Week 1 Problem Solving Assignment Due July 22

week 1 Problem Solving Assignmentdue Tuesday July 22 Befor

Your Name: WEEK 1 PROBLEM SOLVING ASSIGNMENT DUE TUESDAY, JULY 22 (BEFORE THE CLASS) Question 1 A company that makes shopping carts for supermarkets and other stores recently purchased some new equipment that reduces the labor content of the jobs needed to product the shopping carts. Prior to buying the new equipment, the company used five workers, who produced an average of 80 carts per hour. Workers receive $10 per hour, and the machine cost was $40 per hour. With the new equipment, it was possible to transfer one of the workers to another department, and equipment cost increased by $10 per hour while output increased by four carts per hour. a. Compute labor productivity under each system. Use carts per worker per hour as the measure of labor productivity. b. Compute the multifactor productivity under each system. Use carts per dollar cost (labor pus equipment) as the measure. c. Comment on the changes in productivity according to the two measures, and on which one you believe is the more pertinent for this situation. Your Name: Question 2 The following table shows data on the average number of customers processed by several bank service units each day. The hourly wage rate is $25, the overhead rate is 1.0 times labor cost, and material cost is $5 per customer. Unit Employees Customers Processed/Day A 4 36 B 5 40 C 8 60 D 3 20 a. Compute the labor productivity and the multifactor productivity for each unit. Use an eight- hour day for multifactor productivity. b. Suppose a new, more standardized procedure is to be introduced that will enable each employee to process one additional customer per day. Compute the expected labor and multifactor productivity rates for each unit. Your Name: Question 3 A company offers ID theft protection using leads obtained from client banks. Three employees work 40 hours a week on the leads, at a pay rate of $25 per hour per employee. Each employee identifies an average of 3,000 potential leads a week from a list of 5,000. An average of 4 percent actually sign up for the service, paying a one-time fee of $70. Material costs are $1,000 per week, and overhead costs are $9,000 per week. Calculate the multifactor productivity for this operation in fees generated per dollar of input. Your Name: Question 4 Lucky Lumen light bulbs have an expected life that is exponentially distributed with a mean of 20,000 hours. Determine the probability that one of these light bulbs will last. a. At least 24,000 hours. b. No longer than 4,000 hours. c. Between 4,000 hours and 24,000 hours. Your Name: Question 5 A machine can operate for an average of 10 weeks before it needs to be overhauled, a process which takes two days. The machine is operated five days a week. Compute the availability of this machine. (Hint: All times must be in the same units.) Your Name: Question 6 A designer estimates that she can (a) increase the average time between failures of a part by 5 percent at a cost of $450, or (b) reduce the average repair time by 10 percent at a cost of $200. Which option would be more cost-effective? Currently, the average time between failures is 100 hours and the average repair time is 4 hours. Your Name: Question 7 A worker-machine operations was found to involve 3.3 minutes of machine time per cycle in the course of 40 cycles of stopwatch study. The worker’s time averaged 1.9 minutes per cycle, and the worker was given a rating of 120 percent (machine rating is 100 percent). Midway through the study, the worker took a 10-minute rest break. Assuming an allowance factor of 12 percent of work time, determine the standard time for this job.

Paper For Above instruction

Introduction

Productivity measurement is crucial in evaluating the efficiency of processes and operations within a company. It provides a quantitative basis to assess improvements, allocate resources, and set performance targets. The various measures of productivity—labor productivity and multifactor productivity—offer insights into different aspects of operational efficiency. This paper examines several scenarios involving productivity calculations to illustrate their application, interpret their significance, and compare their relevance depending on organizational context.

Question 1: Impact of Equipment on Productivity in Manufacturing

Initially, the company employed five workers who produced 80 carts per hour, making the labor productivity:

• Labor productivity = output / labor input = 80 carts / 5 workers = 16 carts per worker per hour.

Post-equipment acquisition, with four carts per hour more throughput and one worker transferred, the new labor productivity becomes:

• Labor productivity = 84 carts / 4 workers = 21 carts per worker per hour.

The labor productivity has increased, indicating more output per worker. Regarding multifactor productivity (MFP), initially, the total input cost per hour was:

• Labor cost = 5 workers × $10 = $50; Equipment cost = $40, total = $90.

Output per dollar input was:

• Initial MFP = 80 carts / $90 ≈ 0.89 carts per dollar.

After the changes: the equipment cost increased to $50, but output increased to 84 carts, and labor cost reduced to $40 (4 workers × $10). Total input cost per hour:

• Labor = $40; Equipment = $50; Total = $90.

Thus, the multifactor productivity remains:

≈ 0.93 carts per dollar

Compared to labor productivity, which improved from 16 to 21 carts per worker per hour, the multifactor productivity shows a modest increase, highlighting that productivity gains may be more significant at the individual worker level than in overall resource utilization. The measure of labor productivity directly ties to worker efficiency, while multifactor productivity accounts for combined inputs, providing a broader perspective. In this case, for manufacturing productivity assessments, multifactor productivity offers a more comprehensive understanding of resource efficiency, especially when technology or process improvements involve multiple inputs.

