machine n=1–5 AEC: 515,000/510,000/550,000/596,000/644,000; New machine: 750,000/615,000/586,000/583,000/590,000]
Chapter 6: Risk Analysis
Sources of Risk · Sensitivity Analysis · Breakeven Analysis · Scenario Analysis · Decision Tree6.1–6.2 Methods of Describing Risk & Sensitivity Analysis
- 1Explain the Decision tree and its components.
- 2Perform sensitivity analysis of the following project (with an increment of 10%) over the range of (+/–) 40% in (i) initial investment, (ii) net annual revenue, (iii) useful life. Use PW formulation. Draw a sensitivity graph and mention the most sensitive parameter. Given: Initial investment = Rs. 4,50,000; Net annual revenue = Rs. 1,10,000; Salvage value = Rs. 75,000; MARR = 12%.
- 3List out methods of describing risk. Explain Decision tree analysis.
- 4Perform sensitivity analysis over a range of ±30% (10% interval) for the following parameters with given information below: (i) Investment, (ii) Net Annual Revenue, (iii) Life, (iv) MARR. Investment = Rs 10,000; Net Annual Revenue = Rs 3,000; Salvage Value = Rs 1,000 and Life = 4 years. Use MARR = 10% and IRR method. Draw also sensitivity diagram and interpret it.
- 5Perform sensitivity analysis using PW formulation over a range of ±40% in (i) initial investment (ii) Annual revenue (iii) Useful life and (iv) MARR. A project costs Rs. 1,25,000 with annual revenue of Rs. 65,000 and annual cost of Rs. 35,000. Salvage value will be 20% of initial investment. Draw sensitivity diagram and interpret it.
- 6Define project risk. What are the basic methods for describing project risk? A newly established mask company has estimated: Output per annum = 3,00,000 units; Expected sales revenue = Rs. 1,50,00,000; Fixed cost = Rs. 35,00,000; Variable cost = Rs. 66,00,000. (i) Find BEP level of output. (ii) If fixed cost increases to Rs. 40,00,000, find its effect on BEP. (iii) What should be output if profit desired is Rs. 10,00,000 per year?
- 7Perform a sensitivity analysis over a range of ±15% with 5% increment using AW formulation for (i) Initial Investment, (ii) Net annual income, (iii) Useful life, (iv) MARR. Given: Initial cost = Rs 125,000; SV = Rs 25,000; Annual Income = Rs 60,000; Annual Cost = Rs 20,000; Useful Life = 6 Years; MARR = 8%. Which parameter is the most sensitive?
- 8Perform sensitivity analysis over 15% in (i) MARR and (ii) Useful life. Draw sensitivity chart and interpret the result. [First Cost $25,000; Annual Revenue $12,000; Annual Expenses $4,000; Salvage Value $5,000; Useful Life 5 years; MARR 10%]
- 9Perform sensitivity analysis using PW formulation over a range of ±40% in (i) Initial Investment, (ii) Annual Revenue, (iii) Useful Life, (iv) MARR. A project costs Rs. 1,25,000 with annual revenue of Rs. 65,000 and annual cost of Rs. 35,000. Salvage value will be 28% of initial investment. Draw sensitivity diagram and interpret it.
- 10Perform sensitivity analysis (over range of ±15% with 5% increment) with using IRR (AW formulation) among the parameters (i) useful life, (ii) initial Investment, (iii) Revenues. Given: I = Rs 1,00,000; Sv = Rs 22,000; O&M = Rs 12,000; Revenues = Rs 40,000; Life = 6 years; MARR = 10%. Which parameter is more sensitive?
- 11Perform sensitivity analysis over a range of ±30% in (i) Initial Investment, (ii) Annual Revenue, (iii) Life year. Initial Cost = Rs. 5,00,000; Annual revenue = Rs. 1,20,000; SV = Rs. 80,000; Life = 6 years; MARR = 10% per year. Draw sensitivity graph.
- 12Explain about the decision tree analysis. Perform sensitivity analysis of the following project over range of ±30% at an interval of ±10% in (i) Initial Investment, (ii) Net Annual Revenue, and (iii) Useful life. Use PW formulation. [Initial Investment = Rs 1,00,000; Net Annual Revenue = Rs 40,000; SV = Rs 15,000; Useful life = 6 years; MARR = 10%]
- 13A project costs Rs. 125,000 with annual revenue of Rs. 65,000 and annual cost of Rs. 35,000. Salvage value will be 8% of initial investment. Perform sensitivity analysis using PW formulation over a range of ±40% in (i) Initial Investment, (ii) Annual Revenue, (iii) Useful Life, (iv) MARR. Indicate most sensitive and least sensitive parameters.
