Parts Emporium Case Study free essay sample

1. The short answer is that higher inventories do not provide an advantage in any of the nine competitive priority categories. The important point is that firms must have the â€Å"right amount† of inventory to meet their competitive priorities. The only relevant costs considered in this chapter are ordering costs, holding costs, and stockout costs. In the economic order quantity (EOQ) model, costs of placing replenishment orders tradeoff against the costs of holding inventory. Under the assumptions of the EOQ, average inventory is one-half of the order quantity. The number of orders placed per year varies inversely with order quantity. When we consider stockout costs, an additional inventory (safety stock), is held to trade-off costs of poor customer service or costs for expediting shipments from unreliable suppliers. In the lean systems chapter, we see order quantities (lot sizes) that are much smaller than the â€Å"ideal† suggested by the EOQ model. As a result, lean systems average inventory is also much lower. We will write a custom essay sample on Parts Emporium Case Study or any similar topic specifically for you Do Not WasteYour Time HIRE WRITER Only 13.90 / page Are there some other relevant costs of holding inventory that we have not considered in the EOQ model? If there are, a firm that ignores these costs will make the wrong inventory decisions. These wrong decisions will make the firm less competitive. Let’s examine the relationships between inventory and the nine competitive priorities discussed in the operations strategy chapter. We compare competitors H and L. They are similar in all respects except H maintains much higher inventory than does L. 1. Low-cost operations. Costs include materials, scrap, labor, and equipment capacity that are wasted when products are defective. When a process drifts out of control, competitor H’s large lot sizes tend to result in large quantities of defectives. The EOQ does not consider the cost of defectives, and erroneously assumes that setup costs are constant. Small lots cause frequent setups, but the cost per setup decreases due to the learning curve. Competitor L will enjoy competitive advantages with lower setup, materials, labor, equipment, and inventory holding costs. 2. Top quality. Superior features, durability, safety, and convenience result from improved designs. High inventories force competitor H to choose between scrapping obsolete designs or delaying introduction of product improvements until the old inventory is consumed. In either case, L gains a competitive advantage. 3. Consistent quality. Consistency in conforming to design specifications requires consistency in supplied materials, setups, and processes. Small lots made frequently tend to increase consistency. Again, advantage goes to L. 4. Delivery speed. Large lots take longer to produce than small lots. A customer will wait less time for competitor L to set up and produce orders made in small batches. 5. On-time delivery. Contrary to expectations, large inventories do not equate to on-time delivery. It’s more like, lots of inventory equals lots of chaos. Big lots make big scheduling problems. Big lots get dropped, mishandled, and pilfered. Most lean companies experience dramatic improvement in on-time delivery. 6. Development speed. This response is similar to that given for top quality. Low inventories result in getting new designs to the market more quickly. 7. Customization. Lean companies usually don’t claim an advantage in customization. However, large inventories provide no advantage with regard to customization either. It remains unlikely that a customized product will be found in inventory, no matter how large. 8. Variety. Mass customizers compete on service or product variety. They will keep products at raw material or component levels until a customer orders a specific configuration. Inventories are at as low a level as possible. 9. Volume flexibility. Lean (low inventory) companies tend to produce the same quantity of every product every day, but they claim considerable volume flexibility from month to month. On the other hand, a large finished goods inventory can be used to absorb volume fluctuations. In summary, a case can be made that several competitive priorities are not considered in the EOQ model. It is sometimes difficult to place a dollar value on these competitive advantages, but the advantages invariably go to the low-inventory, small lot-size firm. So if the EOQ is too large, what is the â€Å"ideal† lot size? According to the lean philosophy, the â€Å"ideal† lot size is one. 2. The continuous review system requires the determination of two parameters: the order quantity and the reorder point. The ordering cost for each firm will decrease, which means that the economic order quantities will decrease. Because of this, there may be some implications for the logistics system. Smaller, more frequent shipments could require more costly less-than-truckload shipments. In addition, while the order quantities will decrease, the reorder points will also decrease because the lead times will be smaller. The supply chain should experience smaller pipeline inventories as a consequence. If the new information system also reduces the variance of demand or lead times, there can be additional safety stock savings. However, all of these benefits will come at some additional expense for the incorporation of the new system. There will be capital costs for equipment and potential training costs involved. 3. Organizations will never get to the point where inventories are unneeded. Inventories provide many functions and should be managed, not eliminated. It is impossible to eliminate uncertainties in the provision of products or services. In addition, unless materials can be transported instantaneously, there will always be pipeline inventories. Cycle inventories will exist unless we universally get to the point where production of single units is feasible. PROBLEMS 1. Lockwood Industries First we rank the SKUs from top to bottom on the basis of their dollar usage. Then we partition them into classes. The analysis was done using OM Explorer Tutor12. 1—ABC Analysis. Cumulative % Cumulative % SKU # Description Qty Used/Year Value Dollar Usage Pct of Total of Dollar Value of SKUs Class 4 44,000 $1. 00 $44,000 60. 0% 60. 0% 12. 5% A 7 70,000 $0. 30 $21,000 28. 6% 88. 7% 25. 0% A 5 900 $4. 50 $4,050 5. 5% 94. 2% 37. 5% B 2 120,000 $0. 03 $3,600 4. 9% 99. 1% 50. 0% B 6 350 $0. 90 $315 0. 4% 99. 5% 62. 5% C 8 200 $1. 50 $300 0. 4% 99. 9% 75. 0% C 3 100 $0. 45 $45 0. 1% 100. 0% 87. 5% C 1 1,200 $0. 01 $12 0. 0% 100. 0% 100. 0% C Total $73,322 The dollar usage percentages don’t exactly match the predictions of ABC analysis. For example, Class A SKUs account for 88. 7% of the total, rather than 80%. Nonetheless, the important finding is that ABC analysis did find the â€Å"significant few. † For the items sampled, particularly close control is needed for SKUs 4 and 7. 