Mastering Process Analysis: Practice Problems and Solutions
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St Xaviers College**We aren't endorsed by this school
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MGT HUMAN RESO
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Industrial Engineering
Date
Dec 10, 2024
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48
Uploaded by SuperHumanIron2214
SESSION 1 PRACTICE PROBLEMS Textbook Problem 2.2: Consider the baggage check-in of a small airline. Check-in data indicate that from 9 a.m. to 10 a.m., 255 passengers checked in. Moreover, based on counting the number of passengers waiting in line, airport management found that the average number of passengers waiting for check-in was 35. How long did the average passenger have to wait in line? Textbook Problem 2.5 (a): LaVilla is a village in the Italian Alps. Given its enormous popularity among Swiss, German, Austrian, and Italian skiers, all of its beds are always booked in the winter season and there are, on average, 1,200 skiers in the village. On average, skiers stay in LaVilla for 10 days. How many new skiers are arriving—on average—in LaVilla every day? Textbook Problem 2.6: While driving home for the holidays, you can’t seem to get Little’s Law out of your mind. You note that your average speed of travel is about 60 miles per hour. Moreover, the traffic report from the WXPN traffic chopper states that there is an average of 24 cars going in your direction on a one-quarter mile part of the highway. What is the flow rate of the highway (going in your direction) in cars per hour? Textbook Problem 3.1: Consider a process consisting of three resources: Resource Processing Time [min/unit] Number of Workers 1 10 2 2 6 1 3 16 3 What is the bottleneck? What is the process capacity? What is the flow rate if demand is 8 units per hour?
SESSION 1 SOLUTIONS OF THE PRACTICE PROBLEMS Problem 2.2:Use Little’s law to compute the lead time (or flow time), since we know both the flow rate as well as the inventory level: Lead Time (or Flow Time) = Inventory / Flow Rate = 35 passengers / 255 passengers per hour = 0.137 hours = 8.24 minutes Problem 2.5 (a): Flow Rate = Inventory / Lead Time = 1200 skiers / 10 days = 120 skiers per day Problem 2.6: We look at 1 mile of highway as our process. Since the speed is 60 miles per hour, it takes a car 1 minute to travel through the process (flow time). There are 24 cars on ¼ of a mile, i.e. there are 96 cars on the 1 mile stretch (inventory). Inventory =Flow Rate * Lead Time 96 cars=Flow Rate * 1 minute Thus, the Flow Rate is 96 cars per minute, corresponding to 96*60=5760 cars per hour. Problem 3.1 (Single Unit Flow): •Resource 1:210units/ minute = 0.2 units/minute •Resource 2: 16units/ minute = 0.1666 units/minute •Resource 3:316units/ minute = 0.1875 units/minute 1.Resource 2 is the bottleneck and the process capacity is 0.1666 units/minute = 10 units/ hour (least capacity resource or slowest resource is the bottleneck) 2.Flow rate = Minimum {Demand , Capacity} = Minimum {10 units/ hour, 8 units/ hour} = 8 units/ hour = 0.1333 units/minute
SESSION 2 PRACTICE PROBLEMS Textbook Problem 3.1: Consider a process consisting of three resources: Resource Processing Time [min/unit] Number of Workers 1 10 2 2 6 1 3 16 3 What is the utilization of each resource if demand is eight units per hour? Textbook Problem 7.5: Consider the following batch flow process consisting of three process steps performed by three machines:Work is processed in batches at each step. Before a batch is processed at step 1, the machine has to be set up. During a setup, the machine is unable to process any product. 1.Assume that the batch size is 50 parts. What is the capacity of the process? 2.For a batch size of 10 parts, which step is the bottleneck for the process? 3.Using the current production batch size of 50 parts, how long would it take to produce 20 parts starting with an empty system? Assume that the units in the batch have to stay together (no smaller transfer batches allowed) when transferred to step 2 and to step 3. A unit can leave the system the moment it is completed at step 3. Assume step 1 needs to be set up before the beginning of production. 4.Using the current production batch size of 50 parts, how long would it take to produce 20 parts starting with an empty system? Assume that the units in the batch do not have to stay together; specifically, units are transferred to the next step the moment they are completed at any step. Assume step 1 needs to be set up before the beginning of production. 5.What batch size would you choose, assuming that all units of a batch stay together for the entire process? Step1Processing Time:1 Minute/PartSetup Time:20 MinutesStep 2Processing Time:2 Minutes/PartNo SetupStep 3Processing Time:1.5 Minutes/PartNo Setup
SESSION 1 SOLUTIONS OF THE PRACTICE PROBLEMS Problem 3.1 (Single Unit Flow): Capacity of each resource is given by: •Resource 1:210units/ minute = 0.2 units/minute •Resource 2: 16units/ minute = 0.1666 units/minute •Resource 3:316units/ minute = 0.1875 units/minute Utilization of each resource is given by: Resource Capacity Demand Utilization = Demand/ Capacity Resource 1 0.2 units/minute 0.1333 units/minute 0.6666 Resource 2 0.1666 units/minute 0.1333 units/minute 0.8000 Resource 3 0.1875 units/minute 0.1333 units/minute 0.7111
Textbook Problem 7.5: (a)The first step has a process capacity of 50 units / (20 minutes + (1*50) minutes) = .714 units/minute = 42.86 units/hour. The second step has a process capacity of 1 / 2 minutes = 30 units/hour. The third step has a process capacity of 1 / (1.5) minutes = 40 units/hour. Therefore, the process capacity of the entire process = 30 units/hour. (b)The only step that is dependent on the batch size is Step 1. With a batch size of 10 units, the process capacity of the first step becomes 10 units / (20 minutes + (1*10) minutes) = 10 units / 30 minutes = 20 units/hour. Therefore, Step 1 becomes the bottleneck. (c)If the batch size is 50 parts, and the batch must stay together, then in order to process 20 parts, a batch of 50 parts must be processed through Step 1 and Step 2. The total time to process through Step 1 = 20 minutes for set-up + 50 minutes for processing = 70 minutes. The total time to process the batch through Step 2 = 100 minutes. So, the batch takes 170 minutes through the first two steps. At Step 3, we merely need to process the 20 parts. Processing 20 parts through Step 3 takes 30 minutes. Under these conditions it takes a total of 200 minutesto process 20 parts. (d)At Step 1, there is a set-up time of 20 minutes. So, the first part takes a total of 20+1+2+1.5 = 24.5 minutes to complete. The total time to process 20 units = 24.5 + (19 / 20/40) = 62.5 minutes(e)In order to determine an optimal batch size, we set the process capacity of Step 1 (the only step dependent on batch size) equal to the process capacity of the bottleneck capacity. The bottleneck capacity is 30 units/hour, or 1 unit every 2 minutes. So, we solve for B : 1 unit / 2 minutes = B / (20 minutes + B*1 minutes) B = 20 units.
SESSION 3: PRACTICE PROBLEMS An assembly line example Textbook Problem 4.1: Consider a process consisting of three resources in a worker-paced line and a wage rate of $10 per hour. Assume there is unlimited demand for the product. Resource Processing Time [min/unit] Number of Workers 1 10 2 2 6 1 3 16 3 a. How long does it take the process to produce 100 units starting with an empty system? b. What is the average labor content? c. What is the average labor utilization? d. What is the cost of direct labor?