- 14Perform sensitivity analysis of the following project over a range of ±30% in (i) initial investment, (ii) net annual revenue, (iii) useful life year. Draw sensitivity diagram. [Initial investment = Rs. 5,00,000; Net annual revenue = Rs. 1,20,000; Salvage value = Rs. 80,000; Useful life = 6 years; MARR = 10%]
- 15Perform sensitivity analysis over a range of ±30% in (i) initial investment, (ii) annual net revenue, (iii) useful life. Initial investment = Rs. 100,000; Salvage value = Rs 10,000; Annual benefits = Rs 25,000; Annual expenses = Rs 3,000; Useful life = 10 years; MARR = 10%. Draw sensitivity diagram.
- 16Enunciate different methods of analyzing the riskiness of the project. Perform sensitivity analysis to identify the most sensitive parameter over the range of ±15% (interval of 5%) for (i) Net Annual Revenues, (ii) Salvage Value, (iii) Life Span. [Investment = 80,000; Net Annual Revenues = 25,000; Salvage Value = 10,000; Life = 12 years; IRR method]
- 17What are the reasons behind the alternative projects being mutually exclusive to each other? Explain with suitable examples.
6.2.2 Breakeven Analysis
- 1A newly established mask company has estimated: Output per annum = 3,00,000 units; Expected sales revenue = Rs. 1,50,00,000; Fixed cost = Rs. 35,00,000; Variable cost = Rs. 66,00,000. (i) Find BEP level of output. (ii) If fixed cost increases to Rs. 40,00,000, find its effect on BEP. (iii) What should be output if profit is desired Rs. 10,00,000 per year?
- 2The details of production cost and revenue of a project: Total cost = Rs 85,000; Fixed cost = Rs 40,000; Sales = Rs 105,000; Sales volume = Rs 15,000. (i) Find BEP in terms of number of units. (ii) What should be output if profit desired is Rs 50,000? (iii) Find BEP if selling price decreases by 10%.
- 3Define breakeven point and breakeven volume. How does interest rate change affect the project?
- 4Calculate break-even hours of operation per year to become cost equal and recommend economic pump if operated 5 hours daily at full load. [KHASA Pump: 100 hp, cost 5,00,000, efficiency 80%; SARVO Pump: 100 hp, cost 10,00,000, efficiency 90%; Life 5 years; MARR 20%; electricity Rs. 10/kwhr]
- 5Fixed cost = Rs. 60 million; Variable cost/unit = Rs. 50,000; Selling price/unit = Rs. 8,000. Find BEP volume. What would be the effect on profit/loss when Sp increases by 20%?
6.4 Decision Tree and Scenario Analysis
- 1A firm plans to install its manufacturing unit either in Butwal or Birganj. Further added cost for advertisement: Rs 200,000 (Butwal) and Rs 250,000 (Birganj). Where should the firm target to install? [Butwal: High P=0.2, Medium P=0.5, Low P=0.3 with incomes 400,000/300,000/150,000; Birganj: P=0.3/0.4/0.3 with 500,000/350,000/100,000]
- 2Explain the decision tree analysis. What would be the effect on BEP and profit/loss when selling price increases by 20% and fixed cost decreases by 10%? [Variable Cost per unit = $50; Fixed cost = $60,000; Selling price per unit = $80]
- 3Explain the concept of scenario and decision tree analysis. If 20 watt CFL bulb price is Rs. 280 and 100 watt filament bulb price is Rs. 30 at market but their lighting power is equal. Which bulb do you prefer to use in your house when electricity cost is Rs. 12 per unit?
- 4For the improvement of a manufacturing plant, following three alternatives are being considered. Draw decision tree diagram and decide on the best alternative using FW formulation. MARR = 15%. Life = 6 years. [Alternatives A, B, C with investment 1,000,000/600,000/400,000 and different success/probability/annual income scenarios]
- 5If the cost of 25 watt CFL bulb is Rs. 260 whereas the cost of 100 watt Filament bulb is Rs. 35 but these bulbs have equal lighting power. Which bulb do you prefer in your use and why? When electricity cost is Rs. 11 per unit (kw-hr).