2. Stock-Rite Inc. Computing the annual usage value for each SKU and rank ordering them highest to lowest, we get: SKU Annual Value ($) Cumulative Value ($) D205 9,690 9,690 U404 6,075 15,765 A: 55% A104 3,220 18,985 L205 3,035 22,020 B: 22% L104 2,005 24,025 S104 1,604 25,629 X205 1,603 27,232 C: 23% X104 1,500 28,732 One classification might be to group the top two items (i. e. , 25% of the SKUs) in A class accounting for 55% of the total value. The next two SKUs would be classified as B and the last four as C. The dollar usage percentages don’t exactly match the predictions of ABC analysis. For example, Class A SKUs account for only 55% of the total, rather than 80%. Nonetheless, the important finding is that ABC analysis did find the â€Å"significant few. † For the items sampled, particularly close inventory management is needed for SKUs D205 and U404. 3. Yellow Press, Inc. a. Economic order quantity b. Time between orders 4. Babble Inc. a. D =( 400 tapes/month)(12 months/yr) = 4,800 tapes/year b. Time between orders years or 2. 5 months 5. Dot Com a. b. Optimal number of orders/year = (32,000)/400 = 80 orders c. Optimal interval between orders = 300/80 = 3. 75 days d. Demand during lead time = L = (5 days)(32,000/300) = 533 books e. Reorder point = L + safety stock = 533 + 0 = 533 books f. Inventory position = OH + SR – BO = 533 + 400 – 0 = 933 books 6. Leaky Pipe Inc. a. b. Optimal number of orders = (30,000)/(775) = 38. 7 or 39 c. Optimal interval between orders = (300)/(39) = 7. 69 days d. Demand during lead time = L = (4 days)(30,000/300) = 400 units e. Reorder point = L + safety stock = 400 + 0 = 400 units f. Inventory position = OH + SR – BO = 400 +775 – 0 = 1175 units 7. Sam’s Cat Hotel a. Economic order quantity = 90/week D = (90 bags/week)(52 weeks/yr) = 4,680 S = $54 Price = $11. 70 H = (27%)($11. 70) = $3. 16 = 399. 93, or 400 bags. Time between orders, in weeks b. Reorder point, R R = demand during protection interval + safety stock Demand during protection interval = L = 90 * 3 = 270 bags Safety stock = z? dLT When the desired cycle-service level is 80%, . = 15 = 25. 98 or 26 Safety stock = 0. 84 * 26 = 21. 82, or 22 bags c. Initial inventory position = OH + SR – BO = 320 + 0 – 0 320 – 10 = 310.  ­ Because inventory position remains above 292, it is not yet time to place an order. d. Annual holding cost Annual ordering cost When the EOQ is used these two costs are equal. When , the annual holding cost is larger than the ordering cost, therefore Q is too large. Total costs are $789. 75 + $505. 44 = $1,295. 19. e. Annual holding cost Annual ordering cost Total cost using EOQ is $1,263. 60, which is $31. 59 less than when the order quantity is 500 bags. 8. Sam’s Cat Hotel, revisited a. If the demand is only 60 bags per week, the correct EOQ is: D = (60 units/wk)(52 wk/yr) = 3,120 bags = 326. 54, or 327 bags If the demand is incorrectly estimated at 90 bags, the EOQ would be incorrectly calculated (from problem 7) as 400 bags. The total cost, working with the actual demand, is: We can see clearly now that the cost penalty of Sam’s difficulty in foreseeing demand for kitty litter is $21. 31 ($1,053. 20 – $1,031. 89). b. If S = $6, and , the correct EOQ is: = 108. 85, or 109 bags The total cost, working with the actual ordering cost, is If the reduced ordering cost continues to be unseen, the cost penalty for not updating the EOQ is (573. 91 – 343. 96) = $229. 95. 9. A Q system (also known as a reorder point system) = 300 pints/week = 15 pints a. Standard deviation of demand during the protection interval: = 15 = 45 pints b. Average demand during the protection interval: Demand during protection interval = L = 300 * 9 = 2700 pints c. Reorder point R = average demand during protection interval + safety stock Safety stock = z? dLT When the desired cycle-service level is 99%, z = 2. 33. Safety stock = 2. 33 * 45 = 104. 85 or 105 pints R = 2,700 + 105 – 0 = 2,805 pints 10. Petromax Enterprises a. b. Safety stock = z? dLT = = (1. 28)(125) = 277. 13 or 277 units Reorder point= average lead time demand + safety stock = (3)(50,000/50) + 277 = 3,277 units 11. A continuous review system for door knobs. Find the safety stock reduction when lead time is reduced from five weeks to one week. Standard deviation of demand during the (five-week) protection interval is = 85 door knobs. Desired cycle service level is 99% (therefore z = 2. 33). Safety stock required for five-week protection interval: Safety stock = = 2. 33(85) = 198. 05, or 198 door knobs Safety stock required for one-week protection interval ?dLT = = = 85 door knobs = 85/ = 38. 01 door knobs. Safety stock = = 2. 33(38. 01) = 88. 57 or 89 door knobs Safety stock reduction Reduction = 198 – 89 = 109 door knobs. 12. A two-bin system. â€Å"The two-bin system is really a Q system, with the normal level in the second bin being the reorder point R. † Find cycle-service level, given: L = 2 weeks = 5 bolts = 53 bolts/weekR = 120 bolts Safety stock = R – L = 120 – (53*2) = 14 bolts Safety stock = z? dLT = 14 bolts ?dLT = = 5 = 7. 07 bolts z(7. 07) = 14 z = 1. 98 When z = 1. 98, the cycle-service level is 97. 67%. 13. Nationwide Auto Parts a. Protection interval (PI)= P + L = 6 +3 = 9 weeks Average demand during PI= 9 (100) = 900 units Standard deviation during PI= = 60 units b. Target inventory= (P+L) + z? P+L = 900 + (1. 96)(60) = 1,018 c. Order quantity= Target inventory – IP = 1,018 – 350 = 668 units presuming no SR or BO 14. A P system (also known as a periodic review system). Find cycle-service level, given: L = 2 weeks P = 1 week (P + L) = 218 boxes = 40 boxes T = 300 boxes T = Average demand during protection interval + Safety stock T = 218 + z(40) = 300 boxes z = (300 – 218)/40 = 2. 05 When z = 2. 05, cycle-service level is 97. 98 or 98%. 15. A Successful Product Annual Demand, D = (200)(50) = 10,000 units, H = ((0. 20)(12. 50)) = 2. 50 a. Optimal ordering quantity b. Safety stock = = (2. 33)(16) = 74. 56 or 75 units c. Safety stock will now be: (2. 33)(16) = 52. 72 or 53 units % reduction in safety stock= (75 – 53)/75 = 29. 33% d. Safety stock will be= (2. 33)(8) = 37. 28 or 38 units % reduction in safety stock= (75 – 38)/75 = 49. 33% 16. Sam’s Cat Hotel with a P system a. Referring to Problem 7, the EOQ is 400 bags. When the demand rate is 15 per day, the average time between orders is (400/15) = 26. 67 or about 27 days. The lead time is 3 weeks ? 6 days per week = 18 days. If the review period is set equal to the EOQ’s average time between orders (27 days), then the protection interval (P + L) = (27 + 18) = 45 days. For an 80% cycle-service level z = 0. 84 = 41. 08 Safety stock = = 0. 84(41. 08) = 34. 51 or 35 bags T = Average demand during the protection interval + Safety stock T = (15*45) + 35 = 710 b. In Problem 7, the Q system required a safety stock of 22 bags to achieve an 80% cycle-service level. Therefore, the P system requires a safety stock that is larger by (35 – 22) = 13 bags. c. From Problem 7, inventory position, IP = 320. The amount to reorder is T – IP = 710 – 320 = 390. 17. Continuous review system. a. Economic order quantity. or 894 units Time between orders (TBO) = Q/D = 894/20,000 = 0. 0447 years = 2. 32 weeks b. Weekly demand = 20,000/52 = 385 units For a 95% cycle-service level, z = 1. 65 Safety stock: = (1. 65)(100) = 233. 34, or 233 units Now solve for R, as R = L + Safety stock = 385(2) + 233 = 1,003 units c. i. Annual holding cost of cycle inventory ii. Annual ordering cost d. With the 15-unit withdrawal, IP drops from 1,040 to 1,025 units. Because this level is above the reorder point (1,025 1,003), a new order is not placed. 