﻿ Z公司多品种分组联合订货策略研究 Z Company’s Multi-Variety Grouping Combined Ordering Strategy

Management Science and Engineering
Vol. 08  No. 02 ( 2019 ), Article ID: 30582 , 13 pages
10.12677/MSE.2019.82019

Z Company’s Multi-Variety Grouping Combined Ordering Strategy

Qin Zheng

Beijing Jiaotong University, Beijing

Received: May 8th, 2019; accepted: May 24th, 2019; published: May 31st, 2019

ABSTRACT

In the increasingly fierce market competition, low-cost strategy has become the first choice of more and more enterprises. As an important indicator affecting cost, inventory control is more and more valued by enterprises. How to formulate appropriate ordering strategy is the difficulty of inventory control. In the manufacturing enterprises, the multi-variety joint ordering strategy is used to improve the enterprise inventory control method, which plays a certain role in promoting the business development of the enterprise. In this paper, based on the optimal total inventory cost, combined with the order cycle, a multi-item joint ordering model based on the order cycle is established. Through the calculation of the optimal order cycle, the same (similar) order cycle is used to classify the joint order. Finally, taking Z company as an example, the matlab solution is used to verify the practicability of the model.

Keywords:Inventory Control, Ordering Cycle, Multiple Varieties, Ordering Policy

Z公司多品种分组联合订货策略研究

1. 引言

2. 文献综述

3. 多物品联合订购的库存控制模型

3.1. 问题描述

1) 所有物品同时订货

2) 部分物品同时订货

3.2. 多物品分组联合订货模型

1) 参数定义

ei：第i种产品的额外订购费用；

gi：第i种产品的单位购买成本；

h：库存持有成本与购买成本之比；

A：订货成本，与订货次数有关；

Di：第i种产品的月平均需求；

Qi：第i种产品的订货量；

2) 模型的假设条件

a) 当产品的补货量不是整数时，向上取整；

b) 每种产品的需求确定且为固定不变地常数；

c) 以月为周期计算；

d) 不考虑缺货。

${Q}_{i}={D}_{i}T$ (1)

${\stackrel{¯}{I}}_{i}=\frac{{Q}_{i}}{2}=\frac{{D}_{i}T}{2}$ (2)

$AT{C}_{1}\left(T\right)=\frac{A+\underset{i=1}{\overset{n}{\sum }}{e}_{i}}{T}+\underset{i=1}{\overset{n}{\sum }}\frac{{D}_{i}×T×h×{g}_{i}}{2}$ (3)

$T*=\sqrt{\frac{2\left(A+\underset{i=1}{\overset{n}{\sum }}{e}_{i}\right)}{h\underset{i=1}{\overset{n}{\sum }}{D}_{i}{g}_{i}}}$ (4)

$AT{C}_{1}{}^{*}=\sqrt{2\left(A+\underset{i=1}{\overset{n}{\sum }}{e}_{i}\right)h\underset{i=1}{\overset{n}{\sum }}{D}_{i}{g}_{i}}$ (5)

${Q}_{i}={D}_{i}×{m}_{i}×T\text{}i=1,2,...,n$ (6)

${\stackrel{¯}{I}}_{i}=\frac{{Q}_{i}}{2}=\frac{{D}_{i}×{m}_{i}×T}{2}$ (7)

$EATC\left(T,{m}_{1},\cdots ,{m}_{n}\right)=\frac{\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)×{D}_{i}}{{Q}_{i}}+\underset{i=1}{\overset{n}{\sum }}\frac{{D}_{i}×{m}_{i}×T×h×{g}_{i}}{\text{2}}=\frac{\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)}{T}+\underset{i=1}{\overset{n}{\sum }}\frac{{D}_{i}×{m}_{i}×T×h×{g}_{i}}{2}$ (8)

$\frac{\partial EATC\left(T,{m}_{1},\cdots ,{m}_{n}\right)}{\partial T}=-\frac{A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}}{{T}^{2}}+\underset{i=1}{\overset{n}{\sum }}\frac{{D}_{i}×{m}_{i}×T×h×{g}_{i}}{2}$ (9)

