Average Cost (AVCO) Method
Average cost method (AVCO) calculates the cost of ending inventory and cost of goods sold for a period on the basis of weighted average cost per unit of inventory. Weighted average cost per unit is calculated using the following formula:
Weighted Average | = | Total Cost of Inventory |
Unit Cost | Total Units in Inventory |
Like FIFO and LIFO methods, AVCO is also applied differently in periodic inventory system and perpetual inventory system. In periodic inventory system, weighted average cost per unit is calculated for the entire class of inventory. It is then multiplied with number of units sold and number of units in ending inventory to arrive at cost of goods sold and value of ending inventory respectively. In perpetual inventory system, we have to calculate the weighted average cost per unit before each sale transaction.
The calculation of inventory value under average cost method is explained with the help of the following example:
Example
Apply AVCO method of inventory valuation on the following information, first in periodic inventory system and then in perpetual inventory system to determine the value of inventory on hand on Mar 31 and cost of goods sold during March.
Mar 1 | Beginning Inventory | 60 units @ $15.00 per unit |
5 | Purchase | 140 units @ $15.50 per unit |
14 | Sale | 190 units @ $19.00 per unit |
27 | Purchase | 70 units @ $16.00 per unit |
29 | Sale | 30 units @ $19.50 per unit |
Solution
AVCO Periodic
Units Available for Sale | = 60 + 140 + 70 | = 270 | |
Units Sold | = 190 + 30 | = 220 | |
Units in Ending Inventory | = 270 − 220 | = 50 | |
Weighted Average Unit Cost | Units | Unit Cost | Total |
Mar 1 Inventory | 60 | $15.00 | $900 |
Mar 5 Purchase | 140 | $15.50 | $2,170 |
27 Purchase | 70 | $16.00 | $1,120 |
270 | * $15.52 | $4,190 | |
* $4,190 ÷ 270 | |||
Cost of Goods Sold | 220 | $15.52 | $3,414 |
Ending Inventory | 50 | $15.52 | $776 |
AVCO Perpetual
Date | Purchases | Sales | Balance | ||||||
Units | Unit Cost | Total | Units | Unit Cost | Total | Units | Unit Cost | Total | |
Mar 1 | 60 | $15.00 | $900 | ||||||
5 | 140 | $15.50 | $2,170 | 60 | $15.00 | $900 | |||
140 | $15.50 | $2,170 | |||||||
200 | $15.35 | $3,070 | |||||||
14 | 190 | $15.35 | $2,916 | 10 | $15.35 | $154 | |||
27 | 70 | $16.00 | $1,190 | 10 | $15.35 | $154 | |||
70 | $16.00 | $1,120 | |||||||
80 | $15.92 | $1,274 | |||||||
29 | 30 | $15.92 | $478 | 50 | $15.92 | $796 | |||
31 | 50 | $15.92 | $796 |
Average Cost (AVCO) Method的更多相关文章
- 论文解析 "A Non-Local Cost Aggregation Method for Stereo Matching"
传统的使用窗口的方法缺陷主要在 1.窗口外的像素不能参与匹配判断. 2.在低纹理区域很容易产生错误匹配 论文的主要贡献在代价聚类上(左右图像带匹配点/区域的匹配代价计算),目标是图像内所有点都对该点传 ...
- 基于MST的立体匹配及相关改进(A Non-Local Cost Aggregation Method for Stereo Matching)
怀着很纠结的心情来总结这篇论文,这主要是因为作者提虽然供了源代码,但是我并没有仔细去深究他的code,只是把他的算法加进了自己的项目.希望以后有时间能把MST这一结构自己编程实现!! 论文题目是基于非 ...
- Oracle EBS-SQL (CST-1):检查BOM历史成本查询(Average Cost).sql
select msi1.segment1 父件编码, msi1.description 父件描述, msi1.primary_u ...
- 泡泡一分钟: A Linear Least Square Initialization Method for 3D Pose Graph Optimization Problem
张宁 A Linear Least Square Initialization Method for 3D Pose Graph Optimization Problem "链接:https ...
- Odoo13 新变化:存货核算
Odoo13将于2019年10月发布,本次发布也包含了大量的改进,例如,对存货核算的重构. 去掉了 产品历史价格product.price.history ,增加了 stock valuation l ...
- Adding New Functions to MySQL(User-Defined Function Interface UDF、Native Function)
catalog . How to Add New Functions to MySQL . Features of the User-Defined Function Interface . User ...
- UVA 1456 六 Cellular Network
Cellular Network Time Limit:3000MS Memory Limit:0KB 64bit IO Format:%lld & %llu Submit S ...
- 大规模视觉识别挑战赛ILSVRC2015各团队结果和方法 Large Scale Visual Recognition Challenge 2015
Large Scale Visual Recognition Challenge 2015 (ILSVRC2015) Legend: Yellow background = winner in thi ...
- R语言-聚类与分类
一.聚类: 一般步骤: 1.选择合适的变量 2.缩放数据 3.寻找异常点 4.计算距离 5.选择聚类算法 6.采用一种或多种聚类方法 7.确定类的数目 8.获得最终聚类的解决方案 9.结果可视化 10 ...
随机推荐
- bzoj1032 [JSOI2007]祖码Zuma
1032: [JSOI2007]祖码Zuma Time Limit: 10 Sec Memory Limit: 162 MBSubmit: 672 Solved: 335[Submit][Stat ...
- 转载 模板整理 by gc812
http://www.cnblogs.com/gc812/p/5779789.html 上友链,不盗版 CC BY-NC-SA
- HDU 4417 Super Mario(划分树+二分)
题目链接 #include <cstdio> #include <cstring> #include <algorithm> using namespace std ...
- Java_解决java.security.cert.CertificateException: Certificates does not conform to algorithm constraints
找到 jre/lib/security/java.security 将 jdk.certpath.disabledAlgorithms=MD2, DSA, RSA keySize < 2048 ...
- MySQL 用户管理——权限表
权限表 权限表存放在mysql数据库中 user表结构 用户列:Host.User.Password 权限列:*priv 资源控制列:max* 安全列:其余 db表 存储了用户对某个数据库的操作权 ...
- 运行时(iOS)
运行时(iOS) 一.什么是运行时(Runtime)? 运行时是苹果提供的纯C语言的开发库(运行时是一种非常牛逼.开发中经常用到的底层技术) 二.运行时的作用? 能获得某个类的所有成员变量 能获得 ...
- python表达式
算术表达式: 地板除: >>> 10 // 3 3>>> 5 // 2 2>>> 5 // 31 取余: >>> 10 % 31 ...
- Leetcode | Parentheses 相关
Generate Parentheses Given n pairs of parentheses, write a function to generate all combinations of ...
- mongodb复制集配置
#more /opt/mongodb3.0/mongodb_im_conf_47020/mongodb3.0_im_47020.cnf dbpath = /opt/mongodb3.0/mongodb ...
- AMD PerfStudio
用PerfStudio 抓DX11 OIT