SESSION 3: SOLUTIONS OF THE PRACTICE PROBLEMS Textbook Problem 4.1: Resource 1: 2/10 units/ minute=0.2 units/minute Resource 2: 1/6 units/ minute=0.1666 units/minute Resource 3: 3/16 units/ minute=0.1875 units/minute Resource 2 is the bottleneck and the process capacity is 0.1666 units/minute = 10/ hour (a) Time to complete 100 units: •Resource 2 is the bottleneck and the process capacity is 0.1666 units per minute and cycle time is 6 minutes. •The process will take 10+6+16minutes=32 minutes to produce the first unit. •Time to finish 100 units = 32 minutes + 99 * 6 = 626 minutes (b) Labour content: 10 minutes + 6 minutes + 16 minutes = 32 minutes/ unit (c) Labour utilization: Labour time used (in hour) / labour time available (in hour) = (32 * 10)/ (6*60) = 88.89% (d) Direct labour cost = Wages paid (in hour) / Total flow (in hour) = (6 * 10) / 10 = $6/ shirt
Health Care Example Suppose there is a health diagnostic centre which provides services to the arriving patients. Three types of patients arrive, and they undergo a preliminary investigation before further tests are done. There are dedicated preliminary investigation facilities available for each of the three types of customers. •Patient Type 1 compulsorily have to undergo tests in the Coronary Unit and the Radiology departments in the diagnostics centre in that order. •Patient Type 2 generally require some tests in the Radiology Department. However, after Preliminary Investigation, 20% of them will be asked to go the Coronary Unit first for some tests and from there they will proceed to the Radiology Department. •Patient Type 3 undergo Gynaecology related tests. After all tests are done, all the three types of patients visit a common facility in which a pool of doctors is available. One of the doctors meets with the patient, reviews the reports and provides further advice wherever required. Irrespective of the patient type, they meet any of the available doctor for this final review. The following are the Process Times, Number of Resources Available at Each Stage: Department Process Time (min) No. of Resources Available Preliminary Investigation (Patient Type 1)20 5 Preliminary Investigation (Patient Type 2)15 6 Preliminary Investigation (Patient Type 3)10 5 Coronary Unit30 5 Radiology Department 10 5 Gynecology Department 20 8 Final Review 10 10 Based on some historical trends, the diagnostic centre expects the potential daily arrivals of the three patient types as follows: Patient Type 1: 100; Patient Type 2: 200; Patient Type 3: 220 Given this information, analyse the process and identify the bottleneck. Assume health diagnostic centre operates for 10 hours per day.
SESSION 4: SOLUTION OF THE PRACTICE PROBLEM Solution to the Health Care ExampleProcess flow diagram Demand Data •Demand of Type 1= 100/day = 10/hour •Demand of Type 2= 200/day = 20/hour •Demand of Type 3= 220/day = 22/hour Department No. of Resources Available Time available/hr Process Time (min/unit) Job types faced Time used/hr Utilization Preliminary Investigation (Patient Type 1)5 5*60 minutes = 300 minutes 20 Type 1 (100%) 20*10 = 200 minutes =200/300 Preliminary Investigation (Patient Type 2)6 6*60 minutes = 360 minutes 15 Type 2 (100%) 15*20= 300 minutes =300/360 Preliminary Investigation (Patient Type 3)5 5*60 minutes = 300 minutes 10 Type 3 (100%) 10*22= 220 minutes =220/300 Coronary Unit5 5*60 minutes = 300 minutes 30 Type 1 (100%) 30*10 + 30* 0.2* 20 = 420/300
Type 2 (20%) = 420 minutes Radiology Department 5 5*60 minutes = 300 minutes 10 Type 1 (100%) Type 2 (100%) 10*10 + 10*20 = 300 minutes = 300/300 Gynecology Department 8 8*60 minutes = 480 minutes 20 Type 3 (100%) 20*22 = 440 minutes = 440/480 Final Review 10 10*60 minutes = 600 minutes 10 Type 1 (100%) Type 2 (100%) Type 3 (100%) 10*10+ 10*20+ 10*22 = 520 minutes 520/600 The resource with the maximum utilization (420/300 = 1.4) is Coronary Unit. Therefore, it is a bottleneck.
Operations Management (2022) - TutorialSession 1 (Section B)Module 1: Process and Capacity AnalysisProcess Analysis: Identifying BottlenecksConsider a process consisting of three resources in a production line and a wage rate of Rs. 500 perhour. The unit is first processed at the first resource, then processed at the second resource, andis finally processed at the third resource. Each worker handles one unit at a time at a particularstep. Assume demand is unlimited. The processing time data is as follows:ResourceProcess time (min/unit)Number of Workers11022613163Table 1: Details of the process steps•What is bottleneck? What is process capacity? What is flow rate? What is cycle time?•What is average labour content? Labour utilization? Cost of direct labour?Process Analysis with BatchingStepSetup Time (min)Run time (min)12012No setup23No setup1.5Table 2: Details of the process steps•If batch size is 50 parts, what is the capacity of the process?•For a batch size of 10 which step is the bottleneck?•What batch size would you choose to maximize the system’s capacity?1
Capacity Analysis - Multiple Product Scenario (A, B &C)Consider a process consisting of five steps that are operated 8 hours per day. Each worker handlesone unit at a time at a particular step. The process works on three different products A, B and C:StepsNumber ofworkersProcesstime for A(min/unit)Processtime for B(min/unit)Processtime for C(min/unit)12555223453115004103352666Table 3: Details of the process stepsDemand for A is 40 units per day, B is 50 units per day, and C is 60 units per day.•What is the bottleneck?•If the system cannot meet demand then where will you increase capacity?Application of Little’s Law (LaVilla)LaVilla is a village in the Italian Alps.Given its enormous popularity among Swiss, German,Austrian, and Italian skiers, all of its beds are always booked in the winter season and there are,on average, 1200 skiers in the village. On average, skiers stay in LaVilla for 10 days.•How many new skiers are arriving—on average—in LaVilla every day?Application of Little’s Law (Airline)Consider the baggage check-in of a small airline. Check-in data indicate that from 9 a.m. to 10a.m., 255 passengers checked in. Moreover, based on counting the number of passengers waitingin line, airport management found that the average number of passengers waiting for check-in was35.•How long did the average passenger have to wait in line?* * *End of Tutorial 1* * *2
Solutions to Tutorial Session 1: Tutorial Problem 1: Tutorial Problem 1 is same as Session 3 practice problem. Please see the detailed solution in session 3 folder. Tutorial Problem 2:(a) The first step has a process capacity of 50 units / (20 minutes + (1*50) minutes) = .714 units/minute = 42.86 units/hour. The second step has a process capacity of 1 / 2 minutes = 30 units/hour. The third step has a process capacity of 1 / (1.5) minutes = 40 units/hour. Therefore, the process capacity of the entire process = 30 units/hour. (b) The only step that is dependent on the batch size is Step 1. With a batch size of 10 units, the process capacity of the first step becomes 10 units / (20 minutes + (1*10) minutes) = 10 units / 30 minutes = 20 units/hour. Therefore, Step 1 becomes the bottleneck. (c) In order to determine an optimal batch size, we set the process capacity of Step 1 (the only step dependent on batch size) equal to the process capacity of the bottleneck capacity. The bottleneck capacity is 30 units/hour, or 1 unit every 2 minutes. So, we solve for B, i.e., 1 unit / 2 minutes = B / (20 minutes + B*1 minutes), which implies B = 20 units. Tutorial Problem 3:Apply pooled utilization approach learnt in Manzana Insurance case. Here are the Pooled Utilization values: Resource 1 - 0.78, Resource 2 - 0.65, Resource 3 - 1.25, Resource 4 - 0.69 and Resource 5 - 0.94. Hence Resource 3 is bottleneck (highest pooled utilization !). Since Resource 3 utilization (1.25) > 1, demand cannot be met. Hence, you should increase capacity at Resource 3. You will see with 2 workers the Resource 3 utilization will be 0.625 and hence, the bottleneck will shift to Resource 5 ! Tutorial Problem 4: Average Flow Rate = Average Inventory (WIP) / Average Lead Time = 1200 skiers / 10 days = 120 skiers per day Tutorial Problem 5: We use Little’s law to compute the Average Lead Time, since we know both the average flow rate as well as the average inventory level: Average Lead Time = Average Inventory (WIP) / AverageFlow Rate = 35 passengers / 255 passengers per hour = 0.137 hours = 8.24 minutes
SESSION 5 PRACTICE PROBLEMS Textbook Problem 3.2: Consider a process consisting of five resources that are operated eight hours per day. The process works on three different products, A, B, and C: Resource Number of Workers Processing Time for A [min/unit] Processing Time for B [min/unit] Processing Time for C [min/unit] 1 2 5 5 5 2 2 3 4 5 3 1 15 0 0 4 1 0 3 3 5 2 6 6 6 Demand for the three different products is as follows: product A, 40 units per day; product B, 50 units per day; and product C, 60 units per day. Identify the bottleneck. Textbook Problem 3.5: Consider a bagel store selling three types of bagel s that are produced according to the process flow diagram outlined below. We assume the demand is 180 bagels a day, of which there are 30 grilled veggie, 110 veggie only, and 40 cream cheese. Assume that the workday is 10 hours long and each resource is staffed with one worker. Moreover, we assume the following Processing times:Step Processing time [min./bagel] Cut 3 Grilled Stuff 10 Veggies 5 Cream Cheese 4 Wrap 2 Processing times are independent of which bagel type is processed at a resource (for example, cutting a bagel takes the same time for a cream cheese bagel as for a veggie bagel). Where in the process is the bottleneck?
