Lampiran 2
b
t
=
1
α α
−
. S’
t
– S”
t
b
2
= 0,2 1-0,2 . S’
2
– S”
2
= 0,2 0,8 . 14,2-14,04 = 0,25.0,16 = 0,04 F
t+m
= a
t
+ b
t
.m m = 1
Æ 1 periode ke depan F
2+1
= F
3
= d
3
’ = a
2
+ b
2
.1 = 14,36 + 0,041 = 14,4 dt – dt’ = d
3
– d
3
’ = 14 – 14,4 = -0,4 dt – dt’
2
= d
3
– d
3
’
2
= -0,4
2
= 0,16
732 ,
19 22
09 ,
434
1 2
= =
− =
∑
=
n d
d MSE
n t
t t
e. Metode Triple Exponential Smoothing
Tabel L2.59 Peramalan metode triple exponential smoothing item 216 mm x 22 m
t dt
St St
St a
t
b
t
c
t
F
t+m
dt - dt dt - dt
2
1 14
- -
- -
- -
- -
- 2
15 14,2
14,04 14,01
14,49 0,011
0,008 -
- -
3 14
14,16 14,06
14,02 14,31
0,005 0,0032
14,5028 -0,5028
0,25280784 4
24 16,13
14,48 14,11
19,06 0,11
0,0803 14,31392
9,68608 93,82014577
5 19
16,7 14,92
14,27 19,61
0,105 0,0707 19,214032
-0,214032 0,045809697
6 12
15,76 15,09
14,44 16,45
0,018 0,0012 19,754384
-7,754384 60,13047122
7 20
16,61 15,39
14,63 18,27
0,054 0,0281 16,470707 3,5292928
12,45590767 8
21 17,49
15,81 14,86
19,89 0,081
0,0455 18,343264 2,6567356 7,058244133
9 17
17,39 16,13
15,12 18,9
0,045 0,0157 19,993121
-2,993121 8,95877515
10 19 17,71
16,44 15,38
19,18 0,043
0,0128 18,956226 0,0437744 0,001916198
11 20 18,17
16,79 15,66
19,8 0,048
0,0159 19,2335
0,7665001 0,587522397
12 25 19,54
17,34 16
22,59 0,1
0,0536 19,859725 5,1402753 26,42242994
13 21 19,83
17,84 16,37
22,34 0,077
0,0326 22,716607 -1,716607
2,946739673 14 22
20,26 18,32
16,76 22,58
0,069 0,0235 22,435537
-0,435537 0,189692777
15 18 19,81
18,62 17,13
20,7 0,014 -0,0187 22,660383
-4,660383 21,71917152
16 11 18,05
18,51 17,41
16,03 -0,094 -0,0973 20,706391 -9,706391
94,21402169 17 14
17,24 18,25
17,57 14,53 -0,115 -0,1057 15,890797
-1,890797 3,575114136
18 18 17,39
18,08 17,68
15,61 -0,075 -0,0683 14,366159 3,6338407 13,20479853
19 19 17,71
18,01 17,74
16,86 -0,037 -0,0349 15,499324 3,5006764 12,25473506
20 10 16,17
17,64 17,72
13,31 -0,11 -0,0867 16,806239
-6,806239 46,32489079
21 16 16,14
17,34 17,64
14,04 -0,078 -0,056 13,160546
2,839454 8,062498834
22 17 16,31
17,13 17,54
15,07 -0,043 -0,0259 13,931785 3,0682149 9,413942741
23 20 17,05
17,12 17,46
17,25 0,013
0,0171 15,01563
4,9843697 24,84394176
24 18 17,24
17,14 17,39
17,69 0,021
0,022 17,273085 0,7269152
0,528405696 25
- -
- -
- -
- 17,718867
- -
TOTAL 447,0119832
MSE 20,319
Lampiran 2
Perhitungan : Metode Kuadratic Browns Inisialisasi :
2 ,
= α
S’
1
= S”
1
= S”’
1
= d
1
S’
t
= α .d
t
+ 1- α .S’
t-1
S’
2
= 0,2d
2
+ 1-0,2. S’
1
= 0,215 + 0,814 = 14,2 S’
3
= 0,2d
3
+ 1-0,2. S’
2
= 0,214 + 0,814,2 = 14,16 S”
t
= α . S’
t
+ 1- α .S”
t-1
S”
2
= 0,2 S’
2
+ 1-0,2. S”
1
= 0,214,2 + 0,814 = 14,04 S”
3
= 0,2 S’
3
+ 1-0,2. S”
2
= 0,214,16 + 0,814,04 = 14,06 S”’
t
= α . S”
t
+ 1- α .S”’
t-1
S”’
2
= 0,2 S”
2
+ 1-0,2. S”’
1
= 0,214,04 + 0,814 = 14,01 S”’
3
= 0,2 S”
3
+ 1-0,2. S”’
2
= 0,214,06 + 0,814,01 = 14,02 a
t
= 3.S’
t
– 3.S”
t
+ S”’
t
a
2
= 3.S’
2
– 3.S”
2
+ S”’
2
= 314,2 – 314,04 + 14,01 = 14,49 b
t
=
2
1 2
α α
− .[6-5
α .S’
t
– 10-8 α .S”
t
+ 4-3 α .S”’
t
] b
2
= 0,221-0,2
2
. [6-50,2.S’
2
– 10-80,2.S”
2
+ 4-30,2.S”’
2
] b
2
= 0,21,6
2
. [514,2 – 8,414,04 + 3,414,01] = 0,011 c
t
=
2 2
1 α
α −
.[S’
t
– 2.S”
t
+ S”’
t
] c
2
= 0,2
2
1-0,2
2
.[S’
2
– 2.S”
2
+ S”’
2
] c
2
= 0,04 0,64.[14,2 – 214,04 + 14,01] = 0,008 F
t+m
= a
t
+ b
t
.m + 0,5.c
t
.m
2
m = 1 Æ 1 periode ke depan
F
2+1
= F
3
= d
3
’ = a
2
+ b
2
.1 + 0,5.c
2
.1
2
= 14,49 + 0,0111 + 0,50,0081 = 14,5028 dt – dt’ = d
3
– d
3
’ = 14 – 14,5028 = -0,5028 dt – dt’
2
= d
3
– d
3
’
2
= -0,5028
2
= 0,25280784
319 ,
20 22
0119832 ,
447
1 2
= =
− =
∑
=
n d
d MSE
n t
t t
Lampiran 2
f. Metode Double Moving Average
