ImageVerifierCode 换一换
格式:PPT , 页数:28 ,大小:391.38KB ,
资源ID:372944      下载积分:2000 积分
快捷下载
登录下载
邮箱/手机:
温馨提示:
如需开发票,请勿充值!快捷下载时,用户名和密码都是您填写的邮箱或者手机号,方便查询和重复下载(系统自动生成)。
如填写123,账号就是123,密码也是123。
特别说明:
请自助下载,系统不会自动发送文件的哦; 如果您已付费,想二次下载,请登录后访问:我的下载记录
支付方式: 支付宝扫码支付 微信扫码支付   
注意:如需开发票,请勿充值!
验证码:   换一换

加入VIP,免费下载
 

温馨提示:由于个人手机设置不同,如果发现不能下载,请复制以下地址【http://www.mydoc123.com/d-372944.html】到电脑端继续下载(重复下载不扣费)。

已注册用户请登录:
账号:
密码:
验证码:   换一换
  忘记密码?
三方登录: 微信登录  

下载须知

1: 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。
2: 试题试卷类文档,如果标题没有明确说明有答案则都视为没有答案,请知晓。
3: 文件的所有权益归上传用户所有。
4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
5. 本站仅提供交流平台,并不能对任何下载内容负责。
6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。

版权提示 | 免责声明

本文(Inventory Basic Model.ppt)为本站会员(diecharacter305)主动上传,麦多课文库仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对上载内容本身不做任何修改或编辑。 若此文所含内容侵犯了您的版权或隐私,请立即通知麦多课文库(发送邮件至master@mydoc123.com或直接QQ联系客服),我们立即给予删除!

Inventory Basic Model.ppt

1、Inventory Basic Model,How can it be that mathematics, being after all a product of human thought which is independent of experience, is so admirably appropriate to the objects of reality? Albert Einstein,In our EOQ models, R and D are used interchangeable. D is demand, R is throughput. We assume R=D

2、 Everything produced is sold. A toy manufacturer uses 32000 silicon chips annually. The Chips are used at a steady rate during the 240 days a year that the plant operates. Holding cost is 60 cents per unit per year. Ordering cost is $24 per order. a) How much should we order each time to minimize ou

3、r total costs (total ordering and carrying costs)?,Problem 1: Optimal Policy,D = 32000, H = $0.6 per unit per year , S = $24 per order Ordering Quantity = Q # of orders = D/Q = 32000/Q Cost of each order = S = $24 OC = 24*32000/Q,Problem 1: EOQ,Problem 1,Cost of carrying one unit of inventory for on

4、e year = H Average Inventory At the start of cycle we have Q, at the end of the cycle we have 0. Average inventory = (Q+0)/2 = Q/2 Q/2 is also called cycle inventory.,In each cycle we have Q/2 inventory. In all cycles we have Q/2 inventory. Throughout the year we have Q/2 inventory. CC = HQ/2,Proble

5、m 1: Optimal Policy,Cost of carrying one unit of inventory for one year = HAt EOQ (Economic Order Quantity, OC=CC SD/Q = HQ/2 240(32000)/Q= 0.6Q/2 Q2= 2560000 Q = 1600 Q2 = 2DS/H,Problem 1,b) How many times should we order ? D = 32000 per year, EOQ = 1600 each time # of times that we order = D/EOQ D

6、/Q = 32000/1600 = 20 times.c) What is the length of an order cycle ? We order 20 times. Working days = 240/year 240/20 = 12 days. Alternatively 32000 is required for one year (240 days) Each day we need 32000/240 = 133.333 1600 is enough for how long? (1600/133.33) = 12 day,Problem 1,d) Compute the

7、average inventory At the start of cycle we have Q, at the end of the cycle we have 0. Average inventory = (Q+0)/2 = Q/2 Q/2 is also called cycle inventory.,In each cycle we have Q/2 inventory. In all cycles we have Q/2 inventory. Throughout the year we have Q/2 inventory.,Problem 1,d) Compute the av

8、erage inventory Average inventory = (Q+0)/2 = 1600/2 =800e) Compute the total carrying cost. We have Q/2 throughout the year Inventory carrying costs = average inventory (Q/2) multiplied by cost of carrying one unit of inventory for one year (H) Total Annual Carrying Cost = H(Q/2) = 0.6(1600/2) = $4

9、80f) Compute the total ordering cost and total cost. Ordering Cost = 24(32000/1600) = 24(20) = $480 Carrying Cost = H(Q/2) = 0.6(1600/2) = $480 Total Cost = Ordering cost + Carrying cost Total cost = $480+$480 = $960,Problem 1,Note that at EOQ total carrying costs is equal to total ordering costs. H

