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AOR-01 Solved Assignment January 2026 | OPERATIONS RESEARCH | IGNOU BDP

Q1. Which of the following statements are true? Give reasons for your answers.

i)) If an item is ordered frequently, then the risk of running out of stock is least. (200 words)
ii)) In deterministic queuing model, arrival rate must not exceed the service rate. (200 words)
iii)) The critical path of a project network represents the minimum time needed to complete the project. (200 words)
iv)) A necessary and sufficient condition for a basic feasible solution for a minimization LPP to be optimum is that all Z; - C; ≥ 0. (200 words)
v)) If dual has an unbounded solution, primal has a feasible solution. (200 words)

Key Points:

Frequent ordering in inventory may increase stockout risk due to lower safety stock.
Deterministic queuing systems require arrival rate ≤ service rate for stability.
Critical path duration defines the absolute minimum project completion time.
Minimization LPP optimality condition is Zⱼ - Cⱼ ≤ 0 for non-basic variables.

Answer: As an expert IGNOU tutor, I have analyzed the AOR-01 Operations Research course material to provide comprehensive answers to the given statements. Each sub-question is addressed individually, clarifying whether the statement is true or false and providing detailed reasoning based on fundamental concepts of Operations Research.

Q2. Not applicable

a)) A firm produces three product A, B and C. It uses two types of raw materials I and II of which 5000 and 7500 units respectively are available. The raw material requirements per unit of the products are given below:The minimum demand of the three products is 600, 650 and 500 units respectively. Assuming the profits per unit of A, B and C as ₹ 50, ₹50 and ₹80 respectively. Formulate the problem as LPP model in order to determine the number of units of each product which will maximize the profit. (500 words)
b)) Use the graphical method to solve the following LPP: Maximize: z = 2x1 + 3x2 Subject to the constraints: X1 + x2 ≤ 30 X1-X2≥0 x2 ≥3 0 ≤ x1 ≤20 and 0≤ x2 ≤12. (500 words)

Key Points:

LPP formulation requires defining decision variables, objective function, and constraints.
Decision variables represent quantities to be optimized (e.g., units of product to produce).
Objective function is a linear mathematical expression to maximize profit or minimize cost.
Constraints are linear inequalities reflecting resource limitations, demand, or other restrictions.

Answer: Operations Research (OR) provides powerful tools like Linear Programming (LPP) to optimize resource allocation and decision-making in complex business scenarios. LPP models translate real-world problems into a mathematical framework, involving decision variables, an objective function to be maximized or minimized, and a set of constraints representing limitations. The graphical method is a visual technique used to solve LPP problems with two decision variables. It involves plotting constraint i...

Q3. Not applicable

a)) Use simplex method to solve the following LPP: Maximize: z = 2x1-x2+X3 Subject to the constraints: 3x1 + x2 + x3 ≤ 60 X1-X2+2x3 ≤ 10 X1+X2-X3≥0 and X1,X2,X3 ≥ 0. (500 words)
b)) Give the dual of the following LPP: Minimize: z = 2x + 3x2 + 4x3 Subject to the constraints: 2x₁ +3x2+5x3 ≥ 2 3x1 + x2 + 7x3 = 3 x₁+4x2+6x3 ≤ 5 X1,X2 ≥0 and x3is unrestricted. (500 words)

Key Points:

Simplex method solves LPP by iteratively moving from one basic feasible solution to another.
Big-M method handles '≥' or '=' constraints by introducing surplus and artificial variables.
Optimality in maximization simplex is reached when all Cj-Zj values are non-positive.
Duality converts a primal LPP into a dual LPP, reversing objective and transposing coefficients.

Answer: The field of Operations Research (OR) provides powerful analytical tools for decision-making and optimization in various real-world scenarios. Linear Programming (LPP) is a fundamental technique within OR, used to optimize an objective function subject to linear constraints. The Simplex Method is an iterative algorithm designed to solve LPPs, especially those with multiple variables and complex constraints. Duality theory, on the other hand, provides a valuable perspective on LPPs by presenting ...

Q4. Not applicable

a)) Use two-phase method to solve the following LPP: Maximize: z = 3x + 2x2 Subject to the constraints: 2x1 + x2 ≤2 3x₁+4x2 ≥12 X1,X2 ≥0 (500 words)
b)) Use dual simplex method to solve the following LPP: Minimize: z = 3x1 + x2 Subject to the constraints: X1 + x2 ≥1 2x + 3x2 ≥2 X1,X2 ≥0. (500 words)

Key Points:

Two-Phase method uses artificial variables for '≥' or '=' constraints.
Phase I objective minimizes sum of artificial variables (w).
If min w > 0 in Phase I, the original LPP has no feasible solution.
Dual Simplex starts with primal optimal (Cj-Zj ≥ 0) but primal infeasible (negative RHS).

Answer: This response provides a comprehensive solution to the given Linear Programming Problems (LPPs) using the Two-Phase method and the Dual Simplex method, respectively. Each sub-question details the step-by-step application of the specified method, including standard form conversion, tableau construction, pivot operations, and interpretation of results, adhering to the principles of Operations Research as outlined in the course material.

Q5. Not applicable

a)) Given X13 = 50 units, X14 = 20 units, X21 = 55 units, X31 =30units, x2 =35 units and X34 = 25 units. Is it an optimal solution to the transportation problem: If not, modify it to obtain a better feasible solution. (500 words)
b)) In a factory, there are six jobs to perform each of which should go through two machines A and B in the order A B. The processing timings (in hours) for the jobs are given here. You are required to determine the sequence for performing the jobs that would minimize the total elapsed time T, what is the value of T? (500 words)

Key Points:

Transportation Problem: Optimality is checked using the MODI method by calculating opportunity costs (Cij - ui - vj).
Negative Opportunity Costs: Indicate a non-optimal solution and potential for cost reduction.
Improvement Iteration: A closed loop is formed for the most negative cell to reallocate units and reduce total cost.
Johnson's Rule: Used to find the optimal sequence for 'n' jobs on two machines (A then B) to minimize total elapsed time.