- 6A real estate developer seeks to determine the most economical height for a new office building (sold after 5 years). Relevant net annual revenues and salvage values for 2, 3, 4, and 5 floors given. The developer is uncertain about interest rate but certain it is in range 5%–30%. For each building height, find the range of values of interest rate for which it is most economical.
Chapter 7: Depreciation and Corporate Income Taxes
SL · DDB · SOYD · Sinking Fund · MACRS · After-Tax Cash Flow Analysis7.1–7.2 Depreciation Methods (SL, DDB, SOYD, Sinking Fund, MACRS)
- 1An asset with cost basis of Rs. 10,000 has salvage value of Rs. 2,000 at the end of 5 years. Calculate the depreciation amount for each year by SOYD method. If an asset generates uniform annual revenue of Rs. 3,000 and tax rate is 25%, calculate the IRR of after tax cash flow.
- 2Explain briefly advantages of depreciation. What are the major five types of depreciation? Compute depreciation of each year using: (i) Declining Balance Method with switchover to Straight Line; (ii) SOYD Method; (iii) Sinking Fund Method. [Cost basis Rs. 7,000; Salvage value Rs. 2,000; Useful life 5 years; MARR 10%]
- 3Define tax. Explain different types of Taxes.
- 4Prepare a depreciation and book values table for an asset having cost basis of Rs 100,000, salvage value of Rs 5,000, useful life of 4 years, MARR = 10% using: (i) DDB with switchover to straight line; (ii) SOYD; (iii) Sinking fund method.
- 5Calculate after tax NPV of the following cash flow if initial investment is 75,000. Use MACRS method of depreciation assuming 3 year property. Tax rate is 25%. MARR = 15%. [Year 0–4 before tax cash flows: -75,000; 20,000; 25,000; 30,000; 40,000]
- 6Prepare a schedule of charging depreciation and carry out the book value for each year of a machinery having first cost of Rs. 80,000 with salvage value of Rs 10,000 after 5 years using: (i) DDB Method, (ii) SOYD method, (iii) MACRS method.
- 7You purchased a machine at a cost of Rs 320,000 having useful life of 8 years. The estimated salvage value is Rs 50,000. For book purpose straight line method is used, and for tax purpose the machine is depreciated using MACRS over its 5 year class life. Determine the annual depreciation allowance and resulting book value over life for both book and tax purposes.
- 8Using the IRR method, recommend the best project from the following set of mutually exclusive projects with 10-year useful life for all alternatives. Assume MARR = 10%. [Projects A, B, C with different initial investments, annual revenues, salvage values, and annual operating costs]
- 9Find economic service life of an asset. Initial investment = $1800; Operating cost $300 in first year; for remaining years, Operating cost increases by 15% over previous year's OC; salvage value declines each year by 20% from previous year's salvage value. Maximum life = 8 years. MARR = 12% before tax.
- 10Explain the general procedure for after tax economic analysis with suitable examples.
- 11Compute annual depreciation and book value using: (i) Declining Balance Method, (ii) SOYD Method, (iii) Sinking Fund Method. [Cost Basis $7,000; Salvage Value $2,000; Useful Life 5 Years; MARR 10%]
- 12Explain the terms depreciation, corporate tax, personal income tax and book value. Show the depreciations and book values in each year for an equipment using MACRS method. [Investment = 25,00,000; Useful life = 7 years]
- 13Define depreciation. What are the advantages of depreciation concept? Calculate depreciation for each year using declining balance and MACRS. [Vehicle cost Rs. 4,00,000; 5 years useful life; SV Rs. 1,00,000]
- 14Compute the annual depreciation allowances and resulting book value using the double declining balance method with switchover to straight line method. [Cost of asset = Rs. 100,000; Useful life = 5 years; Salvage Value = 20,000]
- 15Write down the causes for depreciation of assets. If a machine costing Rs. 1,00,000 is purchased by expecting salvage value of Rs 20,000 at the end of 6th years, calculate the depreciation amount for each year by SOYD and straight line method.
- 16An asset has installed value of 45,000. Ss = 0. It is classed as a 5 year property. Determine approximate MACRS depreciation schedule. Over 6 years it is estimated to generate revenue of Rs. 23,000 per year with annual operating cost 7,300. Required rate of return = 15% after tax. Tax rate = 40%. Evaluate after tax IRR with annual worth method.