18. Periodic review system a. From Problem 17, or 894 units Number of orders per year = = 20,000/894 = 22. 4 orders per year. weeks P is rounded to 2 weeks. b. For a 95% cycle-service level, z = 1. 65. Therefore Safety stock = 200 units Safety stock = 1. 65(200) = 330 units, T = Average demand during the protection interval + Safety stock T = (385 * 4) + 330 = 1,870 units c. In Problem 17, with a Q system the safety stock is 233 units. Therefore, (330 – 233) = 97 more units of safety stock are needed. 19. Continuous review system a. Economic order quantity b. Safety stock. When cycle-service level is 88%, z = 1. 18. Safety stock = = (1. 18)(12) = 20. 03, or 20 units c. Reorder point R = L + Safety stock = 64(2) + 20 = 148 units. d. If Q = 200 and R = 180, average inventory investment is higher than necessary to achieve an 88% cycle-service level. The larger order quantity increases average cycle stock by 20 units, and the higher reorder point increases safety stock by 32 units. 20. Periodic review system a. From problem 19, EOQ = 160 weeks P is rounded to 3 weeks. b. For an 88% cycle-service level, z = 1. 18. Therefore Safety stock = 26. 83 units. Safety stock = 1. 18(26. 83) = 31. 66, or 32 units T = average demand during the protection interval + Safety stock T = (64 * 5) + 32 = 352 units 21. Wood County Hospital a. D = (1000 boxes/wk)(52 wk/yr) = 52,000 boxes H = (0. l5)($35/box)=$5. 25/box 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 = = (1. 88)(100) = 1. 88(141. 42) = 265. 87, or 266 boxes R = L + Safety stock = 1000(2) + 266 = 2,266 boxes c. In a periodic review system, find target inventory T, given: P = 2 weeks L = 2 weeks Safety stock = = 200 units. Safety stock = 1. 88(200) = 376 units T = Average demand during the protection interval + Safety stock T = 1000(2 + 2) + 376 T = 4,376 units The table below is derived from OM Explorer Solver—Inventory Systems. Notice that the total cost for the Q system is much less than that of the P system. The reason is that the optimal value of P was not used here. The optimal value is weeks. Continuous Review (Q) system Periodic Review (P) System z = 1. 88 Time Between Reviews (P) 2. 00 Weeks ? Enter manually Safety Stock 266 Standard Deviation of Demand d During Protection Interval 200 Reorder Point 2266 Safety Stock 376 Annual Cost $4,258. 32 Average Demand During Protection Interval 4000 Target Inventory Level (T) 4376 Annual Cost $7,614. 00 22. Golf specialty wholesaler a. Periodic Review System or 179 1-irons or 4. 0 weeks When cycle-service level is 90%, z = 1. 28. Weekly demand is (2,000 units/yr)/(50 wk/yr) = 40 units/wk L = 4 weeks Safety stock: z = (1. 28) = 10. 86, or 11 irons T = (P+L) + Safety stock = 40(4+4) + 11 = 331 irons. b. Continuous review system Safety stock = = (1. 28)(3) = 1. 28(3)(2) = 7. 68, or 8 irons R = L + Safety stock = 40(4) + 8 =168 irons 23. Osprey Sports. a. The economic order quantity is = 289. 83, or 290 lures. b. The safety stock and reorder point are = 12. 41 lures The z value for a 97 percent cycle-service level = 1. 88. The safety stock = 1. 88 (12. 41) = 23. 33, or 23 lures The reorder point = + Safety stock = (4)(10) + 23 = 63 lures. c. The total annual cost for this continuous review system is + (H)(Safety stock) = = $312. 83 24. Farmer’s Wife a. The continuous review system is specified by the fixed order quantity and the reorder point. We will use the EOQ for the order quantity. The order quantity is: = 244. 95, or 245 cows. The safety stock is: = 61. 64 cows. The z value for a 90 percent cycle-service level = 1. 28. The safety stock = 1. 28 (61. 64) = 78. 90, or 79 cows. The reorder point = + Safety stock = (30)(8) + 79 = 319 cows b. The system would operate as follows: Whenever the stock of cows drops to 319, order 245 more cows. c. The total annual cost for this continuous review system is + (H)(Safety stock) = = $243. 06 25. Muscle Bound To find the cycle-service level, we must determine the standard deviation of demand during lead time and then use the equation for total annual cost to solve for z. We will use the EOQ for the ordering quantity. The standard deviation of demand during lead time is = 5,078. 14 barbells The economic order quantity is = 3,538. 36, or 3,538 barbells The total annual cost (with z as a variable) is + (H)(Safety stock) = = $16,000 We now solve for z z = = 0. 8785, or 0. 88. This value of z corresponds to a cycle-service level of 81 percent. 26. Georgia Lighting Center. Using the demand data given in the problem statement, we extended text Table 12. 2 below the dashed line in the following way. The beginning inventory for day 7 is the ending inventory for day 6, which is 27 units. The demand for day 7 is 7 units, which leaves 20 units in inventory at the end of day 7. No orders are open to the supplier; consequently the inventory position is 20 units. Because 20 units exceeds the reorder point of 15 units, no new order is placed. Continuing in this manner, the inventory position at the end of day 9 drops below the reorder point; consequently a new order for 40 units is placed. That order will be received three business days later, or day 12. The complete simulation results with Q = 40 and R = 15 are: Open Beginning Orders Daily Ending Inventory Amount Day Inventory Received Demand Inventory Position Ordered 1 19 — 5 14 14 40 2 14 — 3 11 51 — 3 11 — 4 7 47 — 4 7 40 1 46 46 — 5 46 — 10 36 36 — Sat 6 36 — 9 27 27 — Mon7 27 — 7 20 20 — 8 20 — 4 16 16 — 9 16 — 2 14 14 40 10 14 — 7 7 47 — 11 7 — 3 4 44 — 12 4 40 6 38 38 — 13 38 — 10 28 28 — 14 28 — 0 28 28 — 15 28 — 5 23 23 — 16 23 — 10 13 13 40 17 13 — 4 9 49 — 18 9 — 7 2 42 — TOTAL 343 a. The average ending inventory is: or 19 units b. No stockouts occurred during any of the three cycles. myomlab Advanced PROBLEMS 1. Office Supply Shop The screen shot below is taken from OM Explorer Solver – Demand During Protection Interval Simulator. It shows the results of 500 trials. a. Given the simulation, the value of R must yield a service level that meets or exceeds the desired value of 95%. That value of R is 71 pens, which will yield a cycle service level of 96. 4%. b. The average demand during the protection interval is 35 pens. Since the reorder point is 71, the safety stock must be 71 – 35 = 36 pens. The high level of safety stock is necessary because of the high variance in the demand during protection interval distribution and the high variance in lead time. 2. Grocery store. a. The target level (T) should be 150 tubes of Happy Breath Toothpaste. This result comes from OM Explorer Solver – Demand During Protection Interval Simulator. b. Using OM Explorer once again, the cycle-service level for T = 150 would be 97. 8%. Eliminating the variance in supply lead times will significantly increase the cycle service level of the inventory. 3. Floral shop a. The EOQ for the continuous review system would be as follows. The demand during protection interval distribution is shown below. To attain at least a 90% cycle service level, the florist needs to set the reorder point at 166 baskets. b. As the output from OM Explorer Solver – Q-System Simulator shows, the average cost per day is $274. 