Problem 17-2Complete the following tableInstructions Enter y.pdf
Problem 6.15Calculate the annual cash flows for each of the follow.pdf
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Problem 1 Consider the following inventory management scenario: Our bottling plant uses approximately 1000 plastic bottlesper day (normally distributed), while daily usage has a standard deviation of 70. Order lead time is 2 days (constant). 1. What is the average demand during lead time for this item? 2. What is the standard deviation of demand during lead time for this item ? 3. If I use an order point system, what reorder point (ROP) would result in 100 units of safety stock? 4. What is the predicted service level using that ROP value? 5. What ROP value is needed to provide 99% predicted service level? How much safety stock is associated with this policy? Problem 3 Consider the plastic bottles demand pattern from question 1. Suppose I plan to instead set up a periodic review inventory system, and I will order this item every 7 days. There is still a 2 day lead time. 1. The \"order up to\" level T must cover what protection interval ? 2. What is the average demand during the protection interval? 3. What is the standard deviation of demand during the protection interval? 4. Using this periodic review system, what order up to level (T) would result in 100 units of safety stock? 5. What is the predicted service level under this safety stock value? 6. What order up to level is needed to provide 99% predicted service level? How much safety stock is associated with this policy. Solution average demand during lead time= 1000*2+Sqrt 70= 6900 itm Safety Stock: {Z * SQRT (Avg. Lead Time * Standard Deviation of Demand ^2 + Avg. Demand ^2 * Standard Deviation of Lead Time ^2)}. (note ^2 means ‘square’) Reorder Point: Average Lead Time*Average Demand + Safety Stock. In academic texts (std deviation * sqrt(lead time)) is referred to as standard deviation during lead time. However, if I were to take a rolling average of weekly demand over weeks and then take a standard deviation over this rolling average, wouldn\'t that be the same as standard deviation over 4 weeks (lead time)? SS = [( FE)2 x (LTI/FI)beta + ( LT)2 x D2] x Z x (FI/OCI)beta Where: SS = safety stock FE = forecast error LT = lead time interval FI = forecast interval (pick a beta between 0.5 and 0.7) D = average demand during lead time Z = normal distribution service factor based on desired service level OCI = order cycle interval.
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Problem 6.15Calculate the annual cash flows for each of the follow.pdf
- Problem 6.15 Calculate the annual cash flows for each of the following investments: (Round answers to nearest whole dollar, e.g. 5,275.) a) $204,688 invested at 6 percent. b) $30,160 invested at 12 percent. c) $120,207 invested at 10 percent. Please show work I only have one opportunity to get the answer right. Thanks Solution (a) Amount Invested $2,04,688 Interest Rate 6% Annual Cash Flows $12,281 (b) Amount Invested $30,160 Interest Rate 12% Annual Cash Flows $3,619 (c ) Amount Invested $1,20,207 Interest Rate 10% Annual Cash Flows $12,021