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 papersMoney 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
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.
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 programming | 11 | |
| Project appraisal and detailed project reports | 10 | |
| CPM and PERT | 9 | |
| NPV, IRR, benefit-cost ratio and payback | 3 | |
| Break-even analysis | 3 | |
| Replacement analysis | 3 | |
| Time value of money | 1 | |
| Depreciation methods | 1 | |
| Fixed and variable costs | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
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?
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.
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?
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.
In the economic (as opposed to financial) analysis of an irrigation project, how is a government subsidy on pump sets treated?
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.
Which item is normally part of the 'cost of project' in a detailed project report for a food-processing unit?
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.
A project protected by import duties shows a financial IRR of 18 % but an economic IRR of 6 %. What does this most directly suggest?
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.
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?
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.
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?
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.
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)?
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.
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?
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.
Which method for an initial transportation solution uses penalties, the difference between the two lowest costs in each row and column?
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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