Kasus Knapsack
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Transcript of Kasus Knapsack
-
8/17/2019 Kasus Knapsack
1/21
Kasus 2
Jumlah lini pada bagian pengecekan kualitas :
- Jumlah maksimum antrian M = 10
- Jumlah server S = 2
- Waktu antar kedatangan 15 menit
- Waktu pelayanan stasiun 1 adalah 17 menit
- Waktu pelayanan stasiun 2 adalah 26 menit
- Biaya-iaya !
" = iaya server dalam keadaan siuk
= #p 6$000%-&'am
"i = iaya server dalam keadaan menganggur
= #p ($)00%-&'am
"a = Biaya penge*ekan pr+duk
= #p 5$(00%-&'am
", = Biaya antrian penge*ekan pr+duk
= #p$ 1$)00%-&'am
λ=60
15=4 pr+duk&'am
μ1=6017
=3,53 pr+duk&'am
μ2=
60
26=2,31 pr+duk&'am
μ= 60
(17+26)/2=2,791≈3 pr+duk&'am
Mencari Probabilitas Sistem
+
=
∑∑=
−−
=
M
sn
sn s s
n
n
s sn
P
µ
λ
µ
λ
µ
λ
-
1
-
1
1
1
0
0
+
=
∑∑=
−
=
10
2
221
0 2.
(
.
(
-2
1
.
(
-
1
1
n
n
n
n
xn
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8/17/2019 Kasus Knapsack
2/21
++ + + + +
=)210210
2.
($$$
2.
(
2.
(
2.
(
.
(
-2
1
-1
/.&(
-0
/.&(
1
x x x x
= 0%20.
n =( λ
µ)n
n! Po
, jika n ≤ S
n =( λ
µ)n
S ! S
n− s Po% 'ika S n M
1 =(4
3)1
1! 0,203
= 0%271
2 =(4
3)2
2 ! 0,203
= 0%1)0
. =(4
3)3
2 !23−20,203
= 0%113
( =(4
3)4
2 !24−2
0,203= 0%073
5 =(4
3)5
2 !25−2 0,203
= 0%05.
6 =(4
3
)6
2 !26−2 0,203
= 0%0.5
7 =(4
3)7
2 !27−2 0,203
= 0%02.
) =(4
3)8
2 !28−2
0,203= 0%016
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3 =(4
3)9
2 !29−2 0,203
= 0%010
10 =(4
3)10
2 !210−2 0,203
= 0%007
• Mencari Lq, L, Wq, dan W
Lq =
λ/ μ¿
¿sμ λ /¿¿
( λ /s μ)k −s(1+(k −s)(1− λ/ s μ))1−¿ P
0¿
Lq=¿
λ / μ¿¿sμ λ /¿¿
P0¿
¿
=
λ/ μ¿¿
sμ λ /¿¿
( λ /s μ)k −s(1+(k −s)(1− λ/ s μ))1−¿ P
0¿ Lq=¿
4 /3
¿¿2 x34 /¿¿
0,203 ¿¿
= 1%105 40%)52
= 0%3(1
L =
nPn
∑n=0
s−1
¿
¿ Pn
1−∑n=0
s−1
¿
¿
= 0%271 0%3(1 2 1-0%(7(/
= 2%26(
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8/17/2019 Kasus Knapsack
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W q= Lq
λ (1− P( M ))
¿ 0,941
4(1−0,007)
= 0%2.7
W = L
λ (1− P( M ))
¿ 2,264
4(1−0,007)
= 0%57
Total ia!a Sistem
" ia!a Sistem #dle $#%
= "i s 8 9 9: /
= #p$ ($)00%-&'am 2 8 2%26(0%3(1/
= &p '(2)*,+
" ia!a ntrian Produk $-%
; = ", < W: <( )∑ // nn P λ
= #p$ 1$)00%-&'am < 0%2.7/ < 5
= &p .(/00,.
" ia!a Pengecekan Produk $1%
" = "a < W 8 W:/ < ( )∑ // nn P λ
= #p$ 5$(00%-&'am < 0%578 0%2.7/ < 5
= &p /(./.
- ia!a Sistem Sibuk $%
B = " 9 8 9: /
= #p$ 6000%- < 2%26( 8 0%3(1/
= &p /(*'
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Biaya Sistem "/ = ; " B
= &p 20(03*
So4t5are
pabila ser6er 2 dengan nilai 7 ),3 dan 8 7'
Mencari Probabilitas Sistem
+
=
∑∑=
−−
=
M
sn
sn s s
n
n
s sn
P
µ
λ
µ
λ
µ
λ
-
1
-
1
1
1
0
0
+
=
∑∑=
−
=
10
2
221
0 2.