$\frac{{\partial }^{2}EATC\left(T,{m}_{1},\cdots ,{m}_{n}\right)}{\partial {T}^{2}}=\frac{2×\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)}{{T}^{2}}>0$ (10)

$-\frac{A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}}{{T}^{2}}+\underset{i=1}{\overset{n}{\sum }}\frac{{D}_{i}×{m}_{i}×h×{g}_{i}i}{2}=0$

$T\left({m}_{1},\cdots ,{m}_{n}\right)\text{=}\sqrt{\frac{\text{2}×\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)}{h×\underset{i=1}{\overset{n}{\sum }}{D}_{i}×{m}_{i}×{g}_{i}}}$

${T}^{*}\left({m}_{1},\cdots ,{m}_{n}\right)=\sqrt{\frac{2×\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)}{h×\underset{i=1}{\overset{n}{\sum }}{D}_{i}×{m}_{i}×{g}_{i}}}$

${T}^{*}\left({m}_{1},\cdots ,{m}_{n}\right)$ 的值代入式(8)，化简得

$EAT{C}^{*}\left(T,{m}_{1},\cdots ,{m}_{n}\right)=\sqrt{2h\left(A+\underset{i=1}{\overset{n}{\sum }}{e}_{i}\right)\underset{i=1}{\overset{n}{\sum }}{D}_{i}×{m}_{i}×{g}_{i}}$ (11)

$f\left({m}_{1},\cdots ,{m}_{n}\right)=\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)×\underset{i=1}{\overset{n}{\sum }}\left({D}_{i}×{m}_{i}×{g}_{i}\right)$ (12)

$f\left({m}_{1},\cdots ,{m}_{n}\right)$ 式中的mi求导，令偏导等于0得

$-\frac{{e}_{i}}{{m}_{i}^{2}}×\underset{i=1}{\overset{n}{\sum }}\left({D}_{i}×{m}_{i}×{g}_{i}\right)+\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)×{D}_{i}×{g}_{i}=0$

${m}_{i}^{2}=\frac{{e}_{i}×\underset{i=1}{\overset{n}{\sum }}{D}_{i}×{m}_{i}×{g}_{i}}{\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)×{D}_{i}×{g}_{i}},i=1,2,\cdots ,n$ (13)

${m}_{k}^{2}=\frac{{e}_{k}}{{D}_{k}×{g}_{k}}×\frac{\underset{i=1}{\overset{n}{\sum }}{D}_{i}×{m}_{i}×{g}_{i}}{\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)},k=1,2,\cdots ,n$ (14)

$\frac{{m}_{i}}{{m}_{k}}=\sqrt{\frac{{e}_{i}}{{D}_{i}×{g}_{i}}/\frac{{e}_{k}}{{D}_{k}×{g}_{k}}},i\ne k$

$\frac{{e}_{k}}{{D}_{k}×{g}_{k}}<\frac{{e}_{i}}{{D}_{i}×{g}_{i}}$

${m}_{k}<{m}_{i}$

${m}_{k}=\sqrt{\frac{{e}_{k}}{{D}_{k}×{g}_{k}}}×\sqrt{\frac{\underset{i=1}{\overset{n}{\sum }}{D}_{i}×{m}_{i}×{g}_{i}}{\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)}},k=1,2,\cdots ,n$ (15)

$\theta =\sqrt{\frac{\underset{i=1}{\overset{n}{\sum }}{D}_{i}×{m}_{i}×{g}_{i}}{\left(A+\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}\right)}}$ (16)

${m}_{k}=\theta \sqrt{\frac{{e}_{k}}{{D}_{k}×{g}_{k}}},k=1,2,\cdots ,n$ (17)

$\underset{i=1}{\overset{n}{\sum }}{D}_{i}×{m}_{i}×{g}_{i}={D}_{1}×{g}_{1}+\underset{i=2}{\overset{n}{\sum }}\theta \sqrt{\frac{{e}_{i}}{{D}_{i}×{g}_{i}}}×{D}_{i}×{g}_{i}={D}_{1}×{g}_{1}+\underset{i=2}{\overset{n}{\sum }}\theta \sqrt{{e}_{i}×{D}_{i}×{g}_{i}}$ (18)

