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ICAR AICE-JRF/SRF Engineering Economics, Project & Business Management: practice questions

ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering subjects 38–43 · India · National Testing Agency for ICAR

Original practice questions written to the published syllabus · not past papers
10worked here, free
42written to this exam in this subject
9syllabus units

What this subject covers

Money and time in engineering decisions: discounting, costs, break-even, project appraisal, scheduling and starting a business.

In the exams: NCEES FE Engineering Economics 6–9 questions; NCEES PE engineering economics and project management; ICAR AICE subject 41 Unit 10; PRC ABE project management

What ICAR AICE-JRF/SRF asks from it

ICAR AICE-JRF/SRF (PhD) — Agricultural Engineering (subject 41 unit 10)

Project planning and management for processing plants: DPR structure, financial indicators (NPV, IRR, B/C, payback, break-even capacity, DSCR), CPM and PERT computations, and linear programming formulation and duality.

Where the 42 questions fall

How our questions for ICAR AICE-JRF/SRF spread across the units of Engineering Economics, Project & Business Management. Difficulty: 9 easy, 16 medium, 17 hard.

Linear programming11
Project appraisal and detailed project reports10
CPM and PERT9
NPV, IRR, benefit-cost ratio and payback3
Break-even analysis3
Replacement analysis3
Time value of money1
Depreciation methods1
Fixed and variable costs1

10 practice questions, worked

Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.

1 · Replacement analysis · medium

A defender's marginal cost of keeping it one more year is 140,000 and rising each year. The best challenger has a minimum equivalent annual cost of 150,000 at its economic life. What should be done?

  1. Replace the defender now, because its marginal cost is rising every year
  2. Keep the defender until it is fully depreciated in the firm's accounts
  3. Keep the defender for at least one more year and repeat the comparison next year
  4. Replace the defender now, because the challenger is a newer design
Answer and working

C. Keep the defender for at least one more year and repeat the comparison next year

The defender is cheaper for the coming year (140,000 < 150,000), so keeping it saves money now. Because its marginal cost is rising, it will soon exceed 150,000, and the comparison should be repeated each year. Book depreciation and novelty are not decision criteria.

2 · Replacement analysis · hard

A processing plant has 500 lamps that fail at random. Replacing a lamp individually after failure costs much more per lamp than replacing all lamps together. Under the standard group-replacement rule, when should all lamps be replaced?

  1. Only once more than half the lamps have failed, because replacing a whole group earlier throws away good lamps
  2. Never, because replacing lamps that still work cannot reduce the total cost of lighting
  3. Every period, because the cost of replacing all lamps together is lower per lamp
  4. At the end of period t if individual replacements in period t + 1 would cost more than the average cost per period up to t
Answer and working

D. At the end of period t if individual replacements in period t + 1 would cost more than the average cost per period up to t

Group replacement trades away the remaining life of working lamps for a lower cost per lamp and fewer emergency replacements. The best interval minimises average cost per period; the rule stated finds that interval. Replacing every period ignores that working lamps are discarded too often.

3 · Project appraisal and detailed project reports · medium

In the economic (as opposed to financial) analysis of an irrigation project, how is a government subsidy on pump sets treated?

  1. It is added to the project's benefits, because it raises the farmers' net income
  2. It is deducted from the costs and kept, because it is cash the project receives
  3. It is excluded as a transfer, and pump sets are valued at their economic cost
  4. It is counted twice, once as a cost to government and once as a farmer benefit
Answer and working

C. It is excluded as a transfer, and pump sets are valued at their economic cost

Taxes, subsidies and loan transactions move purchasing power between groups without using real resources, so economic analysis removes them and values inputs at their economic or shadow prices. The financial analysis, from the farmer's viewpoint, does include the subsidy.

4 · Project appraisal and detailed project reports · medium

Which item is normally part of the 'cost of project' in a detailed project report for a food-processing unit?

  1. Raw material bought in the second year of commercial operation to keep the line running
  2. Interest paid on the term loan during the fifth year of the loan's repayment schedule
  3. Margin money for working capital, the promoter's share of the working capital needed
  4. Salaries of production staff in the third year after start-up
Answer and working

C. Margin money for working capital, the promoter's share of the working capital needed

The cost of project covers land, buildings, plant and machinery, preliminary and pre-operative expenses, contingencies and margin money for working capital. Raw material, interest and salaries during operation are operating costs shown in the profitability projections, not capital cost.