SESSION 5: SOLUTIONS OF THE PRACTICE PROBLEMS Textbook Problem 3.2. - Solution on Page 457 of the textbook (C&T). Please also try the pooled utilization approach discussed in class and match your answers! Textbook Problem 3.5 - Calculate pooled utilization for each resource. For example, Cut resource faces demand of grilled veggies, veggies and cream cheese. •Demand of grilled veggies = 30/day •Demand of veggies = 110/day •Demand of cream cheese = 40/day Compute the capacity towards each kind of product. •For grilled veggies, capacity is 200/day •For veggies, capacity is 200/day •For cream cheese, capacity is 200/day Pooled Utilization for cutting = 30/200 + 110/200 + 40/200 = 0.9 Similarly, compute the pooled utilization for each resource. Here the values: 1.Pooled Utilization for Cutting= 0.9 2.Pooled Utilization for Grilled stuff= 0.5 3.Pooled Utilization for Veggies on Bagel= 1.16 (70/60) 4.Pooled Utilization for Cream cheese= 0.2666 5.Pooled Utilization for Wrap= 0.6 The resource with highest pooled utilization is bottleneck. So, Veggies on Bagel is bottleneck.
SESSION 6 PRACTICE PROBLEMS Textbook Problem 8.1:Customers send e-mails to a help desk of an online retailer every 2 minutes, on average, and the standard deviation of the interarrival time is also 2 minutes. The online retailer has three employees answering e-mails. It takes on average 4 minutes to write a response e-mail. The standard deviation of the processing times is 2 minutes. a)Estimate the average customer wait before being served. b)How many e-mails would there be, on average, that have been submitted to the online retailer but not yet answered? Textbook Problem 8.2:My-law.com is a recent start-up trying to cater to customers in search of legal services who are intimidated by the idea of talking to a lawyer or simply too lazy to enter a law office. Unlike traditional law firms, My-law.com allows for extensive interaction between lawyers and their customers via telephone and the Internet. This process is used in the upfront part of the customer interaction, largely consisting of answering some basic customer questions prior to entering a formal relationship. In order to allow customers to interact with the firm’s lawyers, customers are encouraged to send e-mails to my-lawyer@My-law.com. From there, the incoming e-mails are distributed to the lawyer who is currently “on call.” Given the broad skills of the lawyers, each lawyer can respond to each incoming request. E-mails arrive from 8 A.M. to 6 P.M. at a rate of 10 e-mails per hour (coefficient of variation for the arrivals is 1). At each moment in time, there is exactly one lawyer “on call,”that is, sitting at his or her desk waiting for incoming e-mails. It takes the lawyer, on average, 5 minutes to write the response e-mail. The standard deviation of this is 4 minutes. a)What is the average time a customer has to wait for the response to his/her e-mail, ignoring any transmission times? Note: This includes the time it takes the lawyer to start writing the e-mail and the actual writing time. b)How many e-mails will a lawyer have received at the end of a 10-hour day? c)When not responding to e-mails, the lawyer on call is encouraged to actively pursue cases that potentially could lead to large settlements. How much time on a 10-hour day can a My-law.com lawyer dedicate to this activity (assume the lawyer can instantly switch between e-mails and work on a settlement)? To increase the responsiveness of the firm, the board of My-law.com proposes a new operating policy. Under the new policy, the response would be highly standardized, reducing the standard deviation for writing the response e-mail to 0.5 minute. The average writing time would remain unchanged. d)How would the amount of time a lawyer can dedicate to the search for large settlement cases change with this new operating policy? e)How would the average time a customer has to wait for the response to his/her e-mail change? Note: This includes the time until the lawyer starts writing the e-mail and the actual writing time.
Textbook Problem 8.3:The airport branch of a car rental company maintains a fleet of 50 SUVs. The interarrival time between requests for an SUV is 2.4 hours, on average, with a standard deviation of 2.4 hours. There is no indication of a systematic arrival pattern over the course of a day. Assume that, if all SUVs are rented, customers are willing to wait until there is an SUV available. An SUV is rented, on average, for 3 days, with a standard deviation of 1 day. a)What is the average number of SUVs parked in the company’s lot?b)Through a marketing survey, the company has discovered that if it reduces its daily rental price of $80 by $25, the average demand would increase to 12 rental requests per day and the average rental duration will become 4 days. Is this price decrease warranted? Provide an analysis! c)What is the average time a customer has to wait to rent an SUV? Please use the initial parameters rather than the information in part (b). d)How would the waiting time change if the company decides to limit all SUV rentals to exactly 4 days? Assume that if such a restriction is imposed, the average interarrival time will increase to 3 hours, with the standard deviation changing to 3 hours. Textbook Problem 8.4:The following situation refers to Tom Opim, a first-year MBA student. In order to pay the rent, Tom decides to take a job in the computer department of a local department store. His only responsibility is to answer telephone calls to the department, most of which are inquiries about store hours and product availability. As Tom is the only person answering calls, the manager of the store is concerned about queuing problems. Currently, the computer department receives an average of one call every 3 minutes, with a standard deviation in this interarrival time of 3 minutes. Tom requires an average of 2 minutes to handle a call. The standard deviation in this processing time is 1 minute. The telephone company charges $5.00 per hour for the telephone lines whenever they are in use (either while a customer is in conversation with Tom or while waiting to be helped). Assume that there are no limits on the number of customers that can be on hold and that customers do not hang up even if forced to wait a long time. a)For one of his courses, Tom has to read a book (The Pole, by E. Silvermouse). He can read 1 page per minute. Tom’s boss has agreed that Tom could use his idle time for studying, as long as he drops the book as soon as a call comes in. How many pages can Tom read during an 8-hour shift? b)How long does a customer have to wait, on average, before talking to Tom? c)What is the average total cost of telephone lines over an 8-hour shift? Note that the department store is billed whenever a line is in use, including when a line is used to put customers on hold.