5%(
.
5%(
-2
1
.
5%(
-
1
1
n
n
n
n
xn
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8/17/2019 Kasus Knapsack
6/21
++ + + + +
=)210210
2.5%($$$
2.5%(
2.5%(
2.5%(
.5%(
-21
-1/.&5%(
-0/.&5%(
1
x x x x
= 0%150
n =( λ
µ)n
n! Po
, jika n ≤ S
n =( λ
µ)n
S! S
n− s Po% 'ika S n M
1 =(4,5
3)1
1! 0,150
= 0%.05
2 =(4,5
3)2
2! 0,150
= 0%22)
. =(4,5
3)3
2 !23−2 0,150
= 0%171
( =(4,5
3)4
2 !24−2 0,150
= 0%12)
5 =(4,5
3)5
2 !25−2 0,150
= 0%036
6 =(4,5
3)6
2 !26−2 0,150
= 0%072
7 =(4,5
3)7
2 !27−2
0,150= 0%05(
) =(4,5
3)8
2 !28−2 0,150
= 0%0(1
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3 =(4,5
3)9
2 !29−2 0,150
= 0%0.0
10 =(4,5
3)10
2 !210−2 0,150
= 0%02.
• Mencari Lq, L, Wq, dan W
Lq =
λ/ μ¿
¿sμ λ /¿¿
( λ /s μ)k −s(1+(k −s)(1− λ/ s μ))1−¿ P
0¿
Lq=¿
λ / μ¿¿sμ λ /¿¿
P0¿
¿
=
λ/ μ¿¿
sμ λ /¿¿
( λ /s μ)k −s(1+(k −s)(1− λ/ s μ))1−¿ P
0¿ Lq=¿
4,5/3
¿¿2 x34,5 /¿¿
0,15 ¿¿
= 2%026 40%7
= 1%(2
L =
nPn
∑n=0
s−1
¿
¿ Pn
1−∑n=0
s−1
¿
¿
= 0%.05 1%(2 2 1-0%(55/
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8/17/2019 Kasus Knapsack
8/21
= 2%)1.
W q= Lq
λ (1− P( M ))
¿ 1,42
4,5(1−0,023)
= 0%.22
W = L
λ (1− P( M ))
¿ 2,813
4,5(1−0,023)
= 0%6(0
Total ia!a Sistem
" ia!a Sistem #dle $#%
= "i s 8 9 9: /
= #p$ ($)00%-&'am 2 8 2%26(0%3(1/
= &p 2(*0'
" ia!a ntrian Produk $-%
; = ", < W: <( )∑ // nn P λ
= #p$ 1$)00%-&'am < 0%.22/ < 5%)5
= &p '('*)
" ia!a Pengecekan Produk $1%
" = "a < W 8 W:/ <( )∑ // nn P λ
= #p$ 5$(00%-&'am < 0%6(08 0%.22/ < 5%)5
= &p .0(020
- ia!a Sistem Sibuk $%
B = " 9 8 9: /
= #p$ 6000%- < 2%)1.-1%(1)/
= &p ('/2
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8/17/2019 Kasus Knapsack
9/21
Biaya Sistem "/ = ; " B
= &p 2)(+*
pabila ser6er 2 dengan nilai 7 3 dan 8 7'
Mencari Probabilitas Sistem
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8/17/2019 Kasus Knapsack
10/21
+
=
∑∑ =−−
=
M
sn
sn s s
n
n
s sn
P
µ λ
µ λ
µ λ
-1
-1
1
1
0
0
+
=
∑∑=
−
=
10
2
221
0 2.
5
.
5
-2
1
.
5
-
1
1
n
n
n
n
xn
++
+
+
+
+
=)210210
2.
5$$$
2.
5
2.
5
2.
5
.
5
-2
1
-1
/.&5
-0
/.&5
1
x x x x
= 0%12)
n =( λ
µ)n
n! Po
, jika n ≤ S
n =( λ
µ)n
S! Sn− s Po
% 'ika S n M
1 =(53)1
1 ! 0,128
= 0%..)
2 =(5
3)2
2! 0,128
= 0%2)2
. =(5
3)3
2 !23−2 0,128
= 0%2.5
( =(5
3)4
2 !24−2 0,128
= 0%136
5 =(,5
3)5
2 !25−2
0,128= 0%16.
6 =(5
3)6
2 !26−2 0,128
= 0%1.6
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7 =(5
3)7
2 !27−2 0,128
= 0%11.
