复制代码 代码如下:
—在仓储管理中经常会碰到的一个问题
一、关于lifo与fifo的简单说明
—fifo: first in, first out.先进先出。
—lifo: last in, first out.后进先出。
–如货物a:本月1日购买10件,单价10元/件,3日购买20件,单价15元/件;10日购买10件,单价8元/件。
–本月15日发货35件。
–按fifo先进先出,就是先购入的存货先发出,所以,先发1日进货的10件,再发3日进货的20件,最后发10日进货的5件,发出成本共为:10*10+20*15+5*8=440元。
–按lifo后进先出,就是后购入的存货先发出,所以,先发10日进货的10件,再发3日进货的20件,最后发1日进货的5件,发出成本共为:10*8+20*15+5*10=430元
二、示例
复制代码 代码如下:
——–
create table stock
(id int not null primary key,
articleno varchar(20) not null,
rcvdate datetime not null,
qty int not null,
unitprice money not null
)
go
—-
insert stock
select 1,’10561122′,’2011-1-1′,15,10 union
select 2,’10561122′,’2011-2-2′,25,12 union
select 3,’10561122′,’2011-3-3′,35,15 union
select 4,’10561122′,’2011-4-4′,45,20 union
select 5,’10561122′,’2011-5-5′,55,10 union
select 6,’10561122′,’2011-6-6′,65,30 union
select 7,’10561122′,’2011-7-7′,75,17 union
select 8,’10561122′,’2011-8-8′,110,8
go
—-此时如果在2011-8-8卖出300件产品,那么应该如何计算库存销售的价值呢?
—-1使用当前的替换成本,2011-8-8时每件产品的成本为8,就是说你这300件产品,成本价值为2400
—-2使用当前的平均成本单价,一共有420,总成本为6530,平均每件的成本为15.55
—-1.lifo (后进先出)
—-2011-8-8 110 *8
—-2011-7-7 75*17
—-2011-6-6 65*30
—-2011-5-5 50*10
—–总成本为 4605
—–2.fifo(先进先出)
—- ‘2011-1-1’,15*10
— ‘2011-2-2’,25*12
—–‘2011-3-3’,35*15
—–‘2011-4-4’,45*20
—–‘2011-5-5’,55*10
—–‘2011-6-6’,65*30
—–‘2011-7-7’,65*17
—-总成本为5480
—成本视图
create view costlifo
as
select unitprice from stock
where rcvdate= (select max(rcvdate) from stock)
go
create view costfifo
as
select sum(unitprice*qty)/sum(qty) as unitprice from stock
go
—–找出满足订单的、足够存货的最近日期。如果运气好的话,某一天的库存数量正好与订单要求的数字完全一样
—–就可以将总成本作为答案返回。如果订单止的数量比库存的多,什么也不返回。如果某一天的库存数量比订单数量多
—则看一下当前的单价,乘以多出来的数量,并减去它。
—下面这些查询和视图只是告诉我们库存商品的库存价值,注意,这些查询与视图并没有实际从库存中向外发货。
create view lifo
as
select s1.rcvdate,s1.unitprice,sum(s2.qty) as qty,sum(s2.qty*s2.unitprice) as totalcost
from stock s1 ,stock s2
where s2.rcvdate>=s1.rcvdate
group by s1.rcvdate,s1.unitprice
go
select (totalcost-((qty-300)*unitprice )) as cost
from lifo as l
where rcvdate=(select max(rcvdate) from lifo as l2 where qty>=300)
go
create view fifo
as
select s1.rcvdate,s1.unitprice,sum(s2.qty) as qty,sum(s2.qty*s2.unitprice) as totalcost
from stock s1 ,stock s2
where s2.rcvdate<=s1.rcvdate
group by s1.rcvdate,s1.unitprice
go
select (totalcost-((qty-300)*unitprice )) as cost
from fifo as l
where rcvdate=(select min(rcvdate) from lifo as l2 where qty>=300)
——–
go
—–
—–在发货之后,实时更新库存表
create view currstock
as
select s1.rcvdate,sum(case when s2.rcvdate>s1.rcvdate then s2.qty else 0 end) as prvqty
,sum(case when s2.rcvdate<=s1.rcvdate then s2.qty else 0 end) as currqty
from stock s1 ,stock s2
where s2.rcvdate<=s1.rcvdate
group by s1.rcvdate,s1.unitprice
go
create proc removeqty
@orderqty int
as
if(@orderqty>0)
begin
update stock set qty =case when @orderqty>=(select currqty from currstock as c where c.rcvdate=stock.rcvdate)
then 0
when @orderqty<(select prvqty from currstock c2 where c2.rcvdate=stock.rcvdate)
then stock.qty
else (select currqty from currstock as c3 where c3.rcvdate=stock.rcvdate)
-@orderqty end
end
—
delete from stock where qty=0
—
go
exec removeqty 20
go
—————
三、使用“贪婪算法”进行订单配货
复制代码 代码如下:
——-还有一个问题,如何使用空间最小或最大的仓库中的货物来满足订单,假设仓库不是顺序排列,你可以按钮希望的顺序任意选择满足订单。
—使用最小的仓库可以为订单的装卸工人带来最小的工作量,使用最大的仓库,则可以在仓库中清理出更多的空间
——-例如:对于这组数据,你可以使用(1,2,3,4,5,6,7)号仓库也可以使用(5,6,7,8)号仓库中的货物来满足订单的需求。
—-这个就是装箱问题,它属于np完全系统问题。对于一般情况来说,这种问题很难解决,因为要尝试所有的组合情况,而且如果数据量大的话,
—-计算机也很难很快处理。
—所以有了“贪婪算法”,这个算法算出来的常常是近乎最优的。这个算法的核心就是“咬最大的一口”直到达到或超越目标。
—
–1. 第一个技巧,要在表中插入一些空的哑仓库,如果你最多需要n次挑选,则增加n-1个哑仓库
insert stock
select -1,’10561122′,’1900-1-1′,0,0 union
select -2,’10561122′,’1900-1-1′,0,0
–select -3,’1900-1-1′,0,0
—-
go
create view pickcombos
as
select distinct (w1.qty+w2.qty+w3.qty) as totalpick
,case when w1.id<0 then 0 else w1.id end as bin1 ,w1.qty as qty1,
case when w2.id<0 then 0 else w2.id end as bin2,w2.qty as qty2
,case when w3.id<0 then 0 else w3.id end as bin3 ,w3.qty as qty3
from stock w1,stock w2, stock w3
where w1.id not in (w2.id,w3.id)
and w2.id not in (w1.id,w3.id)
and w1.qty>=w2.qty
and w2.qty>=w3.qty
—-
—1.使用存储过程来找出满足或接近某一数量的挑选组合
——–
go
create proc overpick
@pickqty int
as
if(@pickqty>0)
begin
select @pickqty,totalpick,bin1,qty1,bin2,qty2,bin3,qty3
from pickcombos
where totalpick=(select min(totalpick) from pickcombos where totalpick>=@pickqty)
end
go
exec overpick 180
———-
select * from stock
drop table stock
drop view lifo
drop view fifo
drop view costfifo
drop view costlifo
drop view currstock
drop proc overpick
drop proc removeqty
drop view pickcombos