SESSION 6: SOLUTIONS OF THE PRACTICE PROBLEMS Textbook Problem 8.1:Solution is available on the back of the textbook. Kindly see page 463. Textbook Problem 8.2: (a)inter-arrival time: 10 emails per hour = 1 email / 6 min a = 6 min, CVa= 1 processing time: p = 5 min, CVp= 4 min / 5 min = 0.8 waiting time = 5 min * [5/6 / (1-5/6)] * [(12+ 0.82)/2] = 20.5 min total response time = waiting time + processing time = 25.5 min(b)emails per day = 10 emails per hour * 10 hours = 100 emails per day(c)idle time = (1- utilization) * 10 hours = 1.66 hours(d)The average amount of time would not change, since utilization is not dependent on the variance or standard deviation, but on the average processing and inter-arrival time. (e)processing time: p = 5 min, CVp= 0.5 min / 5 min = 0.1 waiting time = 5 min * [5/6 / (1-5/6)] * [(12+ 0.12)/2] = 12.63 min total response time = waiting time + processing time = 17.63 minTextbook Problem 8.3: (a)We approach this problem as though the rental car is the “server”. We know that a = 2.4 hours, p = 72 hours, CVa= (2.4/2.4) = 1, CVp= (24/72) = 0.33, and m = 50 cars. To determine the number of cars on the lot, we can look at the utilization rate of our “servers” = (1/a) / (m/p) = (1/2.4) / (50/72) = 60%. Therefore, on average 60% of the cars are in use or 30 cars, so on average 20 carsare in the lot. (b)We assume that the standard deviations DO NOT change. If the average demand is increased to 12 rentals per day, then a = 2 hours. If the average rental duration increases to 4 days, then p = 96 hours. These values raise the utilization rate to (1/2) / (50/96) = 96%. This means that 48 cars are rented on average. With the initial rate average revenue per day = $80*30 cars = $2400. With the proposed rate average revenue per day = $55*48 = $2640. Therefore, the company should make the proposed changes. (c)Usingthe wait time formula, the average wait time = 0.019 hours or 1.15 minutes.(d)Using the wait time formula the average wait time is computed as 0.046 hours. Textbook Problem 8.4: (a)idle time = 1 min / 3 min * 8 hours = 160 min pages read = 160 min * 1 page/min = 160 pages(b)inter-arrival time: 1 call / 3 min a = 3 min, CVa= 3 min / 3 min = 1
processing time: p = 2 min, CVp= 1 min / 2 min = 0.5 waiting time = 2 min * [2/3 / (1-2/3)] * [(12+ 0.52)/2] = 2.5 min (c)The average total time a line is used per customer = average wait time + average processing time. In this case, the average total time per customer = 2.5 + 2 = 4.5 minutes per customer. There are an average of 20 customers per hour, so the average number of minutes per hour = 20*4.5 = 90 minutes. Thus, the total per hour charge = (90/60) * $5 = $7.50 per hour or $60 for 8 hours. Another way to approach the same problem is to look at the average number of callers at any given time = average number of callers on hold at any given time (Iq) + average number of callers talking to Tom at any given time. We can calculate Iq= R * Tqwhere the flow rate = 1 call / 3 minutes and Tq= 2.5 minutes. Thus, Iq= 0.83 calls. The average number of callers talking to Tom must be a number between 0 and 1, and is equal to Tom’s utilization = 0.67. So, the average number of callers at any given time = 0.83 + 0.67 = 1.5 callers. The line charge for 8 hours = $5*8 = $40 per line. Therefore the total cost over an 8-hour shift = 1.5*40 = $60.
SESSION 8 PRACTICE PROBLEMS Problem 1 Sam's Cat Hotel operates 52 weeks per year, 6 days per week, and uses a continuous review inventory system. It purchases kitty litter for $11. 70 per bag. The following information is available about these bags. Demand = 90 bags/week Order cost = $ 54 / order Annual holding cost = 27 percent of cost Desired cycle-service level = 80 percent Lead time= 3 weeks (18 working days) Standard deviation of weekly demand = 15 bags Current on-hand inventory is 320 bags, with no open orders or backorders. a.What is the EOQ? What would be the average time between orders (in weeks)? b.What should ROP be? c.An inventory withdrawal of 10 bags was just made. Is it time to reorder? Problem 2 Wood County Hospital consumes 1,000 boxes of bandages per week. The price of bandages is $35 per box, and the hospital operates 52 weeks per year. The cost of processing an order is $15, and the cost of holding one box for a year is 15 percent of the value of the material. a.The hospital orders bandages in lot sizes of 900 boxes. What extra cost does the hospital incur, which it could save by using the EOQ method? b.Demand is normally distributed, with a standard deviation of weekly demand of 100 boxes. The lead time is 2 weeks. What safety stock is necessary if the hospital uses a continuous review system and a 97 percent cycle-service level is desired? What should be the reorder point?
Problem 3 (EOQ Model with volume discount) The Bucks Grande exhibition baseball team play 50 weeks each year and uses average of 350 baseballs per week. The team orders baseballs from Coopers-Town, lnc., a ball manufacturer noted for six-sigma-level consistency and high product quality. The cost to order baseballs is $100 per order and the annual holding cost per ball is 38 percent of the purchase price. Coopers-Town's price structure is: Order Quantity Price per Unit 1-999 $7.50 1,000-4999 $7.25 5,000 or more $6.50 a.How many baseballs should the team buy per order? b.What is the total annual cost associated with the best order quantity?Kindly note:The above problem is to further enhance your understanding of EOQ Model. Please note a similar problem is also there in Moodle handout (Example 17.3).