) =(5
3)8
2 !28−2 0,128
= 0%03(
3 =(,5
3)9
2 !29−2 0,128
= 0%073
10 =(5
3)10
2 !210−2
0,128 = 0%066
• Mencari Lq, L, Wq, dan W
Lq
=
λ/ μ¿¿sμ λ /¿¿
( λ /s μ)k −s(1+(k −s)(1− λ/ s μ))1−¿ P
0¿
Lq=¿
λ / μ¿¿sμ
λ /¿¿ P
0¿
¿
=
λ/ μ¿¿sμ λ /¿¿
( λ /s μ)k −s
(1+(k −s)(1− λ/ s μ))1−¿ P
0¿
Lq=¿
5/3¿¿
2 x35 /¿¿
0,128 ¿¿
= 5%..) 40%(57
= 2%((1
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8/17/2019 Kasus Knapsack
12/21
L =
nPn
∑n=0
s−1
¿¿ Pn
1−∑n=0
s−1
¿
¿
= 0%..) 2%((1 2 1-0%(66/
= .%)(7
W q= Lq
λ (1− P( M ))
¿ 2,441
5 (1−0,066)
= 0%52.
W = L
λ (1− P( M ))
¿ 3,847
5 (1−0,066)
= 0%)2.
Total ia!a Sistem
" ia!a Sistem #dle $#%
= "i s 8 9 9: /
= #p$ ($)00%-&'am 2 8 .%)(72%((1/
= &p 2(3)
" ia!a ntrian Produk $-%
; = ", < W: <( )∑ // nn P λ
= #p$ 1$)00%-&'am < 0%52./ < 3%15
= &p (+0
" ia!a Pengecekan Produk $1%
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8/17/2019 Kasus Knapsack
13/21
" = "a < W 8 W:/ <( )∑ // nn P λ
= #p$ 5$(00%-&'am < 0%)2.8 0%52./ < 3%15
= &p .)(+3
- ia!a Sistem Sibuk $%
B = " 9 8 9: /
= #p$ 6000%- < .%)(7-2%((1/
= &p ()''
Biaya Sistem "/ = ; " B
= &p ')(/3*
pabila ser6er 2 dengan nilai 7 ) dan 8 7',3
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8/17/2019 Kasus Knapsack
14/21
Mencari Probabilitas Sistem
+
=
∑∑=
−−
=
M
sn
sn s s
n
n
s sn
P
µ
λ
µ
λ
µ
λ
-
1
-
11
1
0
0
+
=
∑∑=
−
=
10
2
221
0 25%.
(
5%.
(
-2
1
5%.
(
-
1
1
n
n
n
n
xn
++
+
+
+
+
=)2102
10
25%.($$$
25%.(
25%.(
25%.(
5%.(
-21
-1/5%.&(
-0/5%.&(
1
x x x x
=1
2,14+(0,65 X 2,318)= 0,2/'
n =( λ
µ)n
n! Po
, jika n ≤ S
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15/21
n =( λ
µ)n
S! Sn− s Po
% 'ika S n M
1 =(4
3)1
1! 0,273
= 0%.6(
2 =(4
3)2
2 ! 0,273
= 0%2(2
. =
(4
3
)3
2 !23−2
0,273 = 0%161
( =(4
3)4
2 !24−2 0,273
= 0%107
5 =(4
3)5
2 !25−2 0,273
= 0%071
6 = (43 )
6
2 !26−2 0,273
= 0%0(73
7 =(4
3)7
2 !27−2
0,273= 0%0.1
) =(4
3)8
2 !28−2 0,273
= 0%021
3 =(4
3)9
2 !29−2
0,273= 0%01(
10 =(4
3)10
2 !210−2 0,273
= 3%(6) >10-.
