case-study-group-mike-and-bike-operation-mgmt

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Fanshawe College *

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Case Study Group Mike and bike operation mgmt operation management (Fanshawe College) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Case Study Group Mike and bike operation mgmt operation management (Fanshawe College) Scan to open on Studocu Studocu is not sponsored or endorsed by any college or university Downloaded by Ly Thanh (lykienthanh586@gmail.com) lOMoARcPSD|33179253

Case Study Report Client: Mike and Bike Submited by Group No: 6 Thi Minh Anh Pham - 1032242 Nhu Quynh Pham - 0963020 Vraj Jignesh Patel - 1071535 Nimrta Kaur - 1036852 Lawrence Kinlin School of Business, Fanshawe College MGMT3069-01-22F OPERATIONS MANAGEMENT Professor: Seyed A. Goosheh Date: November 5, 2022. Downloaded by Ly Thanh (lykienthanh586@gmail.com) lOMoARcPSD|33179253

1. Synopsis and Challenges: (Thi Minh Anh Pham) Mike & Bike is a bike company in Toronto that constructs bikes for clients of large retailers like Walmart and Canadian Tire as well as internet providers. His business also does significant repairs to broken racing bikes and has a superb reputation among professional motorcyclists for its craftsmanship. Mike is happy with his shop's profits, but he occasionally becomes frustrated with the backlog of orders and the difficulty of swiftly and economically delivering the completed jobs. To commit to this contract, Mike and Tome have sought our group’s consultancy to help them with the following analysis and decisions below: - The sufficiency of the product line analysis based on the operational map - The average effective capacity analysis to represent the ability of meeting the demand - Determine the right amount and right time to order before the backlog of critical material. 2. Analysis and Solution 2.1. Operational map ( Như Quynh Pham) 2.2. The sufficiency of the product line analysis Downloaded by Ly Thanh (lykienthanh586@gmail.com) lOMoARcPSD|33179253

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To determine the sufficiency of the assembly and repair process, i will calculate the effective capacity of whole process: Station The effective capacity per hour (bike / h) Inspected crates 6 Assembly job Frame / Wheel 5 Top mount Men’s bike 6 Women’s bike 7.5 Children’s bike 10 Adjustment 12 Repair job Dismantling parts 6 Repaiarability Repair with replacing 6 Repair without replacing 12 Reinstall modules 4 Final check 6 So, according to the flow chart, there will be two bottlenecks that make the process of assembly and repair jobs slow down: Regarding the aassembly service, the bottleneck will located at Frame / Wheel station : 5 bikes / h. As you can see, the previous station: the workers can inspect the scrate with 6 bikes / h and the rest of station of the assembly jobs, they might complete at least 6 bikes / h. So, Frame / Wheel station has the low effective capacity in the process and in the other words, it is considered as bottleneck that makes the operation does not run effectively. In the similar way, the repair job has a bottleneck at the module reinstallation station with 4 bikes / h. Downloaded by Ly Thanh (lykienthanh586@gmail.com) lOMoARcPSD|33179253

As a result, there will be two bottleneck are at Frame / Wheel and the module reinstallation station with only 5 bikes / h and 4 bikes / h respectively. Therefore, one solution is suggested to speed up the productivity, at the those stations is that Mike can hire one more worker to help the other workers in the assembly job. 2.3. The average effective capacity analysis a. Average Capacity ( Thi Minh Anh Pham & Nhu Quynh Pham) According to the information of the case: Throughput time (minute) Assembly service Men’s bikes 47 Women’s bikes 45 Children’s bikes 43 Repair service Repair with replaced parts 50 Repair without replaced parts 55 Before, calculating the average capacity of the store in terms of customers per labour hour, we have to know the throughput time of the assembly process for finishing the Men’s bike, Women’s bike and Children’s bikes. Additionally, the throughput time for the repair service also needs to calculate. For the assembly process, the total number of bikes in customers' orders is 60/((47+45+43)/3), approximately 1.33 bikes/ labour hour. So this means the store labour can cover more than one customer’s order in one hour. For the repair process, the total number of bikes in customer order is 60/((50+55)/2). approximately 1.14 bikes/ labour hour. So this means if the customer's order is for repair service, one store labour can process more than one customer’s order. In conclusion, whether the customer order is the assembly service either or the repair service. The average capacity of the store in terms of customers per labour hour for both the repair and assembly services is: (1.33+1.14)/2, approximately 1.24 bike/ labour hour. Downloaded by Ly Thanh (lykienthanh586@gmail.com) lOMoARcPSD|33179253