10、Q/2 = SD/Q HQ2=2DS If we solve this equation for Q we will have Q2=2DS/H,That is one way to compute EOQ and not to memorize it.,Problem 1,g) Compute the flow time ? Demand = 32000 per year Therefore throughput = 32000 per year Maximum inventory = EOQ = 1600 Average inventory = 1600/2 = 800 RT=I 3200

11、0T=800 T=800/32000=1/40 year Year = 240 days T=240(1/40)= 6 days Alternatively, the length of an order cycle is 12 days. The first item of an order when received spends 0 days, the last item spends 12 days. On average they spend (0+12)/2 = 6 days,Problem 1,h) Compute inventory turns. Inventory turn

12、= Demand divided by average inventory. Average inventory = I = Q/2 Inventory turns = D/(Q/2)= 32000/(1600/2) Inventory turns = 40 times per year.Notes: Cycle inventory is always defined as Max Inventory divided by 2. Cycle inventory = Q/2 If there is no safety stock Average inventory is the same as

13、Cycle inventory = Q/2. If there is safety stock- We will discuss it in ROP lectureAverage inventory = Cycle Inventory +Safety Stock = Q/2 +Is,Victor sells a line of upscale evening dresses in his boutique. He charges $300 per dress, and sells average 30 dresses per week. Currently, Vector orders 10

14、week supply at a time from the manufacturer. He pays $150 per dress, and it takes two weeks to receive each delivery. Victor estimates his administrative cost of placing each order at $225. His inventory carrying cost including cost of capital, storage, and obsolescence is 20% of the purchasing cost

15、. Assume 50 weeks per year.,Problem 2: Other Policies vs. Optimal Policy,a) Compute the total ordering cost and carrying cost under the current ordering policy? Number of orders/yr = D/Q = 1500/300 = 5 (D/Q) S = 5(225) = 1,125/yr. Average inventory = Q/2 = 300/2 = 150 H = 0.2(150) = 30,Flow unit = o

16、ne dress Flow rate D = 30 units/wk 50 weeks per year Ten weeks supply Q = 10(30) = 300 units. Demand 30(50)= 1500 /yr Fixed order cost S = $225 Unit Cost C = $150/unit H = 20% of unit cost. Lead time L = 2 wees,Problem 2,Annual holding cost = H(Q/2) = 30(150) = 4,500 /yr. Total annual costs = 1125+4

17、500 = 5625 b) Without any further computation, is EOQ larger than 300 or smaller? Why?,Problem 2,c) Compute the flow time. Average inventory = cycle inventory = I = Q/2 Average inventory = 300/2 = 150 Throughput? R? R= D, D= 30/week Current flow time RT= I 30T= 150 T= 5 weeks Did we really need this

18、 computations? Cycle is 10 weeks (each time we order demand of 10 weeks). The first item is there for 0 week. The last item is there for 10 weeks. On average (10+0)/2 = 5 weeks.,Problem 2,d) What is average inventory and inventory turns under this policy ? Inventory turn = Demand divided by average

19、inventory. I = Q/2 Inventory turns = D/(Q/2)= 1500/(300/2) = 10 timesInvTurn = R/I T=I/R InvTurn = 1/TWe already computed T T = 5 weeksTurn = 1/T= 1/5 ? Is InvTurn 10 or 1/5 Have we made a mistake? InvTurn = 1/5 per week, year = 50 weeks InvTurn =(1/5)(50) = 10=,Problem 2,e) Compute Victors total an

20、nual cost of inventory system (carrying plus ordering but excluding purchasing) under the optimal ordering policy?,Q* = EOQ =,= 150 units.,The total optimal annual cost will be,225(1500/150) + 30(150/2) = 2250 + 2250 = $4,500,Compared to 5,625, there is about 20% reduction in the total costs. Total

21、cost here is equal to carrying cost there.,Problem 2,f) When do you order (re-order point) ? An order for 150 units two weeks before he expects to run out. That is, whenever current inventory drops to 30 units/wk * 2 wks = 60 units. Which is the re-order point. When to order? When inventory on hand

22、is 60. How much to order? 150. R and Q Strategy.,Central Electric (CE) serves its European customers through a distribution network that consisted of four warehouses, in Poland, Italy, France, and Germany. The network of warehouses was built on the premise that it will allow CE to be close to the cu

23、stomer. Contrary to expectations, establishing the distribution network led to an inventory crisis. CE is considering to consolidate the regional warehouses into a single master warehouse in Austria. The following data is for the sake of analysis of this problem - not real world data. Currently, eac

