PostgreSQL 9.1
经营情况
每个月,都会有一批新的帐户分配给特定的流程。每批可以用月份、账户数和账户总余额来描述。该过程的目标是从客户那里收回部分余额。然后每月单独跟踪每个批次(自批次转移到流程后每月回收的金额)。
目标
我的目标是预测将来会收回多少。
数据定义
create table vintage_data (
granularity date, /* Month when account entered process*/
distance_in_months integer, /* Distance in months from date when accounts entered process*/
entry_accounts integer, /* Number of accounts that entered process in a given month*/
entry_amount numeric, /* Total amount for account that entered process in a given month*/
recovery_amount numeric /* Amount recovered in Nth month on accounts that entered process in a given month */
);
样本数据
insert into vintage_data values('2012-01-31',1,200,100000,1000);
insert into vintage_data values('2012-01-31',2,200,100000,2000);
insert into vintage_data values('2012-01-31',3,200,100000,3000);
insert into vintage_data values('2012-01-31',4,200,100000,3500);
insert into vintage_data values('2012-01-31',5,200,100000,3400);
insert into vintage_data values('2012-01-31',6,200,100000,3300);
insert into vintage_data values('2012-02-28',1,250,150000,1200);
insert into vintage_data values('2012-02-28',2,250,150000,1600);
insert into vintage_data values('2012-02-28',3,250,150000,1800);
insert into vintage_data values('2012-02-28',4,250,150000,1200);
insert into vintage_data values('2012-02-28',5,250,150000,1600);
insert into vintage_data values('2012-03-31',1,200,90000,1300);
insert into vintage_data values('2012-03-31',2,200,90000,1200);
insert into vintage_data values('2012-03-31',3,200,90000,1400);
insert into vintage_data values('2012-03-31',4,200,90000,1000);
insert into vintage_data values('2012-04-30',1,300,180000,1600);
insert into vintage_data values('2012-04-30',2,300,180000,1500);
insert into vintage_data values('2012-04-30',3,300,180000,4000);
insert into vintage_data values('2012-05-31',1,400,225000,2200);
insert into vintage_data values('2012-05-31',2,400,225000,6000);
insert into vintage_data values('2012-06-30',1,100,60000,1000);
计算过程
您可以将数据想象为三角矩阵(要预测 X 值):
distance_in_months 1 2 3 4 5 6
granularity entry_accounts entry_amount
2012-01-31 200 100000 1000 2000 3000 3500 3400 3300
2012-02-28 250 150000 1200 1600 1800 1200 1600 (X-1)
2012-03-31 200 90000 1300 1200 1400 1000 (X0) (X4)
2012-04-30 300 180000 1600 1500 4000 (X1) (X5) (X8)
2012-05-31 400 225000 2200 6000 (X2) (X6) (X9) (X11)
2012-06-30 100 60000 1000 (X3) (X7) (X10) (X12 (X13)
算法
我的目标是预测所有缺失的点(未来)。为了说明这个过程,这是对点 X1 的计算
1) 使用最多 4 的距离获取前三个月的行总数:
2012-01-31 1000+2000+3000+3500=9500 (d4m3)
2012-02-28 1200+1600+1800+1200=5800 (d4m2)
2012-03-31 1300+1200+1400+1000=4900 (d4m1)
2) 使用最多 3 的距离获取前三个月的行总数:
2012-01-31 1000+2000+3000=6000 (d3m3)
2012-02-28 1200+1600+1800=4600 (d3m2)
2012-03-31 1300+1200+1400=3800 (d3m1)
3) 计算距离 3 和距离 4 的加权平均跑步率(由 entry_amount 加权):
(d4m3+d4m2+d4m1)/(100000+150000+90000) = (9500+5800+4900)/(100000+150000+90000) = 20200/340000 = 0.0594
(d3m3+d3m2+d3m1)/(100000+150000+90000) = (6000+4600+3800)/(100000+150000+90000) = 14400/340000 = 0.0424
4)计算距离3和距离4之间的变化
((d4m3+d4m2+d4m1)/(100000+150000+90000))/((d3m3+d3m2+d3m1)/(100000+150000+90000)) =
= (20200/340000)/(14400/340000) =
= 0.0594/0.0424 = 1.403 (PredictionRateForX1)
5) 使用不超过 3 的距离计算预测月份的行总数:
2012-04-30 1600+1500+4000=7100
6) 使用 entry_amount 计算预测月份的费率
7100/180000 = 0.0394
7) 计算 X1 的预测速率
0.0394 * PredictionRateForX1 = 0.05534
8) 计算 X1 的数量
(0.05534-0.0394)*180000 = 2869.2
问题
问题是如何使用 SQL 语句计算矩阵的其余部分(从 x-1 到 x13)。很明显,这将需要某种递归算法。