Answer: This response addresses two distinct sub-questions from the AOR-01 Operations Research course: first, an optimality test and improvement for a transportation problem, and second, a job sequencing problem using Johnson's Rule. For the transportation problem, as crucial cost, supply, and demand data were not provided, a hypothetical but consistent problem setup has been created to demonstrate the Modified Distribution (MODI) method. For the sequencing problem, processing times are assumed to illus...

Q6. Not applicable

a)) The time taken (in hours) by five different machines for completing five different jobs is given below: Find the optimal assignment. (500 words)
b)) A manufacturing company needs 2500 units of a particular item every year. The company buys it at the rate of ₹ 30 per unit. The order processing cost for this item is estimated at 15 and the cost of carrying a item in stock comes to about ₹ 4 per year. The company can manufacture this item internally. In that case it saves 20% of the price of the product. However, it estimates a set-up cost of ₹ 250 per production run. The annual production rate would be 4800 units. However, the inventory holding costs remain unchanged. i) Determine the EOQ and the optimal number of orders placed in a year. ii) Determine the optimum production lot size and the average duration of the production run. (500 words)

Key Points:

Hungarian Algorithm minimizes total cost/time in assignment problems by zero reduction and line covering.
EOQ formula: Q* = √[(2 * D * Co) / Cc] calculates optimal order quantity for purchasing.
Optimal number of orders (N) = Annual Demand (D) / EOQ (Q*).
EPQ formula: Qp* = √[(2 * D * Cs) / (Cc * (1 - D/P))] determines optimal production lot size.

Answer:

Q7. Not applicable

a)) On an average 96 patients per 24-hour a day require the services of an emergency clinic. Also on an average, a patient requires 10 minutes of active attention. Assume that the facility can handle only one emergency at a time. Suppose that it costs the clinic ₹ 100 per patient treated to obtain an average servicing time of 10 minutes, and that each minute of decrease in this average time would cost ₹ 10 per patient treated. How much would have to budgetted by the clinic to decrease the average size of the queue from 1(1/3) patients to 1(1/2) a patient? (500 words)
b)) A project schedule has the following characterizing: i) Construct PERT network. ii) Find the critical path. (500 words)

Key Points:

M/M/1 queuing model calculates average queue length (Lq = λ² / [μ(μ - λ)]).
Service rate (μ) directly impacts queue length; higher μ reduces Lq.
Cost analysis quantifies budget changes for improved service times.
PERT network visualizes project activities and dependencies using nodes and arrows.

Answer:

Q8. Not applicable

a)) A road transport company has one reservation clerk on duty at a time. He handles information of bus schedules and make reservations. Consumers arrive at a rate of 8 per hour and the clerk can service 12 customers on an average per hour. i) What is the average number of customers waiting for the service of the clerk? ii) What is the average time a customer has to wait before getting service? (500 words)
b)) Use branch and bound method to solve the following LPP: Minimize: Z=4x₁+3x2 Subject to the constraints: 5x₁+3x2 ≥30 x₁≤4 X2 ≤6 x1,x2 ≥ 0 and are integers. (500 words)

Key Points:

M/M/1 queuing model calculates waiting line performance for single server, exponential arrivals/service.
Arrival rate (λ) is the average number of customers arriving per unit time.
Service rate (μ) is the average number of customers serviced per unit time.
Average customers waiting in queue (Lq) = λ² / [μ (μ - λ)].

Answer:

Q9. Not applicable

a)) At present a company is purchasing an item ‘x' from outside suppliers. The assumption of unit is 10000 units/year. The cost of the item is ₹ 5 per unit and the ordering cost is estimated to be ₹ 100 per order. The cost of carrying inventory is 25%. If the consumption rate is uniform, determine the economic order quantity. (500 words)
b)) Obtain all the basic solutions to the following system of linear equations: X1 + 2x2 + x3 =4 2x1+x2+5x3 = 5 Which of the solutions are feasible? Justify. (500 words)

Key Points:

Economic Order Quantity (EOQ) minimizes total inventory costs (ordering + carrying).
EOQ formula: EOQ = √[(2 * D * Co) / Cc], where D=Demand, Co=Ordering Cost, Cc=Carrying Cost.
Carrying cost (Cc) is typically a percentage of unit cost per year (Cc = % × P).
Basic solution in linear equations is found by setting (n-m) variables to zero (n=variables, m=equations).

Answer:

Q10. Not applicable

a)) A sales manager wishes to assign four sales territories to four salespersons. The salespersons differ in their ability and skills and consequently the sales expected in each territory are different. The estimates of sales per month for each sales-person in different territories are given below: Find the optimal assignment of the four salespersons to the four different territories and the maximum monthly sales. (500 words)
b)) Find the optimum solution to the following transportation problem: (500 words)

Key Points:

Hungarian Method solves assignment problems by converting to a minimization cost matrix.
Maximization assignment problems are transformed by subtracting elements from the largest value.
Hungarian Method involves row reduction, column reduction, and covering zeros with minimum lines.
Optimal assignment found when number of lines covering zeros equals matrix order.

Answer:

AOR-01 Solved Assignment — January 2026 / July 2026

OPERATIONS RESEARCH | IGNOU BA CHELORS DEGREE PROGRAMME (BDP)

AOR-01—January 2026 / July 2026

OPERATIONS RESEARCH

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AOR-01 Assignment Questions — January 2026

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