Kajian Masalah Transshipment Tidak Seimbang Menggunakan Metode Least Cost - Stepping Stone dan Metode Least Cost - MODI

(1)

Lampiran 1. Nilai baris ui dan kolom vj untuk setiap tabel iterasi dari metode

MODI

Nilai Baris ui dan Kolom vj untuk Tabel 4.28

ui vj

u1=c11−v1=5–14=−9 u2 =c21−v1=0–14=−14 u3 =0

u4 =c44−v4=8−M u5 =c54−u5=15−M

v1=c31−u3=14–0=14 v2=c32−u3=0−0= 0

v3=c43−u4=0−(8−M)=M−8 v4=c34−u3=M–0=M

v5=c55−u5=0−(15−M)=M−15

ui vj

u1=c11−v=5–0 =5 u2=c21−v1=0–0 =0 u3=c31−v1=14–0 =14 u4=c44−v4=8– 4=4 u5=c54−v5=15– 4=11

v1=0

v2=c32−u3=0–14=−14 v3=c43−u4=0–4=−4 v4=c24−u2=4–0=4 v5=c55−u5=0–11=−11

Nilai Baris uidan Kolom vj untuk Tabel 4.30

ui vj

u1=c11−v1=5−0=5 u2=c21−v1=0−0=0 u3=c31−v1=14−0=14 u4=c44−v4=8−4=4 u5=c54−v5=15−4=11

v1=0

v2=c32−u3=0−14=−14 v3=c43−u4=0−4=−4 v4=c24−u2=4−0=4 v5=c55−u5=0−14=−14

ui vj

u1 =c11−v1=5−0=5 u2=c21−v1=0−0=0 u3=c31−v1=14−0=14 u4=c41−v1=3−0=3 u5=c53−v3=6−(−3)=9

v1=0

(2)

ui vj

u1 =c11−v1=5−0=5 u2=c21−v1=0−0=0 u3=c31−v1=14−0=14 u4=c41−v1=3−0=3 u5=c53−v3=6−(−3)=9

v1=0

v2=c32−u3=0−14=−14 v3=c43−u4=0−3=−3 v4=c24−u2=4−0=4 v5=c35−u3=0−14=−14

ui vj

u1 =c11−v1=5−0=5 u2=c21−v1=0−0=0 u3=c33−v3=10−(−3)=13 u4=c41−v1=3−0=3 u5=c53−v3=6−(−3)=9

v1=0

(3)

Lampiran 2. Indeks perbaikan untuk setiap tabel iterasi dari metode MODI

Catatan: Indeks Perbaikan yang berwarna abu-abu pada masing-masing tabel

adalah indeks perbaikan terpilih untuk masuk ke iterasi selanjutnya.

Indeks Perbaikan untuk Tabel 4.28

Sel X12 = k12 = c12−u1−v2 = 3+9 = 12

Sel X13 = k13 = c13−u1−v3 = 3+9−M+8 = 20−M Sel X14 = k14 = c14−u1−v4 = M+9−M = 9

Sel X15 = k15 = c15−u1−v5 = 0+9−M+15 =24−M Sel X22 = k22 = c22−u2−v2 = 14+14−0 = 28 Sel X23 = k23 = c23−u2−v3 = 3+14−M+8 = 25−M

Sel X24 = k24 = c24−u2−v4= 4+14−M = 18−M

Sel X25 = k25 = c25−u2−v5 = 0+14−M+15 = 29−M

Sel X33 = k33 = c33−u3−v3 = 10−0−M+8 = 18−M

Sel X35 = k35 = c35−u3−v5 = 0−0−10 = −10 Sel X41 = k41 = c41−u4−v1 = 3−8+M−14 = M−19 Sel X42 = k42 = c42−u4−v2 = 10−8+M−0 = M+2 Sel X45 = k45 = c45−u4−v5 = 0−8+M−M+15 = 7 Sel X51 = k51 = c51−u5−v1 = M−15+M−14 = 2M−29 Sel X52 = k52 = c52−u5−v2 = M−15+M−0 = 2M−15 Sel X53 = k53 = c53−u5−v3 = 6−15+M−M+8 = −1

Sel X12 = k12 = c12−u1−v2 = 3−5+14 = 12 Sel X13 = k13 = c13−u1−v3 = 3−5+4 = 2 Sel X14 = k14 = c14−u1−v4 = M−5−4 = M−9 Sel X15 = k15 = c15−u1−v5 = 0−5+11 = 6 Sel X22 = k22 = c22−u2−v2 = 14−0+14 = 28 Sel X23 = k23 = c23−u2−v3 = 3−0+4 = 7 Sel X25 = k25 = c25−u2−v5 = 0−0+11 = 11 Sel X33 = k33 = c33−u3−v3 = 10−14+4 = 0 Sel X34 = k34 = c34−u3−v4= M−14−4 = M−18

Sel X35 = k35 = c35−u3−v5 = 0−14+11 = −3

(4)