SESSION 8: SOLUTION OF THE PRACTICE PROBLEM Problem 1 a. Economic order quantity d= 90/week D =(90 bags/week)(52 weeks/yr) = 4,680/yr S = $54 Price = $11.70 H = (27%)($11.70) = $3.16 37.949,15916.3$)54)($680,4(22===HDSEOQ= 399.93, or 400 bags. Time between orders, in weeks QD===4004680008547444..yearsweeksb. Reorder point, R R = demand during protection interval + safety stock Demand during protection interval (or mean demand during lead time) = dL= 90 * 3 = 270 bags Safety stock = zdLTWhen the desired cycle-service level is 80%, . LddLT== 153= 25.98 ~ 26 Safety stock = 0.84 * 26 = 21.82 ~ 22 bags z=084.R=+=27022292
c.Initial inventory position = 320 –10 = 310. Because inventory position remains above 292, it is not yet time to place an order. Problem 2 a. D= (1000 boxes/wk)(52 wk/yr) = 52,000 boxes H= (0.l5)($35/box)=$5.25/box 900545290052,000$5.25$15.00$3,229.16290054552,000$5.25$15.00$2,861.822545QDCHSQCC=+=+==+=The savings would be $3,229.16 –$2,861.82 = $367.34. b. When the cycle-service level is 97%, z= 1.88. Therefore, Safety stock = Lzd= (1.88)(100)2= 1.88(141.42) = 265.87, or 266 boxes R = dL + Safety stock = 1000(2) + 266 = 2,266 boxes Problem 3 Step 1:Calculate the EOQ at the lowest price ($6.50): 22(350)(50)(100)1190.41190.38(6.5)DSEOQorballsH===This solution is infeasible. We cannot buy 1190 balls at a price of $6.50 each. Therefore we calculate the EOQ at the next lowest price ($7.25): 22(350)(50)(100)1127.131127.38(7.25)DSEOQorballsH===This solution is feasible. ()()2 52,000$152545.1 or 545 boxes$5.25DSEOQH===
Step 2:Calculate total costs at the feasible EOQ and at higher discount quantities: 112711275000500021,12717,500[.38($7.25)]($100)$7.25(17,500)21,127$1,552.44$1,552.79$126,875.00$129,980.235,00017,500[.38($6.50)]($100)$6.50(17,500)25,000$6,175.00$350.00$11QDCHSPDQCCCC=++=++=++==++=++3,750.00$120,275.00=a.It is less costly on an annual basis to buy 5,000 baseballs at a time. b.The total annual costs associated with buying 5,000 balls at a time are $120,275.00
SESSION 9 PRACTICE PROBLEMS Textbook Problem 12.1Dan McClure owns a thriving independent bookstore in artsy New Hope, Pennsylvania. He must decide how many copies to order of a new book, Power and Self- Destruction, an exposé on a famous politician’s lurid affairs. Interest in the book will be intense at first and then fizzle quickly as attention turns to other celebrities. The book’s retail price is $20 and the wholesale price is $12. The publisher will buy back the retailer’s leftover copies at a full refund, but McClure Books incurs $4 in shipping and handling costs for each book returned to the publisher. Dan believes his demand forecast can be represented by a normal distribution with mean 200 and standard deviation 80. What order quantity maximizes Dan’s expected profit? Textbook Problem 12.4Flextrola, Inc., an electronics systems integrator, is planning to design a key component for their next-generation product with Solectrics. Flextrola will integrate the component with some software and then sell it to consumers. Given the short life cycles of such products and the long lead times quoted by Solectrics, Flextrola only has one opportunity to place an order with Solectrics prior to the beginning of its selling season. Flextrola’s demand during the season is normally distributed with a mean of 1,000 and a standard deviation of 600. Solectrics’ production cost for the component is $52 per unit and it plans to sell the component for $72 per unit to Flextrola. Flextrola incurs essentially no cost associated with the software integration and handling of each unit. Flextrola sells these units to consumers for $121 each. Flextrola can sell unsold inventory at the end of the season in a secondary electronics market for $50 each. The existing contract specifies that once Flextrola places the order, no changes are allowed to it. Also, Solectrics does not accept any returns of unsold inventory, so Flextrola must dispose of excess inventory in the secondary market. Under this contract, how many units should Flextrola order to maximize its expected profit? Textbook Problem 12.6Teddy Bower is an outdoor clothing and accessories chain that purchases a line of parkas at $10 each from its Asian supplier, TeddySports. Unfortunately, at the time of order placement, demand is still uncertain. Teddy Bower forecasts that its demand is normally distributed with mean of 2,100 and standard deviation of 1,200. Teddy Bower sells these parkas at $22 each. Unsold parkas have little salvage value; Teddy Bower simply gives them away to a charity. How many parkas should Teddy Bower buy from TeddySports to maximize expected profit?
Textbook Problem 12.8Geoff Gullo owns a small firm that manufactures “Gullo Sunglasses.” He has the opportunity to sell a particular seasonal model to Land’s End. Geoff offers Land’s End two purchasing options:•Option 1:Geoff offers to set his price at $65 and agrees to credit Land’s End $53 for each unit Land’s End returns to Geoff at the end of the season (because those units did not sell). Since styles change each year, there is essentially no value in the returned merchandise. •Option 2:Geoff offers a price of $55 for each unit, but returns are no longer accepted. In this case, Land’s End throws out unsold units at the end of the season. This season’s demand for this model will be normally distributed with mean of 200 and standard deviation of 125. Land’s End will sell those sunglasses for $100 each. Geoff ’s production cost is $25. a)How much would Land’s End buy if they chose option 1?b)How much would Land’s End buy if they chose option 2?Teddy Bower Once More! Teddy Bower sources a parka from an Asian supplier for $10 each and sells them to customers for $22 each. Leftover parkas at the end of the season have no salvage value. The demand forecast is normally distributed with mean 2,100 and standard deviation 1,200. Now suppose Teddy Bower found a reliable vendor in the United States that can produce parkas very quickly but at a higher price than Teddy Bower’s Asian supplier. Hence, in addition to parkas from Asia, Teddy Bower can buy an unlimited quantity of additional parkas from this American vendor at $15 each after demand is known.
SESSION 9: SOLUTION OF THE PRACTICE PROBLEM Textbook Problem 12.1See solution of Q12.1(d) on Page 464 at the back of the textbook. Textbook Problem 12.4Find the expected profit maximizing order quantity, first identify the underage and overage costs. The underage cost is 4972121=−=uC, because each lost sale costs Flextrola its gross margin. The overage cost is 225072=−=oC, because each unit of leftover inventory can only be sold for 50. Now evaluate the critical ratio 6901.0492249=+=+uouCCC. Lookup the critical ratio in the Standard Normal Distribution Function Table: 6879.0)49.0(=and 6915.0)50.0(=, so choose z= 0.50. Now convert the z-statistic into an order quantity: 13006005.01000=+=+=zQ. Textbook Problem 12.6To determine the profit maximizing order quantity, begin with the underage cost, 121022=−=uC, and the overage cost, .10010=−=oCThe critical ratio is 5455.0)1210/(12=+. We see from the Standard Normal Distribution Function Tablethat 5438.0)11.0(=and 5478.0)12.0(=, so we choose 12.0=z. Convert that z-statistic back into an order quantity, .244,2120012.02100=+=+=zQTextbook Problem 12.8a)With option 1 Land’s End sales price is $100, purchase cost is $65 and salvage value is $53 (because Geoff buys back unsold glasses for $53). So the underage cost is uC=100 –65 =35 and the overage cost is 125365=−=oC. The critical ratio is /()uouCCC+= 35/47 = 0.7422. From the Standard Normal Distribution Function Tablewe see 7422.0)65.0(=and 7454.0)66.0(=, so we choose .66.0=zThe optimal order quantity is then 5.28212566.0200=+=+=zQ. b)With option 1 Land’s End sales price is $100, purchase cost is $55 and the salvage value is $0. So the underage cost is 4555100=−=uCand the overage cost is 55=oC. The critical ratio is /()uouCCC+= 45/100 = 0.4500. From the Standard Normal Distribution Function Tablewe see 4483.0)13.0(=−and 4522.0)12.0(=−, so we choose .12.0−=zThe optimal order quantity is then 18512512.0200=−=+=zQ. Teddy Bower Once More! The overage cost is oC=10 –0 = 10, because left over parkas must have been purchased in the 1storder at a cost of $10 and they have no value at the end of the season. The underage cost is uC=15 –10 = 5 because there is a $5 premium on units ordered from the American vendor. The critical ratio is 5 / (10 + 5) = 0.3333. From the Std Norm Dist Func Tablewe see that ( 0.44) −=0.3300 and ( 0.43) −=0.3336, so choose z = -0.43. Convert to Q: Q = 2100 –0.43 1200 = 1584.
SESSION 10 PRACTICE PROBLEMSProblem 1:Wood County Hospital consumes 1,000 boxes of bandages per week. The price of bandages is $35 per box, and the hospital operates 52 weeks per year. The cost of processing an order is $15, and the cost of holding one box for a year is 15 percent of the value of the material. If theDemand is normally distributed, with a standard deviation of weekly demand of 100 boxes. The lead time is 2 weeks. If the hospital uses a periodic review system, and a 97 percent cycle-service level is desired, with time interval between the orders(T) equals 2 weeks, what should be the target inventory level, S? Problem 2:Assume a car dealer that faces expected yearly demand of 5,000 cars per year, and that it costs $15,000 to have the cars shipped to the dealership. Holding cost is estimated at $500 per car per year. You estimate the standard deviation of daily demand to be σ =6. Further, the daily demand follows Normal Distribution. If the transportation arrangement requires that you place orders at a regular interval, what should be the time interval between the orders? What would then be your inventory policy? Assume that you would like to implement an in-stock probability of 95% . Assume replenishment lead time to be 10 days.