- 17Define depreciation. What are the causes for it? If a machine costing Rs. 1,50,000 is purchased by expecting salvage value Rs. 40,000 at end of 6th year, calculate depreciation amount for each year by (i) SOYD; (ii) Declining balance.
- 18List out advantages of depreciation. Calculate depreciation amount for each year using declining balance and MACRS methods. [I = 10,00,000; N = 5 years; S = 2,00,000]
- 19Define depreciation and list out important methods of calculating depreciation deductions.
- 20A machine costs Rs 15,000. Its useful life is 5 years and salvage valve is Rs 900. Compute the annual depreciation allowances and resulting book values using double declining balance depreciation methods.
- 21You have purchased a machine at a cost of Rs 1,00,000. Machine has useful life of 8 years with estimated salvage value of Rs 20,000. The sinking fund method depreciation is used for book purposes. For tax purposes, the machine would be depreciated using MACRS over its 7-year class life. Determine the annual depreciation amount over useful life for both book and tax purposes. Assume i = 12%.
- 22Prepare a schedule of charging depreciation and carry out the book value for each year of a machinery having first cost of Rs. 80,000 with salvage value of Rs 10,000 after 5 years using (i) DDB Method (ii) SOYD method (iii) MACRS method. Also, find after tax net present value based on above calculated SOYD depreciation. Net annual revenues = Rs 35,000 for 5 years; 25% tax rate; MARR = 15%.
7.4–7.5 After-Tax Cash Flow and Economic Analysis
- 1Calculate after tax NPV of the following cash flow if initial Investment is 75,000. Use MACRS method of depreciation assuming 3 year property. Tax rate is 25%. MARR = 15%.
- 2A company requires a machine which costs Rs 50,000 and has a salvage value of Rs 10,000 after 5 years and expected to save the annual benefit of Rs 16,000. The Rs 10,000 after 5 years. Calculate the after tax cash flow of a machine if tax rate applicable is 30%. Depreciation is on SOYD.
- 3Define direct tax and indirect tax. Prepare after tax cash flow of a project having following cash flow details: Initial cost = 1,00,000; Useful life = 5 years; Salvage value = 20,000; Annual income = Rs. 20,000 at end of first year and increases by Rs. 2000/year; Tax rate = 20%; Depreciation method = Sum of Year Digit.
- 4Perform after tax cash flow analysis to examine feasibility of a project having investment of Rs 1,00,000 in a machine, with zero salvage value, 5 years useful life, net annual revenues of Rs 20,000 at end of first year then after increases by Rs 10,000 p.a. Use SL depreciation. Tax rate = 25%.
- 5Calculate ATCFs and determine profitability (IRR) when MARR is 15% by using PW method. [Equipment purchased for Rs. 10,00,000; expected to generate income Rs. 3,50,000 per year during 5 years; corporate income tax rate 25% per year; depreciation amounts given for years 1–5]
- 6Define direct tax and indirect tax. Prepare after tax cash flow of a project. Initial cost = 1,00,000; Useful life = 5 years; Salvage value = 20,000; Annual income = Rs. 20,000 at end of first year and increases by Rs. 2000/year; Tax rate = 20%; Depreciation method = SOYD.
- 7A machine cost Rs 20 million with no salvage value. Rs 8 million revenues per year can be gained. Useful life = 4 years. Tax rate = 50%; MARR = 10%. Use straight line depreciation method to evaluate (i) PW (ii) IRR.
- 8A company bought a machine at Rs 25,000 which is expected to produce benefit of Rs 8,000 per year for five years. Its salvage value at end of five years is Rs 10,000. Calculate after tax cash flow if Tax rate is 40% and depreciation is on Sinking Fund method. I = 20%.
Chapter 8: Inflation and Its Impact on Project Cash Flows
Concept of Inflation · Measuring Inflation (CPI) · Constant vs Actual Dollar · Deflation Method · Inflation-Free IRR8.1–8.2 Concept and Measurement of Inflation
- 1Define inflation and list out its causes. Calculate PW of the following cash flow using deflation method. Inflation free rate (i') = 10%. [EOY 0–3: cash flows in actual dollars -45,000; 32,000; 32,000; 32,000; General inflation rates –; 3.5%; 4.2%; 5.5%]
- 2Distinguish between demand-pull and cost-push inflation. What are the effects of inflation in an economy? Suppose that inflation for the first and second year is 5% and 8% respectively. Calculate the average inflation rate for two years if base price is Rs. 100.