Question 2: Analyzing Bank Service Unit Performance

For each bank unit, labor productivity is calculated as:

Customers processed per labor hour = (number of customers processed per day × 8 hours) / number of employees.

• Unit A: (36 × 8) / 4 = 288 / 4 = 72 customers per employee per day
• Unit B: (40 × 8) / 5 = 320 / 5 = 64 customers per employee per day
• Unit C: (60 × 8) / 8 = 480 / 8 = 60 customers per employee per day
• Unit D: (20 × 8) / 3 = 160 / 3 ≈ 53.33 customers per employee per day

Multifactor productivity considers combined inputs, such as wages, materials, and overheads. The total labor cost per employee per day is:

• $25/hour × 8 hours = $200

Material and overhead costs per customer are $5 and $200 (overhead rate is 1.0 times labor cost), respectively. Total costs per customer processed are:

• Labor cost per customer = $200 / customers processed per day.
• Material + overhead per customer = $5 + (Overhead rate of 1 × labor cost per customer).

Calculating overall, the total cost per customer accounts for labor, material, and overhead. For simplicity, assume overhead costs are proportionally allocated, leading to a comprehensive measure of resource utilization. After introducing a procedure that allows each employee to process one additional customer daily, the new productivity metrics increase accordingly, directly reflecting efficiency gains, making a compelling case for process improvement.

Question 3: Evaluating Marketing Efficiency in ID Theft Protection

The weekly effort involves three employees, each working 40 hours, earning $25 hourly, totaling weekly labor costs of:

• 3 employees × 40 hours × $25 = $3,000

Potential leads generated per employee per week: 3,000. Total potential leads:

• 3 employees × 3,000 = 9,000 leads weekly.

Conversion rate: 4% of leads sign up, resulting in:

• Signups = 9,000 × 0.04 = 360.

Fees generated per week:

• 360 signups × $70 = $25,200.

The total input cost includes labor, material, and overhead:

• Material costs = $1,000; Overhead costs = $9,000; Total costs = $3,000 (labor) + $1,000 + $9,000 = $13,000.

Multifactor productivity (MFP) in fees per dollar input is calculated as:

• MFP = $25,200 / $13,000 ≈ 1.94 dollars per input dollar.

This indicates a relatively high return on resource investment, showcasing the operation's efficiency in converting inputs into revenue, especially with the potential for scaling through process optimization.

Question 4: Probability Distribution of Light Bulb Lifetimes

Given exponential distribution with mean 20,000 hours, the probability density function is:

P(t) = (1/μ) e^{-t/μ}, where μ = 20,000 hours.

A) Probability that a bulb lasts at least 24,000 hours:

P(T ≥ 24,000) = e^{-\frac{24,000}{20,000}} = e^{-1.2} ≈ 0.301.

B) Probability that it lasts no longer than 4,000 hours:

P(T ≤ 4,000) = 1 - e^{-\frac{4,000}{20,000}} = 1 - e^{-0.2} ≈ 0.181.

C) Probability that it lasts between 4,000 and 24,000 hours:

P(4,000

These probabilities provide insights into bulb reliability and lifetime expectations.

Question 5: Machine Availability Calculation

Average operating time before overhaul is 10 weeks, with overhaul taking 2 days (which equals 2 × 5 = 10 days, assuming 5 working days per week). The total operational time:

• Operational weeks: 10 weeks
• Operational hours = 10 weeks × 5 days/week × 8 hours/day = 400 hours
• Overhaul time = 2 days × 8 hours/day = 16 hours

Availability = uptime / (uptime + downtime) = 400 / (400 + 16) ≈ 0.9615 or 96.15%.

This high availability indicates a reliable machine with infrequent downtime.

Question 6: Cost-Effectiveness of Reliability Improvements

Current failure parameters: failure every 100 hours, repair time of 4 hours. Option a): increasing time between failures by 5% yields new failure interval:

New failure interval = 100 × 1.05 = 105 hours.

Additional cost: $450.

Option b): reducing repair time by 10% gives new repair time: 4 × 0.9 = 3.6 hours; repair cost proportional to time, so cost savings are 0.4 hours × $25/hr = $10.

The cost per unit of increased reliability is: $450 / (5 hours increase), while savings in repair time per cycle are minimal. Given the larger cost associated with increasing the failure interval, reducing repair time offers a more cost-effective improvement per dollar spent.

Question 7: Standard Time Calculation for Worker-Machine Operation

Average machine time per cycle: 3.3 minutes; worker time: 1.9 minutes; worker rating: 120%. The normal time is:

Normal time = average worker time × rating = 1.9 minutes × 1.2 = 2.28 minutes.

Adding allowances: allowance factor = 12% of work time = 0.12 × 2.28 ≈ 0.27 minutes.

Standard time = normal time + allowances = 2.28 + 0.27 = 2.55 minutes per cycle.

This standard time reflects realistic working conditions inclusive of fatigue and delays.

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