ICAR AICE-JRF/SRF (PhD admission and fellowship) · India · National Testing Agency for ICAR
Original practice questions written to the published syllabus · not past papersMathematical models for decisions: linear programming, transport and assignment, inventory, queues, decision theory, games, simulation and networks.
In the exams: agribusiness management PhD entrance and fellowship examinations, operations research (heavy); agricultural economics PhD entrance and fellowship examinations (present); agricultural scientist eligibility tests, operations research methods (heavy)
ICAR AICE-JRF/SRF (PhD) Agri-Business Management, Unit 8 Operations Research
Heavy unit. LP assumptions, formulation, graphical corner points and special cases (unbounded, infeasible, alternative optima, degeneracy); simplex entering and leaving rules, slack/surplus/artificial variables, Big-M; primal-dual construction, duality theorems, complementary slackness and shadow prices; transportation starting methods (NWC, least cost, VAM), m+n-1 and MODI; Hungarian method; EOQ, reorder point, Q versus P systems; M/M/1 measures and Little's law; maximax, maximin, minimax regret, Laplace, Hurwicz, EMV, EVPI; saddle points and 2x2 mixed games; PERT expected time, variance and completion probability, floats and crashing; Monte Carlo steps; Markov steady state; DEA CCR/BCC and scale efficiency; Bellman's principle.
ICAR AICE-JRF/SRF (PhD) Agricultural Economics
LP as a farm-planning and least-cost ration tool, shadow prices as marginal value products of farm resources, parametric programming for normative supply response, risk programming (MOTAD), DEA for farm efficiency, and decision analysis under risk.
How our questions for ICAR AICE-JRF/SRF spread across the units of Operations Research / Quantitative Techniques. Difficulty: 29 easy, 45 medium, 24 hard.
| Graphical and simplex methods | 8 | |
| Transportation problems | 8 | |
| Linear programming formulation | 7 | |
| Duality | 7 | |
| Assignment problems | 7 | |
| Inventory models: EOQ, Q and P systems | 7 | |
| Two-person zero-sum games | 7 | |
| Markov chains | 7 | |
| Network analysis | 7 | |
| Data envelopment analysis | 6 | |
| Decision criteria: maximax, maximin, regret and Laplace | 6 | |
| Decision trees and expected value of perfect information | 6 | |
| Introduction to dynamic programming | 5 | |
| Queuing models | 5 | |
| Simulation | 5 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
When a >= constraint is converted to standard form for the simplex method (Big-M or two-phase), what is done?
D. A surplus variable is subtracted and an artificial added
For a >= constraint the surplus variable (amount by which the requirement is exceeded) is subtracted. Because the surplus enters with -1 it cannot serve as a starting basic variable, so an artificial variable is added. A slack is added only to <= constraints.
Further reading: Taha, Operations Research: An Introduction
The 'curse of dimensionality' in dynamic programming refers to
B. the explosive growth in the number of states when there are multiple state variables
DP evaluates every state at every stage; with several state variables each taking many values, the number of states multiplies and computation becomes infeasible. DP does not require linear returns (its strength is that it handles non-linear ones), backward recursion is its standard method, and it allows any number of stages.
Further reading: Taha, Operations Research: An Introduction
Three chilling centres supply 50, 60 and 40 thousand litres of milk; three plants need 40, 70 and 40 thousand litres. Unit costs (per thousand litres) from centre 1 to plants 1, 2, 3 are 2, 7, 4; from centre 2: 3, 3, 1; from centre 3: 5, 4, 7. What is the total cost of the initial solution by the least-cost (matrix minimum) method?
A. 410
Cheapest cell first: x23 = 40 at cost 1 (plant 3 filled, centre 2 has 20 left); next cheapest 2: x11 = 40 (plant 1 filled, centre 1 has 10 left); next 3: x22 = 20 (centre 2 empty); next 4: x32 = 40 (centre 3 empty, plant 2 needs 10 more); last x12 = 10 at 7. Cost = 40(1) + 40(2) + 20(3) + 40(4) + 10(7) = 40 + 80 + 60 + 160 + 70 = 410. 610 is the north-west corner cost; 400 is the optimum after MODI improvement. 550 comes from working column by column, taking each plant's cheapest remaining centre.