24、h warehouse manages its ordering independently. Demand at each outlet averages 800 units per day. Assume a year is 250 days. Each unit of product costs $200, and CE has a holding cost of 20% per annum. The fixed cost of each order (administrative plus transportation) is $900 for the decentralized sy

25、stem and $2025 for the centralized system.,Problem 3; Centralization vs. Decentralization,Decentralized: Four warehouses in Poland, Italy, France, and Germany Centralized: One warehouse in Austria The holding cost will be the same in both decentralized and centralized ordering systems. H(decentraliz

26、ed) =20%(200) = $40 per unit per yr. H(centralized) = $40 per unit per yr. The ordering cost in the centralized ordering system is $2025. S(decentralized) = $900 per order. S(centralized) $900 = $2025 per unit per yr. The problem assumes this. It is also realistic, when we deliver centrally, S goes

27、up since the truck travel time in a route to 4 warehouses is longer than a trip to a single warehouse.,Problem 3; Centralization vs. Decentralization,Four outlets Each outlet demand D = 800(250) = 200,000 S= 900 C = 200 H = 0.2(200) = 40 If all warehouses merged into a single warehouse, then S= 2025

28、,Problem 3,=3000,With a cycle inventory of 1500 units for each warehouse.,b) Compute EOQ and cycle inventory in the centralized ordering In this problem, in the centralized system, S = $2025.,=9000,and a cycle inventory of 4500.,Compute EOQ and cycle inventory in decentralized ordering,The total cyc

29、le inventory across all four outlets equals 6000.,Problem 3,c) Compute the total annual holding cost + ordering cost (not including purchasing cost) for both policies TC = S(D/Q) + H(Q/2) Decentralized TC= 900(200000/3000) + 40(3000/2) TC = 60000+60000= 120000 Decentralized: TC for all 4 warehouses

30、= 4(120000)=480000 Centralized TC= 2025(800000/9000) + 40(9000/2) TC= 180000+180000 = 360000 480000 360000; about 25% improvement in the total costs,Problem 3,d) Compute the ordering interval in decentralized and centralized systems.,Decentralized = (3000/200000)(250) = 3.75 days Centralized = (9000

31、/800000)(250) = 2.821 dayse) Compute the average flow time3.75/2 = 1.875 days 2.821/2 = 1.41 days RT = I T= R/I 200000T= 1500 T = 1500/200000 year or 1.875 days 800000T= 4500 T = 4500/800000 year or 1.41 days The same computations,Problem 3: Inventory Turns,f) Compute inventory turns Inventory Turns

32、 = Demand /Average inventory = R/I = InvTurnDemand Average inventory Inventory Turns 200000 1500 200000/1500 = 133.33 800000 4500 800000/4500 = 177.78g) If the lead time is 2 days, when do you order? (re-order point)? Decentralized 2(800) = 1600 units Centralized = 2(4)(800) = 6400 units,Why We are

33、interested in reducing inventory.,Inventory adversely affects all competing edges (P/Q/V/T) Has cost Physical carrying costs Financial costs Has risk of obsolescence Due to market changes Due to technology changes Leads to poor quality Feedback loop is long Hides problems Unreliable suppliers, machi

34、ne breakdowns, long changeover times, too much scrap. Causes long flow time, not-uniform operations,How to Reduce EOQ,To reduce EOQ we may R, S, H Two ways to reduce average inventory - Reduce S - Postponement, Delayed Differentiation - CentralizeS does not increase in proportion of QEOQ increases a

35、s the square route of demand. - Commonality, modularization and standardization is another type of Centralization,Why not Always Centralized,If centralization reduces inventory, why doesnt everybody do it? Higher shipping costLonger response timeLess understanding of customer needsLess understanding

36、 of cultural, linguistics, and regulatory barriers These disadvantages my reduce the demand.,Formula Proof for Total Cost of EOQ,Total cost of any Q? TCQ = SR/Q + HQ/2 Total Cost of EOQ? The same as above, but can also be simplified = 2 + 2 2 = =2 2 2 = 2 = 2 2 = 2 = 2,Formula Proof for Flow Time Under EOQ,Flow time when we order of any Q? Throughput = R, average inventory I = Q/2 RT = Q/2 T = Q/2R Flow time when we order of EOQ? Total Cost of EOQ? The same as above, but can also be simplified I = EOQ/2 = 2 = 2 2 = 2 4 = 2 T = I/R T= = 2 = 22 = 2,

copyright@ 2008-2019 麦多课文库(www.mydoc123.com)网站版权所有
备案/许可证编号:苏ICP备17064731号-1