74. EXPERIENTIAL LEARNING: SWIFT ELECTRONIC SUPPLY, INC. This in-class exercise allows students to test an inventory system of their design against a new demand set. On the day of the simulation, students should come with sufficient copies of Table 1. Table Table1 12. 6 | Simulation Evaluation Sheet Day 1 2 3 4 5 6 7 8 9 10 Beginning inventory position Number ordered Daily demand Day-ending inventory Ordering costs ($200 per order) Holding costs ($0. 05 per piece per day) Shortage costs ($2 per piece) Total cost for day Cumulative cost from last day Cumulative costs to date It is best to precede the simulation with a brief overview of the simulation process and the calculation of costs. The instructor may decide to require students to bring a computer to class and use a spreadsheet of their design to accomplish the tasks embodied in Table 1. Once everyone understands the simulation procedure, the instructor uses the â€Å"actual† demands in TN1, one at a time, and proceeding at a pace such that students have a chance to decide whether or not to order that period, how much to order, and calculate relevant costs. The instructor can stop at any point, using TN2 to benchmark students’ results against any of the four provided systems in this manual. A good idea is to stop at the halfway point in the simulation and ask students what their total costs are. The variance is often quite high. The same benchmarking comparisons can be done at the end of the simulation. The instructor can use the students’ results to discuss differences in the systems tried, the importance of using safety stocks, and the value of perfect information. One of the provided systems in this manual utilizes the Wagner-Whitin (WW) approach, which is optimal for perfect forecasts. The variance in student results will be greater if this exercise is used as a prelude to a discussion of formal inventory systems (such as the Q-system or P-system). Alternatively, the exercise can be used after a presentation of the formal systems to give students a practicum for the theory. TN3 shows the cost structure and system parameters for the EOQ-system, Q-system and P-system. All the relevant case information and derived data are on the left side of the sheet, and key computed parameters for three systems are presented on the right side of the sheet. There are some other points that need to be addressed about TN3 through TN7: â€Å"Average Demand/day† and â€Å"Standard Deviation† come from a statistical analysis of the historical demand data in Table 12. 3. All the ordering quantities are rounded up as integers. Consequently, the associated costs might differ a little from what they actually are. The review time in the EOQ-system is actually up to the student. In TN4 we have used the EOQ divided by average daily demand. TN4 through TN6 show the application of the provided systems for the demand data in TN1. TN7 shows the results from WW system. In all of our reported results, inventory levels at the start of the day are used to make inventory decisions. This is consistent with the daily purchasing routine at Swift. Economic Order Quantity (EOQ) System Under this system, students order the EOQ each and every review period, which using the case data would be 3 days, without any forecasts of future demand or consideration of demand variability. TN4 shows the performance of this system. Students may elect to use varying review periods. If so, their results will differ from TN4. Q-system This system assumes that inventory levels are checked on a daily basis and compared to a â€Å"Reorder Point (RP). † If actual inventory level goes below the RP, an order of EOQ is placed; if above, no order will be placed. In the provided results, the RP is calculated by adding safety stock to average demand during the two-day lead time. The safety stock is designed to meet the 95 percent cycle service level. TN5 shows the results of the Q-system. P-system The inventory level is reviewed every three days, which is determined by dividing EOQ by average demand. The target inventory level is composed of two parts: â€Å"average demand during the protection interval,† which is the review period plus the lead time, and the â€Å"safety stock. † Every review period (three days in the provided results), an order is placed to bring the inventory position up to the target inventory level. TN6 shows the performance of the P-system. Wager-Whitin (WW) System The WW system is based on dynamic programming and assumes all demands are known with certainty. Consequently, it provides an absolute lower bound on the solution found by the students. The WW system assumes that stockouts are to be avoided. It is interesting to show the difference in total costs between the WW solution and another system because it demonstrates the cost of uncertainty. The solution using the WW system is shown in TN7. Also note that the lot sizes are shown in the day in which they must arrive. Actual release dates would be two days earlier. This implies that the first order for 1733 would have been placed in day 0, one day before the actual start of the simulation. TN 1. Actual Demand Data for Simulation CASE: PARTS EMPORIUM * A. Synopsis This case describes the problems facing Sue McCaskey, the new materials manager of a wholesale distributor of auto parts. She seeks ways to cut the bloated inventories while improving customer service. Back orders with excessive lost sales are all too frequent. Inventories were much higher than expected when the new facility was built, even though sales have not increased. Summary data on inventory statistics, such as inventory turns, are not available. McCaskey decides to begin with a sample of two products to uncover the nature of the problems—the EG151 exhaust gasket and the DB032 drive belt. B. Purpose The purpose of this case is to allow the student to put together a plan, using either a continuous review system (Q system) or a periodic review system (P system), for two inventory SKUs. Enough information is available to determine the EOQ and R for a continuous review system (or P and T for a periodic review system). Because stockouts are costly relative to inventory holding costs, a 95% cycle-service level is recommended. Inventory holding costs are 21% of the value of each item (expressed at cost). The ordering costs ($20 for exhaust gaskets and $10 for drive belts) should not be increased to include charges for making customer deliveries. These charges are independent of the inventory replenishment at the warehouse and are reflected in the pricing policy. C. Analysis We now find appropriate policies for a Q system, beginning with the exhaust gasket. Shown here are the calculations of the EOQ and R, followed by a cost comparison between this continuous review system and the one now being used. The difference is what can be realized by a better inventory control system. Reducing lost sales due to back orders is surely the biggest benefit. 1. EG151 Exhaust Gasket a. New plan Begin by estimating annual demand and the variability in the demand during the lead time for this first item. Working with the weekly demands for the first 21 weeks of this year and assuming 52 business weeks per year, we find the EOQ as follows: Weekly demand average = 102 gaskets/week Annual demand (D) = 102(52) = 5304 gaskets Holding cost = $1. 