Further reading: Taha, Operations Research: An Introduction
In the Hungarian method for an n x n assignment problem, an optimal assignment can be made when
D. the minimum number of lines needed to cover all zeros equals n
After row and column reduction, every row and column has a zero, but that alone does not guarantee a one-to-one assignment. Only when the minimum covering lines equal n do n independent zeros exist. The other conditions have no such meaning.
Further reading: Taha, Operations Research: An Introduction
In the Hungarian method, after row and column reduction, the zeros of a 5 x 5 matrix can be covered by only 4 lines. The smallest uncovered element is 3. The next step is to
A. subtract 3 from every uncovered element and add 3 to every element at the intersection of two lines
Subtracting the smallest uncovered value from uncovered cells creates at least one new zero; adding it at intersections keeps all entries non-negative and is equivalent to subtracting from uncovered rows and adding to covered columns. Other covered cells stay unchanged. The reversed version or subtracting from covered cells can create negative entries or lose zeros.
Further reading: Taha, Operations Research: An Introduction
A village input shop reviews stock every 14 days (P system). Lead time is 2 days, average daily demand is 50 units with a daily standard deviation of 8, and z = 1.28 for the target service level. At a review, the inventory position is 300 units. How much should be ordered?
C. 541 units
The P system must protect over P + L = 16 days. Target T = 50 x 16 + 1.28 x 8 x sqrt(16) = 800 + 40.96 = 840.96; order = T - position = 840.96 - 300 = 541 units. 438 protects only the review period and forgets the lead time; 500 omits safety stock; 664 multiplies the standard deviation by 16 instead of sqrt(16).
Further reading: Krajewski et al., Operations Management
A farmer chooses among tomato, chilli and okra. Net returns (thousands) under good, average and poor rainfall are: tomato 90, 50, -20; chilli 70, 55, 10; okra 40, 35, 30. Which crop does the maximax criterion choose?
B. Tomato
Row maxima: tomato 90, chilli 70, okra 40; maximax (optimist) picks the largest, tomato. Okra is the maximin choice and chilli is the minimax regret and Laplace choice.
Further reading: Taha, Operations Research: An Introduction
A farmer chooses among tomato, chilli and okra. Net returns (thousands) under good, average and poor rainfall are: tomato 90, 50, -20; chilli 70, 55, 10; okra 40, 35, 30. Probabilities are 0.3, 0.5 and 0.2. What is the expected value of perfect information?
A. 10 thousand
With perfect information the farmer picks the best crop in each state: 90 (good), 55 (average), 30 (poor). EPPI = 0.3(90) + 0.5(55) + 0.2(30) = 60.5. Max EMV = 50.5 (chilli). EVPI = 60.5 - 50.5 = 10. 60.5 is EPPI itself; 25 subtracts the lowest EMV; 124.5 adds best payoffs without probabilities.
Further reading: Taha, Operations Research: An Introduction
In PERT, the critical activities of a project have variances of 1, 4, 2.25 and 1.44 weeks squared. The standard deviation of project completion time is
D. 2.95 weeks
Variances of independent activities on the critical path add: 1 + 4 + 2.25 + 1.44 = 8.69; sigma = sqrt(8.69) = 2.95 weeks. 8.69 is the variance; 5.7 adds standard deviations (1 + 2 + 1.5 + 1.2), which is wrong; 0.74 divides the standard deviation by the number of activities.
Further reading: Taha, Operations Research: An Introduction
A pack-house project has activities (weeks): A 3 (start); B 5 (after A); C 4 (after A); D 6 (after B); E 3 (after C); F 2 (after D and E). What is the project duration?
C. 16 weeks
Forward pass: A 0-3; B 3-8; C 3-7; D 8-14; E 7-10; F starts at max(14, 10) = 14, ends 16. Duration = 16 weeks along the critical path A-B-D-F (3 + 5 + 6 + 2). 12 is path A-C-E-F; 23 adds all activities as if done in series; 14 is the finish of D, forgetting F.
Further reading: Taha, Operations Research: An Introduction
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