Sel X12 = k12 = c12−u1−v2 = 3−5+14 = 12 Sel X13 = k13 = c13−u1−v3 = 3−5+4 = 2 Sel X14 = k14 = c14−u1−v4 = M−5−4 = M−9 Sel X15 = k15 = c15−u1−v5 = 0−5+14 = 9 Sel X22 = k22 = c22−u2−v2 = 14−0+14 = 28 Sel X23 = k23 = c23−u2−v3 = 3−0+4 = 7 Sel X25 = k25 = c25−u2−v5 = 0−0+14 = 14 Sel X33 = k33 = c33−u3−v3 = 10−14+4 = 0 Sel X34 = k34 = c34−u3−v4= M−14−4 = M−18

Sel X41 = k41 = c41−u4−v1 = 3−4−0 = −1

Sel X42 = k42 = c42−u4−v2 = 10−4+14 = 20 Sel X45 = k45 = c45−u4−v5 = 0−4+14 = 10 Sel X51 = k51 = c51−u5−v1 = M−11−0 = M−11 Sel X52 = k52 = c52−u5−v2 = M−11+14 = M+3

Sel X53 = k53 = c53−u5−v3 = 6−11+4 = −1

Sel X55 = k55 = c55−u5−v5 = 0−11+14 = 3

Sel X12 = k12 = c12−u1−v2 = 3−5+14 = 12 Sel X13 = k13 = c13−u1−v3 = 3−5+3 = 1 Sel X14 = k14 = c14−u1−v4 = M−5−4 =M−9 Sel X15 = k15 = c15−u1−v5 = 0−5+14 = 9 Sel X22 = k22 = c22−u2−v2 = 14−0+14 = 28 Sel X23 = k23 = c23−u2−v3 = 3−0+3 = 6 Sel X25 = k25 = c25−u2−v5 = 0−0+14 = 14 Sel X33 = k33 = c33−u3−v3 = 10−14+3 = −1 Sel X34 = k34 = c34−u3−v4= M−14−4 = M−18 Sel X42 = k42 = c42−u4−v2 = 10−3+14 = 21 Sel X44 = k44 = c44−u4−v4 = 8−3−4 = 1 Sel X45 = k45 = c45−u4−v5 = 0−11+14 = 3 Sel X51 = k51 = c51−u5−v1 = M−11−0 = M−11 Sel X52 = k52 = c52−u5−v2 = M−11+14 = M+3

Sel X53 = k53 = c53−u5−v3 = 6−11+3 = −2

(5)

Sel X12 = k12 = c12−u1−v2 = 3−5+14 = 12 Sel X13 = k13 = c13−u1−v3 = 3−5+3 = 1 Sel X14 = k14 = c14−u1−v4 = M−5−4 = M−9 Sel X15 = k15 = c15−u1−v5 = 0−5+14 = 9 Sel X22 = k22 = c22−u2−v2 = 14−0+14 = 28 Sel X23 = k23 = c23−u2−v3 = 3−0+3 = 6 Sel X24 = k24 = c24−u2−v4 = 0−0+14 = 14

Sel X33 = k33 = c33−u3−v3 = 10−14+3 = −1

Sel X34 = k34 = c34−u3−v4= M−14−4 = M−18 Sel X42 = k42 = c42−u4−v2 = 10−3+14 = 21 Sel X44 = k44 = c44−u4−v4 = 8−3−4 = 1 Sel X45 = k45 = c45−u4−v5 = 0−3+14 = 11 Sel X51 = k51 = c51−u5−v1 = M−9−0 = M−9 Sel X52 = k52 = c52−u5−v2 = M−9+14 = M+5 Sel X54 = k54 = c54−u5−v4 = 15−9−4 = 2 Sel X55 = k55 = c55−u5−v5 = 0−9+14 = 5

(6)

Lampiran 3. Tahapan Penyelesaian Fisibel Awal dengan Metode Least Cost Untuk Posisi 2

Iterasi Ke 1 Tujuan Persediaan

TB TC TD TF TG

(7)

TB TC TD TF TG

(8)

(9)

Lampiran 4. Tahapan Uji Optimalitas dengan Metode Stepping Stone

Untuk Posisi 2

TB TC TD TF TG

Indeks Perbaikan Iterasi Ke 0

(10)

(11)

(12)

(13)

(14)

TB TC TD TF TG

S

u

m

b

er

SB 0

70

14 3 4 0

70

SA

5 3 3 M 0

20 20

SC

14 0 10 M 0

10 90

70 10

SD

3 10 0 8 35

0

70

5 30

SE

M M 6 15 0

30 30

Permintaan 95 70 70 35 10

Sel X12= 14−0+10−0+3−0 = 27 Sel X34= M−10+0−3+0−4 = M−17 Sel X13 = 3−0+3−0 = 6 Sel X42= 10−0+10−0 = 20

(15)

Lampiran 5. Tahapan Penyelesaian Fisibel Awal dengan Metode Least Cost Untuk Posisi 3

TC TB TD TF TG

(16)

TC TB TD TF TG

(17)

(18)

Lampiran 6. Tahapan Uji Optimalitas dengan Metode Stepping Stone

Untuk Posisi 3

TC TB TD TF TG

(19)

(20)

(21)

(22)

TC TB TD TF TG

S

u

m

b

er

SA

3 5 20

3 M 0

20

SB

14 0 3 4 35

0

70

0 35

SC

0 14 10 M 0

10 90

70 10

SD

10 3 0 8 0

0 70

0 ₄₀ ₃₀

SE

M M 6 15 0

0 30

30

Permintaan 70 95 70 35 10