• Mencari Lq, L, Wq, dan W
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16/21
Lq =
λ/ μ¿
¿sμ λ /¿¿
( λ /s μ)k −s(1+(k −s)(1− λ/ s μ))1−¿ P
0¿
Lq=¿
λ / μ
¿¿sμ λ /¿¿
P0¿
¿
=
λ/ μ¿¿
sμ λ /¿¿
( λ /s μ)k −s(1+(k −s)(1− λ/ s μ))1−¿ P
0¿
Lq=¿
4 /3 ,5
¿¿2 x3 ,54 /¿¿
0,273 ¿¿
= 0%55 0%3) 8 0%0.)/
= 0%51)1
L =
nPn
∑n=0
s−1
¿
¿ Pn
1−∑n=0
s−1
¿
¿
= 0%.6( 0%51)1 2 1-0%6.7/
=1%60)
W q= Lq
λ (1− P( M ))
4 (1−9,468 X 10−3)
¿ 0,5181
¿ ¿
= 0%1.07
W = L
λ (1− P( M ))
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8/17/2019 Kasus Knapsack
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¿ 1,608
4(1−9,468 X 10−3)
= 0%(05)
( )∑ // nn P λ = 5%. diulatkan 5
Total ia!a Sistem
" ia!a Sistem #dle $#%
= "i s 8 9 9: /
= #p$ ($)00%-&'am 2 8 1%60)0%51)1/
= &p )('+,"
" ia!a ntrian Produk $-%
; = ", < W: <( )∑ // nn P λ
= #p$ 1$)00%-&'am < 0%1.07/ < 5
= &p .(./+,"
" ia!a Pengecekan Produk $1%
" = "a < W 8 W:/ <( )∑ // nn P λ
= #p$ 5$(00%-&'am < 0%(05)8 0%1.07/ < 5
= &p /()2/,"
- ia!a Sistem Sibuk $%
B = " 9 8 9: /
= #p$ 6000%- < 1%60) 8 0%51)1/
= &p +(3'*,"
Biaya Sistem "/ = ; " B
= &p .*(3.0
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18/21
pabila ser6er 2 dengan nilai 7 ) dan 8 7)
Mencari Probabilitas Sistem
+
=
∑∑=
−−
=
M
sn
sn s s
n
n
s sn
P
µ
λ
µ
λ
µ
λ
-
1
-
1
1
1
0
0
+
=
∑∑=
−
=
10
2
221
0 2(
(
(
(
-2
1
(
(
-
1
1
n
n
n
n
xn
++
+
+
+
+
=)210210
2(
($$$
2(
(
2(
(
2(
(
(
(
-2
1
-1
/(&(
-0
/(&(
1
x x x x
=1
2+(0,5 x2)= 0,''
n =( λ
µ)n
n! Po
, jika n ≤ S
n =( λ
µ)n
S ! Sn− s Po
% 'ika S n M
1 =(4
4)1
1! 0,33
= 0%..
2 =(4
4)2
2 ! 0,33
= 0%165
. =(4
4)3
2 !23−2 0,33
= 0%0)25
( =(4
4)4
2 !24−2
0,33= 0%0(125
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5 =(4
4)5
2 !25−2 0,33
= 0%020
6 =(4
4)6
2 !26−2 0,33
= 0%010
7 =(4
4)7
2 !27−2 0,33
= 5%156 < 10-.
) =(4
4)8
2 !28−2
0,33 = 2%57 < 10-.
3 =(4
4)9
2 !29−2 0,33
= 1%2) < 10-.
10 =(4
4)10
2 !210−2 0,33
= 6%(( >10-.
• Mencari Lq, L, Wq, dan W
Lq =
λ/ μ¿¿sμ λ /¿¿
( λ /s μ)k −s(1+(k −s)(1− λ/ s μ))1−¿
P0¿ Lq=¿
λ / μ¿¿sμ λ /¿¿
P0¿
¿
=
λ/ μ¿¿sμ λ /¿¿
( λ /s μ)k −s(1+(k −s)(1− λ/ s μ))1−¿ P
0¿
Lq=¿
4 /4¿¿
2 x 44 /¿¿
0,33 ¿¿
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8/17/2019 Kasus Knapsack
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= 0%..0%336 8 0%0156/
= 0%.2.5
L =
nPn
∑n=0
s−1
¿
¿ Pn
1−∑n=0
s−1
¿
¿
= 0%.. 0$.2.5 21-0%66/
=1%...
W q= Lq
λ (1− P( M ))
4 (1−6,44 x 10−3)
¿ 0,3235
¿ ¿
= 0%0)1
W = L
λ (1− P( M ))
¿ 1 ,33
4(1−6,44 x10−3)
= 0%..
( )∑ // nn P λ = .%37 diulatkan (
Total ia!a Sistem" ia!a Sistem #dle $#%
= "i s 8 9 9: /
= #p$ ($)00%-&'am 2 8 1%..0%.2.5/
= &p )/+,"
" ia!a ntrian Produk $-%
; = ", < W: < ( )∑ // nn P
λ
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= #p$ 1$)00%-&'am < 0%0)1/ < (
= &p 3',2
" ia!a Pengecekan Produk $1%
" = "a < W 8 W:/ <( )∑ // nn P λ
= #p$ 5$(00%-&'am < 0%..8 0%0)1/ < (
= &p 3('/,"
- ia!a Sistem Sibuk $%
B = " 9 8 9: /
= #p$ 6000%- < 1%.. 8 0%.2.5/
= &p +(0'*,"
Biaya Sistem "/ = ; " B
= &p .+(/+