2.4. Determine the right amount and right time to order To define when and how many items Mike needs to order before running out of vital parts in the repair jobs, there will be some measurements to calculate: - Actual annual cost before applying EOQ method - EOQ method to determine the mount to order suitably - Actual annual cost after applying EOQ method - Reorder point that allows Mike to point out when he must order 2.4.1. Actual annual ordering cost before calculating EOQ method (Nhu Quynh Pham & Vraj Jignesh Patel) The backlog level of parts at the Mike & Bike are different. In this case, i will take spacers like an example to analyze and specifically calculate the actual annual ordering cost before EOQ method. Mike and Bike use around 20 spacers every week. Each of order is placed every three weeks with the transportation cost is $60. The cost to keep these parts in an oil container is $10 once a year. Due to the Mike & Bike placed an order from the supplier every three week So, the annual cycle inventory cost is calculated by the formula: C = (Q/2)*H + (D/Q)*S with Q - Lot size 60 H - Holding cost per unit $10 D - Annual demand 20*(50 weeks)= 1000 (one year include 52 weeks and the Mike and Bike off on two week of Christmas event) S - Ordering cost per order $60 So, the annual cycle inventory cost is C = (60/2)*10 + (1000/60)*60 = $1,300.00 (1) 2.4.2. EOQ method Downloaded by Ly Thanh (lykienthanh586@gmail.com) lOMoARcPSD|33179253

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The method to determine the optimal order quantity to meet the demand while minimize the expenses used in this analysis is that EOQ method. The formula of EOQ method: EOQ = square root of ((2*D*S)/H) = square root of ((2*1000*60)/10) = 109.54 = 110 units per order 2.4.3. The annual cycle-inventory cost after EOQ method Lot size of this time is 110 units per order Q - Lot size 110 H - Holding cost per unit $10 D - Annual demand 20*(50 weeks)= 1000 S - Ordering cost per order $60 So, the annual cycle-inventory cost of 110 units C = (Q/2)*H + (D/Q)*S = (110/2)*10 + (1000/110)*60 = $1,095.46 (2) From (1) and (2), they show that the annual cycle-inventory cost of 110 units based on the EOQ method is better than annual cycle-inventory cost before applying EOQ method that helps Mike & Bike shop saves approximately $204.54 if Mike & Bike orders 110 spacers per order. 2.4.4. Reorder point According to the foreman of Mike & Bike shop, Mike pressures to not fall behind 5% finish jobs, so Mike wants the level of service is 95% that is approximately 1.64 according to Z-Factor. Moreover, 5 spacers are altered when they have to wait their order and receive new job, so the deviation is 5. Finally, there will be some specifications below: Level service policy 95% Z- factor 1.64 Standard deviation 5 Downloaded by Ly Thanh (lykienthanh586@gmail.com) lOMoARcPSD|33179253

Lead time 1 week Demand 20 spacer/week Standard deviation during the lead time = Deviation Demand * √ of Lead Time = 5 √1= 5 Safety stock = Z-Factor *Standard deviation during the lead time = 1.64*5 = 8.2 = 8 spacers Reorder point = dL + safety stock = 1* 20 + 8 = 28 spacers Therefore, If Mike & Bike does not want to run out of critical material to provide for repair service, Mike should order 110 spacers per order whenever the inventory position reaches 28 spacers in the warehouse. Safety stock helps Mike & Bike shop have available spacers against the variability of demand as well as maintain 95 percent of customer service. So instead of placing orders every 3 weeks now Mike & Bike can base on the EOQ method to re- define the time between orders TBO=EOQ/D= (110/1000)*50 weeks= 5.5 weeks 3.Conclusion and recommendations (Thi Minh Anh Pham) In order to address the recurring inventory issues, our specialists carefully considered Mike and Bike's production concerns before analysing the data and calculating the values above. To prevent stockouts of the parts required for their repair services, our advisors advised Mike and Tom to place fewer orders with the necessary quantities of inventory. Accordingly, we advise Mike to buy spacers in batches of 110. If this strategy is used, orders can be placed every five and a half weeks, which will save Mike $54.54 yearly because the business won't need to place orders every three weeks. For the determined demand Mike and Bike's shop experiences, we suggest that a reorder point of twenty-eight units is appropriate. Furthermore, since the one-week delivery window is still in operation, a safety stock of eight spacers is maintained on hand in case demand, lead time, or supply should unexpectedly grow. References: Downloaded by Ly Thanh (lykienthanh586@gmail.com) lOMoARcPSD|33179253

Ritzman, L., Krajewski, L., Maljoltra, M., & Klassen, R. (2016). Foundations of Operations Management . Fourth Canadian Edition. Pearson Canada Inc Downloaded by Ly Thanh (lykienthanh586@gmail.com) lOMoARcPSD|33179253

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