OIDD101 - Spring 2016 - R4 exercise - solutions

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Nov 24, 2024

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OIDD 101 - Spring 2016 Recitation 4 Exercise - Solutions Section # (201,202,…,214): _____________ Instructor (circle): Drobnicki Fu Goel Lee Squire Tschupp PennCard Last Name: PennCard First Name: PennID: Place your answers in the boxes provided. Joe’s Beer, Bait, & Tackle Co. is a small chain of fishing tackle stores in northern Minnesota. Joe needs to purchase malt for his micro-­‐brew production. His supplier charges $50 per delivery (no matter how much is delivered) and $1.25 per gallon. Joe’s annual holding cost per unit is 20% of the dollar value of the unit. Joe uses 4000 gallons of malt per week. Assume 52 weeks per year. Q1. Suppose Joe orders 20,000 gallons of malt per order. On average, what is the dollar value of his malt inventory? Q2. Calculate the weekly holding cost (in $s) for one gallon of malt. (HINT: Given we are trying to calculate the weekly holding cost that would enter into the EOQ equation, fixed costs are excluded.) Weekly holding cost per gallon (h) = (20%/52)*($1.25) = $0.004808 $0.004808 $12500 Average inventory = 20000/2=10000 gallons Average dollar value of inventory = 10000 * 1.25 =$12500
Q3. Use the EOQ model to find the number of gallons of malt per order which minimizes the total (ordering + inventory holding) cost per unit. Q4. Suppose the Joe orders 32,000 gallons of malt with each order. How many orders per year will he place on average? Q5. Cornucopia is a small chain (15 stores) that specializes in locally grown and organic food. They sell granola in large bins so that customer can purchase the quantity they wish and bring the granola home in their own containers. Granola is sold for $6 per lbs and weekly demand is 900 lbs. Their supplier charges $25 per delivery and $4 per lb. They will deliver in any quantity ordered. Cornucopia’s annual holding cost is 20% of the value of the inventory. (Recall, this means that holding $10 in inventory for one year would cost 0.2 x $10 = $2.) Say Cornucopia orders 450 lbs with each order. What is the total holding and ordering costs (in $s) incurred per unit? Assume 52 weeks per year. K = setup cost = $50 h = weekly holding cost per gallon = $0.004808 R = demand flow rate = 4000 gallons per week EOQ = sqrt (2 x K x R / h) = 9121 gallons 9121 Q = 32000 gallons R = 4000 gallons per week = 4000 x 52 weeks per year = 208000 per year Number of orders per year = R / Q = 208000 / 32000 = 6.5 6.5 K = setup cost = $25 R = demand flow rate = 900 per week h = 4 x 20%/52 = $0.01538/lb per week C(Q) = K * R / Q + (1/2) Q * h = $53.46 Ordering and holding costs per unit is C(Q)/R = 0.059 0.059
Q6. Consider the following production process with two stages: Step A can produce at the rate of 45 gallons per hour. Step B, when in production, produces at the rate of 50 gallons per hour. Step A can operate continuously, but Step B must shutdown for 60 minutes after producing 300 gallons. How many gallons per hour can this process produce? Q7. Now suppose the process described in Q6 is modified. In particular, the inventory buffer between steps A and B is removed. Hence, step A cannot produce while step B is shutdown. How many gallons per hour can the process now produce? A B A B 42.9 Step B produces 300 gallons in (300gallons/50gallons)+1hour=7hrs The rate of Step B 300/7=42.9 gallons/hr, which is less than the rate of Step A. The rate of the process is the rate of Step B, which is 42.9 gallons/hr. 39.1 As the buffer between A and B is removed, A cannot produce if B is shut down. As a result, the process alternates between production of 300 gallons and a shutdown for 1 hour. When in production, the process produces at the rate of A. Thus, it takes 300/45+1 = 7.67 hours to product 300 gallons. The capacity of this process is 300gallons/7.67hour = 39.13 gallons/hr.
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Q8. A technician assembles lens. Starting from parts, the technician can assemble a lens at the rate of 3 per hour. 80% of the newly assembled lens pass a calibration test and are then considered completed units. However, 20% of the newly assembled lens fail the calibration test and must be recalibrated by the technician. The technician requires 10 minutes to recalibrate a lens. 100% of recalibrated lens pass the subsequent calibration test and are then considered completed units. Assuming the technician prioritizes recalibration work (i.e., if there are lens to recalibrate, the technician recalibrates the lens rather than assembles new lens from parts), how many completed lens does the technician produce per hour (units/hr)? 2.72 At a rate of 3 per hour, the technician requires 60 / 3 = 20 mins per lens. 10 mins are required to recalibrate. So the average processing time is 0.8 x 20 + 0.2 * 30 = 22 min. The technician can therefore complete 60 / 22 = 2.72 lens per hour