- 3What is Actual and Constant Dollar? The cost of apple per kg is Rs 150. The cost increases to 6%, 8% and 9% per kg in first year, second year and third year respectively. Determine the average inflation rate for 3 years.
- 4What is inflation? Explain constant dollar and actual dollar. Explain deflation method of analysis using suitable example.
- 5Why is inflation important to engineers for economic analysis? Suppose that the 1st year inflation rates for first year and second year are 5% and 8% respectively. Calculate the average inflation rate of two years if base price is Rs 100.
- 6What is inflation? List out the impact of inflation. Calculate the rate of inflation when CPI moves from 100 to 250 over three years.
- 7Define constant dollar and actual dollar amount. Suppose you borrowed Rs. 1,20,000 from a bank to buy a bike and you have promised to pay Rs. 6000 per month for two years. What is the inflation free interest rate you are supposed to pay if average inflation rate is 0.75% per month.
- 8Define constant dollar amount and actual dollar amount. Suppose you borrowed Rs. 100,000 from a bank to buy a bike and you have promised to pay Rs. 5500 per month for two years. What is the inflation free interest rate you are supposed to pay if average inflation rate is 0.75% per month?
- 9Define inflation. Calculate IRR if MARR = 12% and inflation rate is 8%. [Year 0–4 Constant Dollar cash flows: -6000; 1500; 2000; 2500; 3000]
- 10Define inflation. Calculate inflation free IRR if MARR = 12% and general inflation rate is 8% per year. [Year 0–4 Constant $ cash flows: -6000; 1500; 2000; 2500; 3000]
- 11Draw the concept of inflation using suitable example. If the real interest rate is given as 5% and general inflation rate as 8%, calculate MARR. Explain meaning of constant dollar and actual dollar.
- 12A series of five constant dollar (real dollar) income beginning with $5,000 at end of first year are increasing at rate of 7% per year for five years. Inflation free interest rate = 5% and inflation = 8%. Is it feasible investment if investment cost is $20,000?
8.3–8.4 Equivalence Calculation Under Inflation (Actual/Constant Dollar Analysis)
- 1A series of four annual constant dollar payments beginning with $7,000 at the end of first year is growing at the rate of 8% per year. If the market interest rate is 13% per year and the general inflation rate is 7% per year, find the present worth of series.
- 2Calculate the equivalent present worth of the project from the following cash flow. Assume inflation free interest rate = 5% and inflation = 10% respectively. [EOY 0–5: -7,50,000; 3,20,000; 3,75,000; 3,28,000; 2,90,000; 5,80,000 in Actual $]
- 3Evaluate the PW of the following project: [Initial investment = Rs. 1,00,000 constant dollars; Annual sales income = Rs. 40,000; Annual labour cost = Rs. 3,000; Annual material X = Rs. 2,000; Annual material Y = Rs. 1,000; SV = 20% of initial investment. Inflation rates for sales income/labour/material X/material Y/SV: 5%/8%/0%/6%/3% respectively. Market interest rate = 20%; Life = 4 years]
- 4Define inflation. Calculate inflation free IRR if MARR = 12% and general inflation rate is 8% per yr. [Year 0–4 Const. $: -6000; 1500; 2000; 2500; 3000]
- 5The annual fuel cost required to operate a small solid waste treatment plant are projected to be Rs. 200,000 without considering any future inflation. Annual inflation free interest rate I' = 6% and general inflation rate f = 5%. If the plant has remaining useful life of 4 years, what is the present equivalent of its fuel costs? Use actual dollar analysis.
- 6Define inflation. What are its causes? Find rate of inflation per year when price of a product has increased from Rs. 5,00,000 to Rs. 6,30,000 over the period of 3 years.
- 7Define Actual dollar and Constant dollar. Calculate NPV if i is 6% and inflation rate is 8% of the following cash flow. [Year 0–3 Actual dollar cashflow: -1000; +500; +500; +500]
Compiled from TU IOE Engineering Economics (CE 655) exam papers: 2065–2081 | All questions matched to syllabus chapters
