ASRB NET and ARS / SMS / STO recruitment · India · Agricultural Scientists Recruitment Board (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)
How our questions for ASRB NET and ARS / SMS / STO recruitment spread across the units of Operations Research / Quantitative Techniques. Difficulty: 2 easy, 24 medium, 8 hard.
| Transportation problems | 4 | |
| Inventory models: EOQ, Q and P systems | 4 | |
| Queuing models | 3 | |
| Linear programming formulation | 2 | |
| Graphical and simplex methods | 2 | |
| Duality | 2 | |
| Data envelopment analysis | 2 | |
| Introduction to dynamic programming | 2 | |
| Assignment problems | 2 | |
| Decision criteria: maximax, maximin, regret and Laplace | 2 | |
| Decision trees and expected value of perfect information | 2 | |
| Simulation | 2 | |
| Markov chains | 2 | |
| Network analysis | 2 | |
| Two-person zero-sum games | 1 |
Choose an answer, then open the working. Each explanation says why the right answer is right and why the others are not.
A farmer has 3 units of irrigation water for two crops. Returns (thousands) from 0, 1, 2 and 3 units are: crop A 0, 6, 10, 12; crop B 0, 5, 8, 11. What is the maximum total return?
D. 15 thousand
Enumerate the stage decisions: 0 to A and 3 to B gives 0 + 11 = 11; 1 and 2 gives 6 + 8 = 14; 2 and 1 gives 10 + 5 = 15; 3 and 0 gives 12 + 0 = 12. The best is 2 units to A and 1 to B, return 15. 12 puts all water on A; 14 is the (1, 2) split; 23 adds the two 3-unit returns, which would need 6 units of water.
Further reading: Taha, Operations Research: An Introduction
A dairy farm uses dynamic programming to decide each year whether to keep or replace each cow. In this model the state variable is best described as
D. the cow's current age or lactation number and productivity
In a replacement DP the stage is the period (year or lactation), the state is the condition of the asset (age, lactation, yield level), and the decision is keep or replace. The keep/replace choice is the decision, not the state; lifetime profit is the value function being maximised; a fixed price is a parameter.
Further reading: Hardaker et al., Coping with Risk in Agriculture
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 north-west corner rule?
B. 610
North-west corner: x11 = 40 fills plant 1; x12 = 10 empties centre 1; x22 = 60 empties centre 2 and fills plant 2; x33 = 40 finishes. Cost = 40(2) + 10(7) + 60(3) + 40(7) = 80 + 70 + 180 + 280 = 610. 410 is the least-cost method's starting cost and 400 is the true optimum, both of which use the cost matrix; the north-west corner rule ignores costs, so it is usually the most expensive start. 360 is not the cost of any feasible allocation.
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
A retailer orders fresh coriander daily. Each bunch sold earns a margin of 8; each unsold bunch is a loss of 2. Daily demand is normal with mean 500 and standard deviation 50 bunches. The optimal order quantity is about
D. 542 bunches
Newsvendor critical ratio = Cu/(Cu + Co) = 8/(8 + 2) = 0.8, and z(0.8) = 0.842. Q = 500 + 0.842 x 50 = 542. 458 uses the ratio Co/(Cu + Co) = 0.2 (orders below the mean when shortage is the costlier error); 500 ignores the cost asymmetry; 466 uses Co/Cu = 0.25 as the service level.
Further reading: Krajewski et al., Operations Management
Two rival dairies compete on promotion. The payoff matrix (market share gain of Dairy A, rows are A's strategies) is: row 1: 5, 3, 4; row 2: 6, 4, 7; row 3: 2, 1, 8. What is the value of the game?
C. 4
Row minima: 3, 4, 1; maximin = 4 (row 2). Column maxima: 6, 4, 8; minimax = 4 (column 2). Maximin = minimax = 4, so there is a saddle point at row 2, column 2, and the value is 4 with pure strategies. 3 is row 1's minimum; 6 is column 1's maximum; 5 is not a saddle element.
Further reading: Taha, Operations Research: An Introduction
A sugar mill's cane yard has several stages (weighbridge, unloading, empty-truck weighing) with non-exponential service times and shift-dependent arrivals. The most suitable tool to evaluate alternative yard layouts is
C. discrete-event simulation
Analytical queuing formulas need specific assumptions (Poisson arrivals, exponential service, a single stage) that this system violates. Simulation can represent several stages, any distributions and changing arrival patterns. The Hungarian method and graphical LP address different problems altogether.
Further reading: Hillier and Lieberman, Introduction to Operations Research
Farmers buy seed from company X or company Y. Each season 85 percent of X's buyers stay with X and 15 percent switch to Y; 75 percent of Y's buyers stay and 25 percent switch to X. Current shares are X 40 percent and Y 60 percent. Next season, X's share will be
C. 49 percent
X's next share = (X keeps 85 percent of 40) + (Y loses 25 percent of 60 to X) = 0.34 + 0.15 = 0.49, that is 49 percent. 79 adds both retention rates, mixing X and Y; 34 counts only retained X buyers; 21 adds only the switchers.
Further reading: Taha, Operations Research: An Introduction
A cow's health state each month is Healthy (H) or Sick (S). P(H to H) = 0.6, P(H to S) = 0.4, P(S to H) = 0.3, P(S to S) = 0.7. A cow is healthy now. What is the probability that she is sick two months from now?
A. 0.52
Two-step probability = P(H to H) P(H to S) + P(H to S) P(S to S) = 0.6(0.4) + 0.4(0.7) = 0.24 + 0.28 = 0.52, the (H, S) element of P squared. 0.4 is the one-step probability; 0.16 is 0.4 squared (sick, then 'sick to sick' taken as 0.4); 0.28 counts only the path through S and misses the path that stays healthy for one month.
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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