85 per gasket per year (or 0. 21 ? 0. 68 ? $12. 99) Ordering cost = $20 per order gaskets Turning to R, the Normal Distribution appendix shows that a 95% cycle-service level corresponds to a z = 1. 65. We then use the EG151 data to find the standard deviation of demand. Standard deviation in weekly demand () = 2. 86 gaskets Standard deviation in demand during lead time R= Average demand during the lead time + Safety stock = 2(102) + 1. 65(4. 04) = 210. 66, or 211 gaskets b. Cost comparison After developing their plan, students can compare its annual cost with what would be experienced with current policies. Cost Category Current Plan Proposed Plan Ordering cost $707 $313 Holding cost (cycle inventory) 139 314 TOTAL $846 $627 The total of these two costs for the gasket is reduced by 26 percent (from $846 to $627) per year. The safety stock with the proposed plan may be higher than the current plan, if the reason for the excess back orders is that no safety stock is now being held (inaccurate inventory records or a faulty replenishment system are other explanations). We cannot determine the safety stock level (if any) in the current system. The extra cost of safety stock for the proposed system is minimal, however. Only seven gaskets are being proposed as safety stock, and their annual holding cost is just another $1. 85(7) = $12. 95. Surely the lost sales due to back orders are substantial with the current plan and will be much less with the proposed plan. One symptom of such losses is that 11 units are on back order in week 21. A lost sale costs a minimum of $4. 16 per gasket (0. 32. ? $12. 99). If 10 percent of annual sales were lost with the current policy, this cost would be $4. 16(0. 10)(5,304) = $2,206 per year. Such a loss would be much reduced with the 95% cycle-service level implemented with the proposed plan. 2. DB032 Drive Belt a. New plan The following demand estimates are based on weeks 13 through 21. Weeks 11 and 12 are excluded from the analysis because the new product’s start-up makes them unrepresentative. We find the EOQ as follows: Weekly demand average = 52 belts/week Annual demand (D) = 52(52) = 2704 belts Holding cost $0. 97 per belt per year (or 0. 21 ? 0. 52 ? $8. 89) Ordering cost $10 per order gaskets Turning now to R, where z remains at 1. 65, we use the data in the DB032 table to find: Standard deviation in weekly demand () = 1. 76 belts Standard deviation in demand during lead time belts R= Average demand during the lead time + Safety stock = 3(52) + 1. 65(3. 05) = 161. 03, or 161 belts b. Cost comparison After developing their plan, students again can compare the cost for the belts with what would be experienced with current policies. Cost Category Current Plan Proposed Plan Ordering cost $ 27 $115 Holding cost (cycle inventory) 485 114 TOTAL $512 $229 With the belt, the total of these two costs is reduced by 55 percent. The safety stock with the proposed plan

Eleanor, Queen of Castile (1162 - 1214)

Eleanor, Queen of Castile (1162 - 1214) Eleanor Plantagenet, born in 1162,   was the wife of Alfonso VIII of Castile, daughter of Henry II of England and  Eleanor of Aquitaine, sister of kings and a queen; mother of several queens and a king. This Eleanor was the first of a long line of Eleanors of Castile. She was also known as  Eleanor Plantagenet, Eleanor of England, Eleanor of Castile, Leonora of Castile, and Leonor of Castile. She died on October 31st, 1214.   Early Life Eleanor was named for her mother, Eleanor of Aquitaine. As a daughter of Henry II of England, her marriage was arranged for political purposes. She was paired with King Alfonso VIII of Castile, betrothed in 1170 and married sometime before September 17, 1177, when she was fourteen. Her full siblings were William IX, Count of Poitiers; Henry the Young King; Matilda, Duchess of Saxony; Richard I of England; Geoffrey II, Duke of Brittany;  Joan of England, Queen of Sicily; and John of England. Her older half-siblings were  Marie of France  and  Alix of France Eleanor as Queen Eleanor was granted control in her marriage treaty of lands and towns so that her own power was nearly as much as her husbands. The marriage of Eleanor and Alfonso produced a number of children. Several sons who were, in turn, expected heirs of their father died in childhood. Their youngest child, Henry or Enrique, survived to succeed his father. Alfonso claimed Gascony as part of Eleanors dowry, invading the duchy in his wifes name in 1205, and abandoning the claim in 1208.    Eleanor wielded considerable power in her new position.   Ã‚  She was also a patron of many religious sites and institutions,  including Santa Maria la Real at Las Huelgas where many in her family became nuns.   She sponsored troubadours to court.  She helped arrange the marriage of their daughter  Berenguela  (or Berengaria) to Leons king. Another daughter, Urraca, was married to the future king of Portugal, Alfonso II; a third daughter, Blanche or Blanca, was married to the future King Louis VIII of France; a fourth daughter, Leonor, married the king of Aragon (though their marriage was later dissolved by the church). Other daughters included Mafalda who married her sister Berenguelas stepson and Constanza who became an  Abbess. Her husband appointed her as ruler with their son upon his death, and also appointed her executor of his estate.   Death Although Eleanor thus became the regent for her son Enrique on her husbands death, in 1214 when Enrique was only ten, Eleanors grief was so great that her daughter Berenguela had to handle the burial of Alfonso. Eleanor died on October 31st, 1214, less than a month after Alfonsos death, leaving Berenguela as her brothers regent. Enrique died at age 13, killed by a falling roof tile. Eleanor was the mother of eleven children, but only six survived her: Berenguela  (1180 - 1246) - she married Conrad II of Swabia but the marriage contract was annulled. She married Alfonso IX of Leon, but that marriage was dissolved on grounds of consanguinity.   She became regent for her brother Enrique (Henry) I, and became Queen of Castile in her own right when he died in 1217. She abdicated right after that, and her son Ferdinand III of Castile brought together Castile and Leon.Sancho (1181 - 1181) - briefly heir to Castile, died at three monthsSancha (1182 - 1185)Enrique (1184 - 1184?) - heir during his very short life - there is some doubt that this child existed.Urraca - Urraca of Castile, Queen of Portugal (1187 - 1220), married to Afonso II of Portugal.Blanca -  Blanche of Castile, Queen of France (1188 - 1252), married the future Louis VIII of France, crowned Queen in 1223. She served as regent of France after Louis died and before their son was of age.Fernando (1189 - 1211). Died of a fever, heir to the throne at that time.Mafalda (11 91 - 1211). Betrothed to Ferdinand of Leon, stepson of her sister Berenguela. Constanza (1195 or 1202 - 1243), became a nun at Santa Maria la Real at Las Huelgas.Leonor - Eleanor of Castile (1200 or 1202 - 1244): married James I of Aragon but separated 8 years later, with consanguinity as the grounds.Enrique I of Castile (1204 - 1217). He became king in 1214 when his father died; he was only 10. He died three years later, struck by a tile that fell from a roof.