SESSION 10: SOLUTION OF THE PRACTICE PROBLEMProblem 1In a periodic review system, find target inventory S, given:T = 2 weeks L= 2 weeks safety stock (ss) = KSL* Sigma * SQRT (T+L) , where SQRT means square root SL = 97% KSL=1.88 𝑠𝑠 = 1.88 (100)√2 + 2= 376 units S= Average demand during the protection interval + Safety stock S= 1000(2 + 2) + 376 S= 4,376 units Problem 2We estimate the time between the order as follows: T = Q (from EOQ formula)/ D Now, D= 5000/yr , S = 15000 and h = 500. Q (EOQ) = √2𝐷𝑆ℎ= 548 T = Q (from EOQ formula)/ D = 40 days S = mean demand during T+L period + safety stock safety stock (ss) = KSL* Sigma * SQRT (T+L), where SQRT means square root. L= 10, T= 40, Sigma = 6, SL =95% and KSL= 1.65 𝑠𝑠 = 1.65 (6)√10 + 40S= (10+40)(5000/365) + ss Kindly substitute and find the value of S.
Operations Management (2022) - Tutorial Session 2(Section B)Queuing Theory & Inventory Management (Sessions 6 to 10)Deterministic Demand: How much to order?A supplier to Toyota stamps out parts using a press.Changing a part type requires the supplier tochange the die on the press. This changeover currently takes four hours. The supplier estimates thateach hour spent on the changeover costs $250. Demand for parts is 1,000 per month. Each part coststhe supplier $100, and the supplier incurs an annual holding cost of 25%•Determine the optimal production batch size for the supplier.Stochastic Demand - Q Model (< Q, ROP >policy)Assume a car dealer that faces expected yearly demand of 5,000 cars per year, and that it costs $15,000to have the cars shipped to the dealership. Holding cost is estimated at $500 per car per year. Youestimate the standard deviation of daily demand to beσ= 6. Assume that the daily demand followsNormal Distribution. When should you re-order under continous review policy model if you want to be95 sure you don’t run out of cars? What should be the order size? Assume replenishment lead time tobe 10 days.Stochastic Demand - P Model (< S, T >policy)Assume a car dealer that faces expected yearly demand of 5,000 cars per year, and that it costs $15,000to have the cars shipped to the dealership. Holding cost is estimated at $500 per car per year. Youestimate the standard deviation of daily demand to beσ= 6. Further, the daily demand follows NormalDistribution. If the transportation arrangement requires that you place orders at a regular interval, whatshould be the time interval between the orders? What would then be your inventory policy? Assumethat you would like to implement an in-stock probability of 95%. Assume replenishment lead time tobe 10 days.Capacity Analysis under UncertaintyFirst Service Bank’s office in a suburb of London has retail operations with two tellers operating duringthe workday. The first teller specializes in day-to-day operations of the bank (related to check book en-tries, savings account information, bank drafts etc.), while the second teller deals only with applicationsfor credit and loans. Service times for both tellers are exponential, the first teller takes 10 minutes on1
average to serve a customer, while the second teller takes 30 minutes on average to serve a customer.For both the tellers, assume that coefficient of variations for both inter-arrival and processing times =1.The demand for the first teller is 5 customers per hour on average, and is 3 customers every twohours for the second teller on average.1. Compute the mean time spent in the system (queue + server) for a day-to-day operations customer.Also find the average number of day-to-day operations customer in the system.2. Compute the mean time spent in the system for a credit and loans customer. Also find the averagenumber of credit and loans customer in the system.Newsvendor Model Application: Book publishing problemA Yamuna publisher sells books to Ganga Ram book store at $10 each. Ganga Ram sells book storesells the book at $20. The demand of book store is normally distributed with a mean of 10,000 and astandard deviation of 2,500. Book store places a single order with the publisher. Any unsold books atthe end of selling period is sold at discount $5 and any unsold books sell at this price. Assume the unitproduction cost of Yamuna publisher to be $2.•How many books should Ganga Ram book store order?Newsvendor Model Application:DVD rental supply chainsettingA Fox-mount movie studio produces and distributes a movie DVD at unit cost of $2. The studio sellthe DVD to Murugan records at $4. Murugan records finally sells the DVD to customers at $20. Theconsumer demand is uncertain and follows Normal distribution with mean 6000 and standard deviationof 2,000. Any unsold DVD are discounted to $0.5, and (all DVD sell at this price). Further, Muruganrecords also agrees to share 35% of the revenue with the Fox-mount studio (keeping 65% for itself).•How may DVDs should Murugan records order?Please also practice Tutorial 1 problems towards Sessions 1 to 5. All the best for Midterm exam !!!2
Solutions to Tutorial 2 Problems Problem 1: EOQ problem: Changeover time = 4hrs resulting in a fixed cost, S = 4 × 250 = $1,000. Annual demand, S = 1000/mo × 12 = 12,000 units / yr. Unit cost, C = 100 Holding cost = H = $25 / unit / yr. The optimal production batch size can be solved using EOQ formulla = 980 Problem 2: Q Model Problem: To compute the optimal order quantity we use the EOQ formula. D= 5000/yr , S (Ordering Cost) = 15000 and H = 500. Q (EOQ) = 548 The replenishment lead time (L) = 10 weeks. Safety stock for 95% level of service = KSL * Sigma * SQRT (L), where SQRT means square root L= 10, Sigma = 6, SL =95% and KSL = 1.65. The safety stock is given by 31 Reorder point (ROP)= (mean demand during time L) + Safety stock = 5000/365*10 + 31 = 168. Problem 3: <S,T> model problem or P system problem T = Q (from EOQ formula)/ D Now, D= 5000/yr , S (Ordering Cost) = 15000 and H = 500. Q (EOQ) = 548 T = Q (from EOQ formula)/ D = 40 days S (Order upto level) = mean demand during T+L period + safety stock safety stock = KSL * Sigma * SQRT (T+L), where SQRT means square root. L= 10, T= 40, Sigma = 6, SL =95% and KSL = 1.65 substitute and find values of S (Order upto level). You get the value of S (Order upto level) equals to 755. Problem 4: (a) For day to day operations : Demand = 5/hr, Capacity = 6/hr, m =1, Ca = 1, Cs = 1. Use Queuing formula, you will get the waiting time in system to be 1 hour for day to day operations. Apply Little's law you get average number of persons in system to be 5/hr * 1 hr = 5.
(b) For credit and loan operations : Demand = 1.5/hr, Capacity = 2/hr, m =1, Ca = 1, Cs = 1. Use Queuing formula, you will get the waiting time in system to be 2 hour for credit and loan operations. Apply Little's law you get average number of persons in system to be 1.5/hr * 2 hr = 3. Problem 5: Apply Newsvendor Formulla: Cu = $10 ; Co = $5 , F(Q) = 0.6667, z = 0.43, Q = 11075. Problem 6: Apply Newsvendor Formulla: Cu = $9 ; Co = $3.5 , F(Q) = 0.72, z = 0.58, Q = 7160.
