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


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


3/21

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(


4/21

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 /(*'


5/21

Biaya Sistem "/ = ; " B

= &p 20(03*

So4t5are

pabila ser6er 2 dengan nilai 7 ),3 dan 8 7'


+

=

∑∑=

−−

=

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


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


7/21

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.


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/


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


= "i s 8 9 9: /

= #p$ ($)00%-&'am 2 8 2%26(0%3(1/

= &p 2(*0'


; = ", < W: <( )∑ // nn P λ

= #p$ 1$)00%-&'am < 0%.22/ < 5%)5

= &p '('*)


" = "a < W 8 W:/ <( )∑ // nn P λ

= #p$ 5$(00%-&'am < 0%6(08 0%.22/ < 5%)5

= &p .0(020


B = " 9 8 9: /

= #p$ 6000%- < 2%)1.-1%(1)/

= &p ('/2


9/21


= &p 2)(+*

pabila ser6er 2 dengan nilai 7 3 dan 8 7'



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


11/21

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


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


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


= "i s 8 9 9: /

= #p$ ($)00%-&'am 2 8 .%)(72%((1/

= &p 2(3)


; = ", < W: <( )∑ // nn P λ

= #p$ 1$)00%-&'am < 0%52./ < 3%15

= &p (+0



13/21

" = "a < W 8 W:/ <( )∑ // nn P λ

= #p$ 5$(00%-&'am < 0%)2.8 0%52./ < 3%15

= &p .)(+3


B = " 9 8 9: /

= #p$ 6000%- < .%)(7-2%((1/

= &p ()''


= &p ')(/3*

pabila ser6er 2 dengan nilai 7 ) dan 8 7',3


14/21


+

=

∑∑=

−−

=

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


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-.



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 ))


17/21

¿ 1,608

4(1−9,468 X 10−3)

= 0%(05)

( )∑ // nn P λ = 5%. diulatkan 5

Total ia!a Sistem


= "i s 8 9 9: /

= #p$ ($)00%-&'am 2 8 1%60)0%51)1/

= &p )('+,"


; = ", < W: <( )∑ // nn P λ

= #p$ 1$)00%-&'am < 0%1.07/ < 5

= &p .(./+,"


" = "a < W 8 W:/ <( )∑ // nn P λ

= #p$ 5$(00%-&'am < 0%(05)8 0%1.07/ < 5

= &p /()2/,"


B = " 9 8 9: /

= #p$ 6000%- < 1%60) 8 0%51)1/

= &p +(3'*,"


= &p .*(3.0


18/21

pabila ser6er 2 dengan nilai 7 ) dan 8 7)


+

=

∑∑=

−−

=

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


19/21

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-.


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 ¿¿


20/21

= 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 )/+,"


; = ", < W: < ( )∑ // nn P

λ


21/21

= #p$ 1$)00%-&'am < 0%0)1/ < (

= &p 3',2


" = "a < W 8 W:/ <( )∑ // nn P λ

= #p$ 5$(00%-&'am < 0%..8 0%0)1/ < (

= &p 3('/,"


B = " 9 8 9: /

= #p$ 6000%- < 1%.. 8 0%.2.5/

= &p +(0'*,"


= &p .+(/+

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