The Easy Guide to the 30-60-90 Triangle

The Easy Guide to the 30-60-90 Triangle SAT / ACT Prep Online Guides and Tips Acute, obtuse, isosceles, equilateral†¦.When it comes to triangles, there are many different varieties, but only a choice few that are â€Å"special.† These special triangles have sides and angles which are consistent and predictable and can be used to shortcut your way through your geometry or trigonometry problems. And a 30-60-90 triangle- pronounced â€Å"thirty sixty ninety†- happens to be a very special type of triangle indeed. In this guide, we’ll walk you through what a 30-60-90 triangle is, why it works, and when (and how) to use your knowledge of it. So let’s get to it! What Is a 30-60-90 Triangle? A 30-60-90 triangle is a special right triangle (a right triangle being any triangle that contains a 90 degree angle) that always has degree angles of 30 degrees, 60 degrees, and 90 degrees. Because it is a special triangle, it also has side length values which are always in a consistent relationship with one another. The basic 30-60-90 triangle ratiois: Side opposite the 30 ° angle: x Side opposite the 60 ° angle: x * √3 Side opposite the 90 ° angle: 2x For example, a 30-60-90 degree triangle could have side lengths of: 2, 2√3, 4 7, 7√3, 14 √3,3, 2√3 (Why is the longer leg 3?In this triangle, the shortest leg (x) is √3, so for the longerleg, x√3 = √3 *√3 = √9 = 3. And the hypotenuse is 2 times the shortestleg, or 2√3) And so on. The side opposite the 30 ° angle is always the smallest, because 30 degrees is the smallest angle. The side opposite the 60 ° angle will be the middle length, because 60 degrees is the mid-sized degree angle in this triangle. And, finally, the side opposite the 90 ° angle will always be the largest side (the hypotenuse) because 90 degrees is the largest angle. Want to get better grades and test scores? We can help. PrepScholar Tutors is the world's best tutoring service. We combine world-class expert tutors with our proprietary teaching techniques. Our students have gotten A's on thousands of classes, perfect 5's on AP tests, and ludicrously high SAT Subject Test scores. Whether you need help with science, math, English, social science, or more, we've got you covered. Get better grades today with PrepScholar Tutors. Though it may look similar to other types of right triangles, the reason a 30-60-90 triangle is so special is that you only need three pieces of informationin order to find every othermeasurement. So long as you know the value oftwo angle measures and one side length (doesn’t matter which side), you know everything you need to know about your triangle. For example, we can use the30-60-90 triangle formula tofill in all the remaining information blanks of the triangles below. Example 1 We can see that this is a right triangle in which the hypotenuse is twice the length of one of the legs. This means this must be a 30-60-90 triangle and the smaller given sideis opposite the 30 °. The longer leg must, therefore, be opposite the 60 ° angle and measure 6 *√3, or 6√3. Example 2 We can see that this must be a 30-60-90 triangle because we can seethat this is a right triangle with one given measurement, 30 °. The unmarked angle must then be 60 °. Since 18 is the measure opposite the 60 ° angle, it must be equal to x√3. The shortest leg must then measure $18/√3$. (Note that the leg lengthwill actually be $18/{√3} *{√3}/{√3} = {18√3}/3 = 6√3$ because adenominator cannot contain a radical/square root). And the hypotenuse will be $2(18/√3)$ (Note that, again, you cannot have a radical in the denominator, so the final answer will really be 2 times the leg length of 6√3= 12√3). Example 3 Again, we are given two angle measurements (90 ° and 60 °), so the third measure will be 30 °. Because this is a 30-60-90 triangle and the hypotenuse is 30, the shortest leg will equal 15 and the longer leg will equal 15√3. No need to consult the magic eight ball- these rules always work. Why It Works: 30-60-90 Triangle Theorem Proof But why does this special triangle work the way it does? How do we know these rules are legit? Let’s walk through exactly how the 30-60-90 triangle theorem works and prove why these side lengths will always be consistent. First, let’s forget about right triangles for a second and look at an equilateral triangle. An equilateral triangle is a triangle that has all equal sides and all equal angles. Because a triangle’s interior angles always add up to 180 ° and $180/3 = 60$, an equilateral triangle will always have three 60 ° angles. Now let's drop down a height from the topmost angle to the base of the triangle. We've now created two right angles and two congruent (equal) triangles. How do we know they’re equal triangles? Because we dropped a height from an equilateral triangle, we’ve split the base exactly in half. The new triangles also share one side length (the height), and they each have the same hypotenuse length. Because they share three side lengths in common (SSS), this means the triangles are congruent. Note: not only are the two triangles congruent based on the principles of side-side-side lengths, or SSS, but also based on side-angle-side measures (SAS), angle-angle-side (AAS), and angle-side-angle (ASA). Basically? They're most definitely congruent. Now that we’ve proven the congruencies of the two new triangles, we can see that the top angles must each be equal to 30 degrees (because each triangle already has angles of 90 ° and 60 ° and must add up to 180 °). This means we have made two 30-60-90 triangles. And because we know that we cutthe base of the equilateral triangle in half, we can see that the side opposite the 30 ° angle (the shortest side) of each of our 30-60-90 triangles is exactly half the length of the hypotenuse. So let us call our original side length x and our bisected length $x/2$. Now all that leaves us to do is to find our mid-side length that the two triangles share. To do this, we can simply use the pythagorean theorem. $a^2 + b^2 = c^2$ $(x/2)^2 + b^2 = x^2$ $b^2 = x^2 - ({x^2}/4)$ $b^2 = {4x^2}/4 - {x^2}/4$ $b^2 = {3x^2}/4$ $b = {√3x}/2$ So we're left with: $x/2, {x√3}/2, x$ Now let's multiply each measure by 2, just to make life easier and avoid all the fractions. That way, we're left with: x, x√3, 2x We can see, therefore, that a 30-60-90 triangle will always have consistent side lengths of x, x√3, and 2x (or $x/2$, ${√3x}/2$, and x). Luckily for us, we can prove 30-60-90 triangle rules true without all of...this. When to Use30-60-90 Triangle Rules Knowing the30-60-90 triangle rules will be able to save you time and energy on a multitude of different math problems, namely a wide variety of geometry and trigonometry problems. Geometry Proper understanding of the 30-60-90 triangles will allow you to solve geometry questions that would either be impossible to solve without knowing these ratio rules, or at the very least, would take considerable time and effort to solve the "long way." With thespecial triangle ratios, you can figure out missing triangle heights or leg lengths (without having to use the pythagorean theorem), find the area of a triangle by using missing height or base length information, and quickly calculate perimeters. Any time you need speed to answer a question, remembering shortcuts like your 30-60-90 rules will come in handy. Trigonometry Memorizing and understanding the 30-60-90 triangle ratio will also allow you to solve many trigonometry problems without either the need for a calculator or the needto approximate your answers in decimal form. A 30-60-90 triangle has fairly simple sines, cosines, and tangents for each angle (and these measurements will always be consistent). Sine of 30 ° will always be $1/2$. Cosine of 60 ° will always be $1/2$. Though the other sines, cosines, and tangents are fairly simple, these are the two that are the easiest to memorize and are likely to show up on tests. So knowing these rules will allowyou to find these trigonometry measurements as quickly as possible. Tips for Remembering the30-60-90 Rules You know these30-60-90 ratio rules are useful, but how do you keep the information in your head? Remembering