SESSION 13 PRACTICE PROBLEMMaterial Requirements Planning (MRP) problemThe MPS start quantities for product A calls for the assembly department to begin final assemblyaccording to the following schedule: 100 units in week 2; 200 units in week 4; 120 units in week 6;180 units in week 7; and 60 units in week 8. Develop a material requirements plan for the next 8weeks for items B,C, and D. One unit of A is composed of one unit of B and two units of C. Eachunit of C is composed of one unit of D. The inventory records data is as follows:ItemsData CategoryABCDLot sizing rulelot-for-lot3 periodslot-for-lotlot-for-lotLead Timezero1 week2 weeks3 weeksScheduled ReceiptsNoneNone200 (in week 1)NoneBeginning on-hand inventory0200425Table 1: Inventory Records Data•Develop MRP plan for items A, B, C and D.1
SESSION 12: SOLUTION OF THE PRACTICE PROBLEMFor Part A, here are the planned order releases: •Period 1 = 0 , Period 2 = 100, Period 3 = 0, Period 4 = 200, Period 5 = 0, Period 6 = 120, Period 7 = 180 and Period 8 = 60 For Part B, here are the planned order releases: •Period 1 = 280, Period 2 = 0, Period 3 = 0, Period 4 = 0, Period 5 = 360, Period 6 = 0, Period 7 = 0 and Period 8 = 0 For Part C, here are the planned order releases: •Period 1 = 0, Period 2 = 400, Period 3 = 0, Period 4 = 240, Period 5 = 360, Period 6 = 120, Period 7 = 0 and Period 8 = 0 For Part D, here are the planned order releases: •Period 1 = 215, Period 2 = 360, Period 3 = 120, Period 4 = 0, Period 5 = 0, Period 6 = 0, Period 7 = 0 and Period 8 = 0
SESSION 15 PRACTICE PROBLEMS1.A quality assurance manager is assessing the capability of a process that fills pressurized grease in an aerosol can. The design specification calls for 60 pounds per inch (psi) pressure in each can with upper and lower specification (tolerance) limits as 65 and 55 psi, respectively. It is found that the process pressure follows Normal Distribution with average of 61 psi and std deviation of 2 psi. Calculate process capability. What is the probability of producing a defect? 2.The state and city police departments are trying to analyze crime rates so they can shift their patrols from decreasing-rate areas to areas where rates are increasing. The city and country have been geographically segmented into areas containing 5000 residences. The police recognize that not all crimes and offences are reported for various reasons. Every month, because of this, the police are contacting by phone a random sample of 1000 of the 5000 residences for data on crime. Here are past 12 months’ data: Month Crimes Sample Size Crime Rate January,2015 7 1000 0.007 February,2015 9 1000 0.009 March,2015 7 1000 0.007 April,2015 7 1000 0.007 May,2015 7 1000 0.007 June,2015 9 1000 0.009 July,2015 7 1000 0.007 August,2015 10 1000 0.01 September,2015 8 1000 0.008 October,2015 11 1000 0.011 November,2015 10 1000 0.01 December,2015 8 1000 0.008 Construct a p chart. If the next 3 months in 2016 show following crime (in sample size 1000), what is your comment? Jan = 10, Feb = 12, March = 11.
3.The following table contains the measurements of the key length dimensions from a fuel injector. These samples of size 5 were taken at 1-hour intervals. Construct a X-bar chart and R chart for the length of fuel injector. What is observation about the process? Sample number 1 2 3 4 5 Mean Range 1 .486 .499 .493 .511 .481 .494 .030 2 .499 .506 .516 .494 .529 .509 .035 3 .496 .500 .515 .488 .521 .504 .033 4 .495 .506 .483 .487 .489 .492 .023 5 .472 .502 .526 .469 .481 .490 .057 6 .473 .495 .507 .493 .506 .495 .034 7 .495 .512 .490 .471 .504 .494 .041 8 .525 .501 .498 .474 .485 .497 .051 9 .497 .501 .517 .506 .516 .507 .020 10 .495 .505 .516 .511 .497 .505 .021 11 .495 .482 .468 .492 .492 .486 .027 12 .483 .459 .526 .506 .522 .499 .067 13 .521 .512 .493 .525 .510 .512 .032 14 .487 .521 .507 .501 .500 .503 .034 15 .493 .516 .499 .511 .513 .506 .023 16 .473 .506 .479 .480 .523 .492 .050 17 .477 .485 .513 .484 .496 .491 .036 18 .515 .493 .493 .485 .475 .492 .040 19 .511 .536 .486 .497 .491 .504 .050 20 .509 .490 .470 .504 .512 .497 .042
SESSION 15: SOLUTION OF THE PRACTICE PROBLEMProblem 1 LSL = 55 and USL = 65 Mean of process = 61 and sigma = 2 The process is not mean centered, we compute the value of Cpk: Cpk = min{(Mean of process - LSL)/(3*sigma), (USL –Mean of process)/(3*sigma)} = {(61-55)/3(2), (65-61)/3(2))} = 0.67 Defect: = pressure less than 55 or greater that 65 psi. For pressure < 55 psi Z= (x –X_bar)/sigma = (55-61)/2 = -3, => find probability from z-table. Again, for pressure > 65 psi Z = (x –X_bar)/sigma = (65 - 61)/2 = 2, => find probability from z-table. Prob {defect} = Prob{ less than 55 } + Prob{ greater that 65 psi } = 0.0241 = 2.4%
Problem 2: Find the pvalues: Month Crimes Sample Size p value January 7 1000 0.007 February 9 1000 0.009 March 7 1000 0.007 April 7 1000 0.007 May 7 1000 0.007 June 9 1000 0.009 July 7 1000 0.007 August 10 1000 0.01 September 8 1000 0.008 October 11 1000 0.011 November 10 1000 0.01 December 8 1000 0.008 SUM: 100 12000 )1000(12100=p= 0.008333 1000)008333.1(008333.)1(−=−=nppSp= 0.00287 UCL = p+ 3pS= .008333 + 3 (.00287) LCL = p- 3pS= .008333 - 3(.00287) Plot pvalues in UCL and LCL. The process appears in control. Therefore, it can be stated that the crime rate has not increased. However, there appears to be a gradual increasing trend in the crime rate in later data, including the new Jan –Mar figures given in the problem. That may warrant investigation into the cause, or at the least should be watched to confirm if there is indeed a trend that continues in future periods.
Problem 3: =X0.499, R= 0.037 Control limits for X-barchart: UCL, LCL = RAX2+, = 0.499 + .58(.037) = 0.520, 0.478 Control limits for Rchart: UCL = RD4= 2.11(.037) = 0.078 LCL = RD3= 0(.037) = 0.00 Here are the X-barand Rcharts: Process appears to be in statistical control.
SESSION 16 PRACTICE PROBLEMS Tire Manufacturing Scenario Given the large number of steps in building a tire, errors tended to accumulate. As a result, even if each step produced only 1 % defects, at the end of 20 steps only 81.2% of the product would be of good quality. A tire plant produced about 10,000 tires per hour; such a process would result in about 1,880 defective tires per hour. Thus, it was very important that each stage had a high level of quality. Another quality issue was related to the settings on various machines. Over time, these settings tended to vary because of wear and tear on the machines. In such a situation, a machine would produce defective products even if the machine had the correct setting. To detect such situations, the quality manager implemented a SPC (statistical process control) program. At the extruder, the rubber for the tires had thickness specifications of 400 +/- 10 thousandth of an inch (thou). The Quality Manager and her staff had analyzed many samples of output from the extruder and determined that if the extruder settings were accurate, the output produced by the extruder had a thickness that was normally distributed with a mean of 400 thou and a standard deviation of 4 thou. For some questions you will have to use Excel Spreadsheet (because normal distribution tables do generally go beyond 3 or 4 sigma) 1.If the extruder setting is accurate, what proportion of the rubber extruded will be within specifications? 2.The Quality Manager (QM) had asked operators to take a sample of 10 sheets each hour from the extruder and measure the thickness of each sheet. Based on the average thickness of this sample, operators will decide whether the extrusion process is in control or not. For a three-sigma control limits, what upper and lower control limits should QM specify to the operators? 3.If a bearing is worn out, the extruder produces a mean thickness of 403 thou when the setting is 400 thou. Under this condition, what proportion of defective sheets will the extruder produce? Assuming the control limits in question 2, what is the probability that a sample taken from the extruder with the worn bearings will be out of control? 4.Now consider a case where the extrusion is a six-sigma process. In this case the extruder output should have a standard deviation of 1.667 thou. What proportion of the rubber extruded will be within specifications in this case? 5.Assuming the operators will continue to take samples of 10 sheets each hour to check if the process is in control, what control limits should the QM set for the case when extrusion when the extrusion is a six sigma process?
