Description

Quantitative Methods – I

Sep 2026 Examination

 

Q1 A manufacturing company produces high-precision micro-components and packages them in lots of 400 units. Historical data show that each unit is defective with probability 0.004, independently of other units. To maintain quality standards, a lot is accepted only if it contains at most 2 defective units. Lots with 3 or more defectives are rejected and sent for rework at a cost of Rs.1,800 per lot. Quality engineers have observed that rework is successful only when the rejected lot originally contains 3 or 4 defectives. If the lot contains 5 or more defectives, rework fails and the company incurs an additional scrap loss of Rs.6,500. Management wants to estimate the expected quality-related cost associated with this policy and assess the risk of scrap in large shipments. Questions: 1. Using an appropriate probability distribution and justifying any approximation used, calculate the expected total quality cost per lot. 2. If a shipment contains 120 lots, determine the probability that the number of lots ultimately ending in scrap exceeds its expected value by at least 3 lots. (10 Marks)

Ans 1.

Introduction

Quality control decisions in manufacturing depend on how defect rates translate into real costs. Here a company packages micro components in lots of four hundred units, where each unit has a small independent chance of being defective. Since the defect probability is low and the lot size is large, the binomial distribution can be approximated using the Poisson distribution, which simplifies the analysis considerably. The company faces a two stage cost structure, where rejected lots are first sent for rework at a fixed cost, and among these some cannot be repaired and lead to an additional scrap loss. Understanding the expected cost per lot and the probability of scrap exceeding its average across a large shipment helps

 

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Q2 A precision engineering company manufactures metal shafts for an automotive supplier. The shaft diameter (in millimetres) is assumed to follow a normal distribution with unknown process mean and standard deviation. Recent quality reports indicate that: 18% of the shafts are rejected because their diameter is below 49.20 mm. 9% of the shafts are rejected because their diameter is above 51.05 mm. Under the supply contract: The buyer pays Rs.480 for every accepted shaft. A Rs.900 penalty is charged for every rejected shaft. To encourage process consistency, the buyer offers a performance bonus of Rs.1,20,000 on a production batch of 40,000 shafts if the central 60% of the diameter distribution lies entirely within the engineering tolerance band [49.60 mm, 50.80 mm]. The Operations Manager wants to assess the financial performance of the process. Questions: 1. Estimate the process parameters using the given rejection rates. 2. Determine whether the process qualifies for the Rs.1,20,000 bonus. 3. Calculate the expected net revenue from the batch of 40,000 shafts, taking into account: Revenue from accepted shafts, Penalties for rejected shafts, and The contingent bonus, if applicable. (10 Marks)

Ans 2.

Introduction

Shaft diameter in this precision engineering process is assumed to follow a normal distribution with unknown mean and standard deviation. The company only knows the proportion of shafts rejected for being too thin and too thick, and from these rejection rates the underlying process parameters can be estimated using standard normal properties. Once the parameters are known, the operations manager can check whether the process is consistent enough to earn a performance bonus and can assess the expected net revenue from

 

Q3(A) A manufacturing company operates two CNC production lines, Line A and Line B, to produce precision shafts. Following a recent recalibration of Line A, management claims that the process variability of Line A is no greater than that of Line B, indicating improved process consistency. To evaluate this claim, a quality engineer collected random samples from both lines on a particular production day. The sample information is shown below: Line A, Sample Size 12, Sample Variance 18.4. Line B, Sample Size 15, Sample Variance 9.6. Assume that shaft diameters from both production lines are normally distributed. Management wishes to determine, at the 5% significance level, whether the recalibrated Line A has maintained variability at or below that of Line B. Questions: 1. Formulate the appropriate null and alternative hypotheses to test the claim that Line A does not have higher variance than Line B. 2. Using an F-test for comparing two population variances, compute the test statistic. 3. Determine the critical value and decision rule at the 5% significance level. 4. Based on the test results, conclude whether the company’s claim regarding the variability of Line A is supported by the data. 5. Briefly discuss the managerial implications of the conclusion for quality-control monitoring and process improvement. (5 Marks)

Ans 3A.

Introduction

Line A has recently been recalibrated and management claims its process variability is now no greater than that of Line B. This claim can be formally tested using an F test for comparing two population variances, using the sample variances collected from both lines.

Concept and Application

Setting up the Hypotheses

 

Q3(B) A manufacturing company is analyzing the factors affecting its weekly maintenance cost. The operations analyst believes that maintenance cost Y (in Rs. lakh) is linearly related to machine running intensity X (measured in hundred machine-hours). However, before the data reach the analyst, the accounting department makes two adjustments: 1. A fixed environmental compliance charge of Rs.0.80 lakh is added to every week’s maintenance cost. 2. Machine usage is recorded in actual machine-hours rather than in hundred machine-hours. As a result, the following processed records are available for eight weeks: Week 1: X 820, Y 6.72. Week 2: X 910, Y 7.43. Week 3: X 760, Y 6.14. Week 4: X 1040, Y 8.35. Week 5: X 980, Y 7.96. Week 6: X 870, Y 7.07. Week 7: X 1110, Y 8.88. Week 8: X 1010, Y 8.00. Before conducting the analysis, the analyst must convert the data to the intended variables by expressing: X in hundred machine-hours, and Y after removing the fixed compliance charge of Rs.0.80 lakh. Questions: 1. Transform the recorded data into the appropriate variables for analysis. 2. Using a simple linear regression model, obtain the regression equation of corrected maintenance cost on corrected machine running intensity. 3. Compute and interpret the Pearson correlation coefficient between the corrected variables. 4. If the plant is expected to operate for 950 machine-hours next week, estimate the corrected maintenance cost using the fitted regression model. 5. Briefly comment on the strength of the relationship and its usefulness for maintenance budgeting and operational planning. (5 Marks)

Ans 3B.

Introduction

Weekly maintenance cost is believed to depend on machine running intensity, but the recorded data include a fixed compliance charge and are measured in actual machine hours rather than the intended units, so correcting these adjustments is necessary before any regression analysis.

Concept and Application

Why Data Transformation Matters