the30-60-90 triangle rules is a matter of remembering the ratio of 1: √3: 2, and knowing that the shortest side length is always opposite the shortest angle (30 °) and the longest side length is always opposite the largest angle (90 °). Some people memorize the ratio by thinking,â€Å"x, 2x, x√3,† because the â€Å"1, 2, 3† succession is typically easy to remember. The one precaution to using this technique is to remember that the longest side is actually the 2x, not the x times √3. Another way to remember your ratios is to use a mnemonic wordplay on the 1: root 3: 2 ratio in their proper order. For example, â€Å"Jackie Mitchell struck out Lou Gehrig and ‘won Ruthy too,’†: one, root three, two. (And it's a true baseball history fact to boot!) Play around with your own mnemonic devices if these don’t appeal to you- sing the ratio to a song, find your own â€Å"one, root three, two† phrases, or come up with a ratio poem. You can even just remember that a 30-60-90 triangle is half an equilateral and figure out the measurements from there if you don't like memorizing them. However it makes sense to you to rememberthese 30-60-90 rules, keep those ratios your head for your future geometry and trigonometry questions. Memorization is your friend, however you can make it happen. Want to get better grades and test scores? We can help. PrepScholar Tutors is the world's best tutoring service. We combine world-class expert tutors with our proprietary teaching techniques. Our students have gotten A's on thousands of classes, perfect 5's on AP tests, and ludicrously high SAT Subject Test scores. Whether you need help with science, math, English, social science, or more, we've got you covered. Get better grades today with PrepScholar Tutors. Example 30-60-90 Questions Now that we've looked at the hows and whys of 30-60-90 triangles, let's work throughsome practice problems. Geometry A construction worker leans a 40-foot ladder up against the side of a building at an angle of 30 degrees off the ground. The ground is level and the side of the building is perpendicular to the ground. How far up the building does the ladder reach, to the nearest foot? Without knowing our 30-60-90 special triangle rules, we would have to use trigonometry and a calculator to find the solution to this problem, since we only have one side measurement of a triangle. But because we know that this is a special triangle, we can find the answer in just seconds. If the building and the ground are perpendicular to one another, that must mean the building and the ground form a right (90 °) angle. It’s also a given that the ladder meets the ground at a 30 ° angle. We can therefore see that the remaining angle must be 60 °, which makes this a 30-60-90 triangle. Now we know that the hypotenuse (longest side) of this 30-60-90 is 40 feet, which means that the shortest side will be half that length. (Remember that the longest side is always twice- 2x- as long as the shortest side.) Because the shortest side is opposite the 30 ° angle, and that angle is the degree measure of the ladder from the ground, that means that the top of the ladder hits the building 20 feet off the ground. Our final answer is 20 feet. Trigonometry If, in a right triangle, sinÃŽËœ = $1/2$ and the shortest leg lengthis 8. What is the length of the missing side that is NOT the hypotenuse? Because you know your 30-60-90 rules, you can solve this problem without the need for either the pythagorean theorem or a calculator. We were told that this is a right triangle, and we know from our special right triangle rules that sine 30 ° = $1/2$. The missing angle must, therefore, be 60 degrees, which makes this a 30-60-90 triangle. And because this is a 30-60-90 triangle, and we were told that the shortest side is 8, the hypotenuse must be 16 and the missing side must be 8 * √3, or 8√3. Our final answer is 8√3. The Take-Aways Remembering the rules for 30-60-90 triangles will help you to shortcut your way through a variety of math problems. But do keep in mind that, while knowing these rules isa handy tool to keep in your belt, you can still solve mostproblems without them. Keep track of the rules of x, x√3, 2x and 30-60-90 in whatever way makes sense to you and try to keep them straight if you can, but don't panic if your mind blanks out when it's crunch time. Either way, you've got this. And, if you need more practice, go ahead and check out this 30-60-90 triangle quiz. Happy test-taking!

Dada in the world war one Essay Example | Topics and Well Written Essays - 1750 words

Dada in the world war one - Essay Example With the future of Germany an uncertainty, many avant-garde artists, including George Ehrenfried Grosz, were inspired by the Russian Revolution and formed the Berlin Dada Club (McCloskey 45). They believed classic Marxism was the best solution to the turmoil in Germany (McCloskey 46). They thought the major social, economic, and political turmoil in Germany stemmed from the bourgeoisies oppression of the proletariat and dedicated their art to ending the war and revolutionary change. George Ehrenfried Grosz (1893-1959) was born in Berlin, Germany, and emerged as one of the leading German political artists between World War I. His artwork during the 1920s reflected the appalling conditions in Berlin that resulted from the German loss of World War I, which included: food shortages, staggering inflation, the sight of war cripples begging in the streets, rampant prostitution, and widespread violence. As the German military began to accept the loss of the war, Grosz became increasingly involved with left wing activities and publishing his revolutionarily motivated art. His sentiments were fueled by a close relationship he developed with two brothers, Weiland Herzfelde and John Herzfelde (or John Heartfield as he called himself). Grosz had met Heartfield while in the infantry and they both decided to Americanize their names as protest against the German military (George Groszs given name was Georg Gross) (McCloskey 20). The Herzfelde brothers had been influenced by th eir parents to have strong socialist views and revolutionary attitudes. Their father was a socialist writer and their mother a textile union organizer. Grosz and the Herzfelde brothers, the historian Beth Lewis pointed out, "shared a common conviction that the war was not only despicable, but that it was lost" (Lewis 42). Grosz also believed that the common soldier had an interest in not wanting the war to end. He reasoned that the accepted violence in the

Mental Causation Essay Example | Topics and Well Written Essays - 1500 words

Mental Causation - Essay Example The problem of mental causation is one that has been described to be an intuitive one, focusing on the possibility that mental events cause physical events and physical events cause mental events. The major problem put forward by mental causation has to do with how mental events have causal effects on physical events, given the fact that the body, which is responsible for physical events is deemed to be independent of the mind. The main issue with mental causation has to do with a non-correspondence in explanation that seems to underline the understanding that the body is only physical and material. This is because if the body is only physical and material, then it should be controlled only by physical and material motivations for the body to act. For example, a person should decide to buy a certain color of a shoe because the leg has the material urge to wear that shoe. But once the problem of mental causation is brought into the argument, a new position is taken that it is possible for the mind, rather than the leg to be the source of motivation as to why a person would want to buy the said color of the shoe. This makes the problem of mental causation a highly dualistic situation, whereby the mind and the body are seen to relate together and affect each other.The whole problem of mental causation has been claimed by some to be a dualist philosophy rather than a generalized philosophy of mind. This is because as the name implies, dualism takes the position that the mind and the body are not identical.