SESSION 16: ANSWERS OF THE PRACTICE PROBLEMTire Manufacturing Scenario For some questions you will have to use Excel Spreadsheet (because normal distribution tables do generally go beyond 3 or 4 sigma) 1.If the extruder setting is accurate, what proportion of the rubber extruded will be within specifications? 2.The Quality Manager (QM) had asked operators to take a sample of 10 sheets each hour from the extruder and measure the thickness of each sheet. Based on the average thickness of this sample, operators will decide whether the extrusion process is in control or not. For a three-sigma control limits, what upper and lower control limits should QM specify to the operators? 3.If a bearing is worn out, the extruder produces a mean thickness of 403 thou when the setting is 400 thou. Under this condition, what proportion of defective sheets will the extruder produce? Assuming the control limits in question 2, what is the probability that a sample taken from the extruder with the worn bearings will be out of control? 4.Now consider a case where the extrusion is a six-sigma process. In this case the extruder output should have a standard deviation of 1.667 thou. What proportion of the rubber extruded will be within specifications in this case? 5.Assuming the operators will continue to take samples of 10 sheets each hour to check if the process is in control, what control limits should the QM set for the case when extrusion when the extrusion is a six sigma process? Answers: 1.98.8% of output within specs 2.403.795 and 306.205 3.73.51% chance that a sample will be within control limits; 3.77 samples (on average) will be tested before a out-of-control sample is detected 4.0.999999996 (use excel spreadsheet) 5.401.581 and 398.419
Operations Management (2022) - Tutorial Session 3Section BOperations Planning, Quality Management, and some RecapMaterial Requirements Planning (MRP) problemThe MPS quantities for product A is given by: 100 units in week 2; 200 units in week 4; 120 units in week 6;180 units in week 7; and 60 units in week 8. Develop a material requirements plan for the next 8 weeks foritems B,C, and D. One unit of A is composed of one unit of B and two units of C. Each unit of C is composedof one unit of D. The inventory records data is as follows:ItemsData CategoryABCDLot sizing rulelot-for-lot3 periodslot-for-lotlot-for-lotLead Timezero1 week2 weeks3 weeksScheduled ReceiptsNoneNone200 (in week 1)NoneBeginning on-hand inventory0200425Table 1: Inventory Records Data•Develop MRP plan for items A, B, C and D.A problem on Process Capability Index and Defect rates estimation!A quality assurance manager is assessing the capability of a process that fills pressurized grease in an aerosolcan.The design specification calls for 60 pounds per inch (psi) pressure in each can with upper and lowerspecification (tolerance) limits as 65 and 55 psi, respectively. The process output follows Normal distributionsuch that pressure is 61 psi on average with std deviation 2 psi.•Calculate the process capability index.•Calculate the sigma measure of the process.•What is the probability of producing a defect?A problem on Process Control!A production manager at a tire manufacturing plant has inspected the number of defective tires in twentyrandom samples with twenty observations each.Following are the number of defective tires found in eachsample:1
Sample NumberNumber of Defective Tires132231425163738291102113122132141151162174183191201•Construct a control chart with this information? Is the tire manufacturing process in control?An Inventory Mangement Problem!Centura Health hospital incurs a cost of $130 regardless of the quantity purchased each time it places an orderfor and receives IV Starter Kits. Hence, fixed cost per order is $130. The unit cost of each kit is $3. The weeklydemand of 600 translates to an annual demand of 31200 per year, assuming 52 weeks per year. It is estimatedthat the Centura’s inventory holding cost is $0.90 per unit per year.•What is the optimal order size for Centura to order IV Starter Kits?•What is the annual holding cost for Centura?•What is the annual ordering cost for Centura?All the best for End Term exam- Do very well !!!2
•Tutorial problem 1: For Part A, here are the planned order releases: Period 1 = 0 , Period 2 = 100, Period 3 = 0, Period 4 = 200, Period 5 = 0, Period 6 = 120, Period 7 = 180 and Period 8 = 60. For Part B, here are the planned order releases: Period 1 = 280 , Period 2 = 0, Period 3 = 0, Period 4 = 0, Period 5 = 360, Period 6 = 0, Period 7 = 0 and Period 8 = 0. For Part C, here are the planned order releases: Period 1 = 0 , Period 2 = 400, Period 3 = 0, Period 4 = 240, Period 5 = 360, Period 6 = 120, Period 7 = 0 and Period 8 = 0. For Part D, here are the planned order releases: Period 1 = 215 , Period 2 = 360, Period 3 = 120, Period 4 = 0, Period 5 = 0, Period 6 = 0, Period 7 = 0 and Period 8 = 0. •Tutorial Problem 2: LSL = 55, USL = 65, X_bar (mean of the process) = •61, sigma (SD of the process)= 2, Cpk = min{ (X_bar - LSL)/(3*sigma), (USL –X_bar)/(3*sigma) } = {(61-55)/3(2), (65-61)/3(2))} = 0.67. Sigma Measure= min{ (X_bar - LSL)/ sigma), (USL –X_bar)/ sigma) } = 2 Defect: = pressure less than 55 or greater that 65 psi. For pressure < 55 psi, Z= (x –X_bar)/sigma = (55-61)/2 = -3, => find probability from z-table. Again, for pressure > 65 psi Z = (x –X_bar)/sigma = (65 - 61)/2 = 2, => find probability from z-table. Prob {defect} = Prob{ less than 55 } + Prob{ greater that 65 psi } = 0.0241 = 2.4% •Tutorial Problem 3: Use p chart formullas, p_bar = 0.1, UCL = 0.301, LCL= 0, and draw the control chart diagram. You will see, process is in control. •Tutorial Problem 4: Use EOQ formulla, Q = 3002, annual holding cost = $1351, annual ordering cost = $1351.
MPSPeriodI23Y5b98100150stywoodShutterLT=1IParstructure-TIW.s/4)Frame(2)LT=1L=2&ReportPeriodI23YI6T8Item:ShutterGrossrequent100150feedtime:Iweekscheduledreceipton-handIrv:0on-handsafetystock:0Netrequent100150&&:Lot-for-LotI↑lannedorderreceiptI100⑯150↑lannedorderrelease100150&ReportItem:frame100X2150X2Grossrequent=200=300On-handIrv:0I-PeriodI23YI6T8feedtime:Iwakscheduledreceipton-handsafetystock:ONetrequent200300&:L-For-Lot↑lannedorderreceipt200300a-↑lannedorderrelease200300①
&ReportPeriod123456T8Item:WoodSection,100x4150x4=600feadtime:twakregantceiptcot400on-handIrv:0on-handTo7010safetystock:0Netrequent330600&&:Lot-for-Lot↑lannedorderreceipt.330600↑lannedorderrelease330600&ReportPeriod45678Item:Frame150x2Grossregant=300week-feedtime:2Scheduledreceipt-on-handIrv:0on-hand140-NetSoonareaabooggointe=320MRPReport-Period123456T8#ene=woodsection18x4*ant10=600feedtime:IweekscheduledreceiptTOon-handIrv:0on-handTo70to2020202050safetystock:0&Netrequent330580&:Lotsize:70↑lannedorderreceipt350638↑lannedorderrelease350630