5 · Project appraisal and detailed project reports · medium

A project protected by import duties shows a financial IRR of 18 % but an economic IRR of 6 %. What does this most directly suggest?

  1. That the financial analysis has an arithmetic error, because the two rates must be equal
  2. That the economic analysis is wrong, because it should always give the higher of the rates
  3. That protection makes it pay its owner while it uses resources inefficiently for society
  4. That the project should be financed at 12 %, the average of its financial and economic rates
Answer and working

C. That protection makes it pay its owner while it uses resources inefficiently for society

Financial and economic IRRs differ when market prices differ from economic values. Protection raises the price the owner receives above the border price, inflating financial returns; at border prices the project earns little. Neither rate must exceed the other in general.

6 · CPM and PERT · medium

Activities and durations in days, with predecessors: A 4 (none), B 6 (none), C 5 (after A), D 3 (after A), E 4 (after B and D), F 2 (after C and E). What is the project duration?

  1. 24 days
  2. 11 days
  3. 12 days
  4. 13 days
Answer and working

D. 13 days

Forward pass: A 0–4, B 0–6, C 4–9, D 4–7, E starts at max(6, 7) = 7 and ends 11, F starts at max(9, 11) = 11 and ends 13. 24 days adds all durations as if in series; 11 days is path A–C–F and 12 days path B–E–F, both shorter than the critical path A–D–E–F.

7 · CPM and PERT · medium

An activity has a normal time of 10 days at a cost of 50,000 and a crash time of 6 days at a cost of 70,000. What is its cost slope?

  1. 5,000 per day
  2. 2,000 per day
  3. 7,000 per day
  4. 11,667 per day
Answer and working

A. 5,000 per day

Cost slope = (crash cost − normal cost)/(normal time − crash time) = (70,000 − 50,000)/(10 − 6) = 5,000 per day. 2,000 divides by the normal time; 7,000 divides the whole crash cost by 10; 11,667 divides crash cost by 6 days.

8 · CPM and PERT · hard

To shorten a project by one day at least cost, which activity should be crashed first: critical activity P (cost slope 4,000 per day), critical activity Q (2,500 per day), critical activity R (6,000 per day) or non-critical activity S (1,000 per day)?

  1. S, the activity with the lowest cost slope of all
  2. R, the critical activity with the highest cost slope
  3. P, the critical activity with the middle cost slope
  4. Q, the critical activity with the lowest cost slope
Answer and working

D. Q, the critical activity with the lowest cost slope

Only shortening a critical activity reduces project duration; crashing S spends money and saves nothing. Among critical activities, crash the one with the lowest cost slope first, then recompute the critical path because another path may become critical.

9 · Linear programming · easy

In a transportation problem, total supply from the depots is 900 t and total demand at the markets is 750 t. What is done before solving?

  1. Add a dummy market with a demand of 150 t and zero transport cost
  2. Add a dummy depot with a supply of 150 t and zero transport cost
  3. Reduce every depot's supply in proportion until supply equals 750 t
  4. Solve as it is, because the method works with unequal totals
Answer and working

A. Add a dummy market with a demand of 150 t and zero transport cost

Supply exceeds demand by 150 t, so a dummy destination with demand 150 t and zero cost absorbs the surplus; goods 'shipped' to it are simply not shipped. A dummy source is used when demand exceeds supply.

10 · Linear programming · medium

Which method for an initial transportation solution uses penalties, the difference between the two lowest costs in each row and column?

  1. Vogel's approximation method
  2. The north-west corner rule
  3. The modified distribution (MODI) method
  4. The stepping-stone method
Answer and working

A. Vogel's approximation method

Vogel's method allocates first to the row or column with the largest penalty, which gives a starting solution close to optimal. The north-west corner rule ignores costs; MODI and stepping-stone are used to test and improve a solution, not to start one.

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Read the subject

Engineering Economics, Project & Business ManagementThe full study manual, free ICAR AICE-JRF/SRFFormat, sections and how it has changed

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