In Broad Daylgith Essay Example for Free

In Broad Daylgith Essay â€Å"In Broad Daylight† is about Mu Ying, nicknamed Old Whore. She has affairs with different men and is publicly denounced and paraded before the community by the Red Guards who travel from another city and happen to know her bad name. Her dwarf peddler husband Meng Su tries to rescue her from the public humiliation, only to be humiliated himself by the Red Guards, the spectators and his wife as well. Finally, he is found crushed by a train, and Mu Ying lies alone at bus stop, deranged. Written from the point view of a naive boy, nicknamed White Cat, Ha Jin intends to portray through untainted and authentic lens a Chinese woman with a self-awakening feminist consciousness who stands up for her sexuality. Turning point number one in the story is when the questioning of Mu Ying has started and has to confess her crimes of adultery with three other men. She then comes to the point where she explains the feeling of wanting a man holding her with his strong arms very detailed. After Mu Ying describes this feeling of sexual need, a woman who is the mother of Bare Hips speaks from the front of the crowd and says â€Å"You have your own man, who doesn’t lack an arm or a leg. It’s wrong to have others’ men and more wrong to pocket their money† (Jin 156). And on this moment Mu Ying is still recovering from a punch of the Red Guards and still replies with a smirk on her face looking down on her husband â€Å"I have my own man? † (Jin 156). â€Å"My man is nothing. He is no good, I mean in bed. He always comes before I feel anything† (Jin 156). As a central focus of the public denunciation, Mu appeared to be rather calm when she was caught at home. She neither protested nor said a word, but followed the Red Guards quietly. In her eyes, these Red Guards were only a group of children. She did not expect that the join forces of the Red Guards and the revolutionary masses in the town would be tremendous enough to put her in destruction; more importantly, she did not think that her behavior had violated any rule or law. When her husband appealed to the Red Guards, she stared at him without a word, and a faint smile passed the corners of her mouth. In her yes, the behavior of her impotent husband is pedantic and ridiculous. When the Red Guard asked her why she â€Å"seduced men and paralyze heir revolutionary will with your bourgeois poison† (156), she responded rather calmly with a rhetorical question, â€Å"I’ve never invited any man to my home, have I? † (Jin 156). When several women hissed in the crowd, she even tried to persuade them by citing her own experience: â€Å"Sisters,† she spoke aloud. â€Å"All right, it was wrong to sleep with them. But you all know what it feels like when you want a man, don’t you? Don’t you once in a while have that feeling in your bones? Contemptuously, she looked at the few withered middle-aged women standing in the front row, then closed her eyes. â€Å"Oh, you want that real man to have you in his arms and let him touch every part of your body. For that man alone you want to blossom into a woman, a real woman—† (Jin 156). As a group of juveniles, White Cat and his companions know little about the adult life. This limitation makes their participation a journey of discovery. When adults burst out laughing at Mu Ying’s assertion of her husband’s impotence, the teenagers appeared to be puzzled. The dialogue between them shows this point clearly: â€Å"What’s that? What’s so funny? † Big Shrimp asked Bare Hips. â€Å"You didn’t get it? † Bare Hips said impatiently. â€Å"You don’t know anything about what happens between a man and a woman. It means that whenever she doesn’t want him to come close to her he comes. Bad timing. † â€Å"It doesn’t sound like that,† I said (Jin 157). Obviously, Bare Hips does not know any more than Big Shrimp though his impatient tone tries to conceal this ignorance. Ignorant as he is, Bare Hips makes so bold as to cry at Mu Ying, â€Å"Shameless Old Whore! † (Jin 154). Impossible to perceive, the innocent children are acting the role of accomplice in the public denunciation against Mu. Their thoughts and behaviors manifest the influence they have taken from their parents. In this sense, the innocent teenagers have degenerated from lovely angels to dreadful demons. This is a turning point because when the red guards pulled Mu Ying out of her house and started patrolling her through the city streets to get her to the school yard to sentence her, her husband came running from a street corner begging to let Mu Ying go â€Å"Please don’t take her away. It’s my fault. I haven’t disciplined her well. Please give her a change to be a new person. I promise, she won’t do it again† (Jin 155). And even though I have a feeling Mu Ying does not take the Red guard’s judging protocol as a serious punishment at that point, once she has confessed everything it all gets to her. That is when she gets the feeling that she really has humiliated and hurt her husband, because she is looking for him after she got hit with an ink bottle, only to find that he suddenly left the scene. That is the point she says she does not want to be punished after all and promises to better herself. What I saw as turning point number two becomes clear in the last paragraph, which means I did not see that really as the conclusion. It is the scene where Mu Ying is lying alone at the bus stop, saying: â€Å"Take me home. Oh, help me. Who can help me? Where are you? Why don’t you come and carry me home† (Jin 156). It is a perfect example of â€Å"You don’t know what you have until it is gone. † The second train that used its steam horn killed the husband of Mu Ying and made her a widow and truly alone. What I mean with that is that Mu Ying already felt alone because her man was not good enough for her. This feeling of loneliness combined with her husband’s underachieving partly lead to her committing adultery. Now with her husband dead, she finally exactly knows what it actually means to be alone. But moreover the story clearly shows that her husband saved her live once, after she was raped and now he again tries to get her â€Å"off the hook,† by taking the blame. Her grudge against men because of this rape is probably the other part of why she turned to adultery. It is almost shocking to observe that Mu Ying herself is not punished as the â€Å"Old Whore,† where Meng Su did not only try to take the blame, but also gets himself punished like in Old China. While the Red Guards only make her walk down every street saying the words â€Å"I am an evil monster† (Jin 160), instead of burning her alive, Meng Su is beheaded like in the old days. I strongly felt a parallel between the â€Å"blare of horns† announcing the beheading of Mu Ying’s husband. In Ha Jin’s story, the rough death of Meng Su, the husband, constitutes an â€Å"unscheduled event,† which brings White Cat and his companions to the violence of the adult world. The public denunciation of Mu Ying was no longer a thrilling scene, but something that touched them to their souls, evoking their introspection or maybe disillusion about the world. Bare Hips’s vomiting is a strong signal, indicating the shocking effect that the violence may have brought to him. After the shocking experience, they are no longer innocent adolescences, but adults struggling at the verge of understanding. This whole story turned out like a gruesome, tragic circle of misery and unanswered love to me. I think Meng Su loved Mu Ying dearly, but couldn’t take her grudge against men away that she got from the rape by the Russians. Imaginably, as a victim of the gang rape, she must have experienced a hard time of being treated with disdain. Instead of being hit to death by the accident, she has walked out of the shadow of the concept of chastity, and began to enjoy the pleasure of the flesh as well as economic benefits brought about by men, the invader of her virginity. The bitter time she has experienced has actually hardened her heart and paved the way for her further self-liberation, both physically and spiritually. Meng Su couldn’t cope with not saving her from the shame that the Red Guard trial put on her and killed himself, leaving his beloved Mu Ying really alone and helpless at the bus stop.

The Meaning Behind Alien Abductions :: essays research papers

www.angelfire.com/mt/montalk Souls can only incarnate inoto bodies with DNA responsive to and compatible with that soul. Because souls are shaped by experiences and thoughts, different soul types will reincarnate into different body types of the same species. As humans, their personality is influenced and restricted by the characteristics of their soul. Spiritual evolution is limited by one's biological, genetic, and physical composition. For instance, a dog cannot grow emotionally as we do because their brains are biologically incapable of perceiving the same range of emotions. Dogs also do not feel equal self sacrifice and devotion to truth that some of us humans enjoy, and we are therefore superior to them. Lastly, since all souls, once out of the physical body and deep into the stages of the afterlife, are essentially the same being, there is no place for evolution in the spiritual realm; all evolution occurs on the physical plane of existence. This is why planet earth is so valued by the powers that be. It is the perfect training, evolving, and learning grounds for positive souls with the privilege of physical presence. In this region of the solar system, without an earth, there is no evolution, and whatever pain a soul is in continues indefinitely until that soul can reincarnate to evolve out of its pain. Because souls evolve by earthly hardships, new types of souls can only develop by modification of the older ones over millions of years of evolution. Thus, if a sudden jump occured in the genetic sophistication of a species, a spiritual void would be left as the older souls would be outdated and incompatible with the newly created species. The new species would be born as still births unless some source of soul were available that is suitable with the DNA of that species. The human race was a result of drastic genetic engineering by a coalition of several alien species/races several hundred thousand years ago, each group putting its own spin on the basic human template. Most of the aliens themselves were products of undisturbed evolution upon their homeworlds. Since human evolution was catalyzed, the souls of our ancient primate ancestors could not reincarnate in the newer advanced bodies. Thus existed a spiritual void which was immediately filled with a new source. Considering it was aliens who created humans, these aliens must have had some reason to profit from the task. In fact, most alien bodies were too primitive to allow further spiritual