$\underset{i=1}{\overset{n}{\sum }}\frac{{e}_{i}}{{m}_{i}}={e}_{1}+\frac{1}{\theta }×\underset{i=2}{\overset{n}{\sum }}\sqrt{{e}_{i}×{D}_{i}×{g}_{i}}$ (19)

${\theta }^{2}=\frac{{D}_{1}×{g}_{1}+\theta ×\underset{i=2}{\overset{n}{\sum }}\sqrt{{e}_{i}×{D}_{i}×{g}_{i}}}{A+{e}_{1}+\frac{1}{\theta }×\underset{i=2}{\overset{n}{\sum }}\sqrt{{e}_{i}×{D}_{i}×{g}_{i}}}$

$\theta =\sqrt{\frac{{D}_{1}×{g}_{1}}{A+{e}_{1}}}$ (20)

${m}_{i}=\sqrt{\frac{{D}_{1}×{g}_{1}}{A+{e}_{1}}×\frac{{e}_{i}}{{D}_{i}×{g}_{i}}},i=1,2,...,n$ (21)

4. 算例验证

4.1. 问题提出

Z公司是一家以现代农业装备生产为主的制造企业，主营包括汽车车身、车架、车轿等核心零部件的设计、制造。当前Z公司已形成年产各种类型汽车六十万辆，电动车二十万辆，农业机械设备十万台，汽车配件三十余万套的规模。以Z公司汽车事业部2018年10月至2019年1月四个月时间里15种按月交付物品的需求量与订货量汇总表。

Table 1. Summary of material requirements, orders and stocks for October-November 2018

Table 2. Summary of material requirements, orders and stocks from December 2018 to January 2019

4.2. 模型求解

Table 3. Z company order cycle related data

T0 = 16 + 8 + 16 + 8 + 8 = 56 (天)

T1 = 56 − (8 − 1) = 49 (天)

T2 = 56 − [(8 − 1) + (8 − 1)] = 42 (天)

Table 4. Solving results of order cycle correlation coefficient

Figure 1. Order cycle cost distribution

Table 5. Data related to 15 parts procurement

Figure 2. Distribution of order cycle

${T}^{*}\left({m}_{1},\cdots ,{m}_{k}\right)=\sqrt{\frac{2×\left[A+\underset{k=1}{\overset{n}{\sum }}\frac{{e}_{k}}{{m}_{k}}+\underset{j=1}{\overset{i-1}{\sum }}{c}_{j}×\left({b}_{j}-{a}_{j}\right)+{c}_{i}×{T}_{i-1}\right]}{h×\underset{k=1}{\overset{n}{\sum }}\left({D}_{k}×{m}_{k}×{g}_{k}\right)}}=0.0406\left(年\right)14\left(天\right)$

Table 6. Correlation coefficient solution results

$EAT{C}^{*}\left({m}_{1},\cdots ,{m}_{k}\right)=\sqrt{2×\left[A+\underset{k=1}{\overset{n}{\sum }}\frac{{e}_{k}}{{m}_{k}}+\underset{j=1}{\overset{i-1}{\sum }}{c}_{j}×\left({b}_{j}-{a}_{j}\right)+{c}_{i}×{T}_{i-1}\right]×h×\underset{k=1}{\overset{n}{\sum }}\left({D}_{k}×{m}_{k}×{g}_{k}\right)}-{c}_{i}=284,490\left(元\right)$

4.3. 结果对比

4.4. Z公司多品种联合订货策略方案

Figure 3. Order cycle distribution of joint order

Table 7. Optimal ordering plan for 15 parts of Z company

5. 总结

Z Company’s Multi-Variety Grouping Combined Ordering Strategy[J]. 管理科学与工程, 2019, 08(02): 143-155. https://doi.org/10.12677/MSE.2019.82019

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