KE TEMPAT TUJUAN SECARA OPTIMAL

Presentasi berjudul: "KE TEMPAT TUJUAN SECARA OPTIMAL"— Transcript presentasi:

1 KE TEMPAT TUJUAN SECARA OPTIMAL
METODE TRANSPORTASI ADALAH SUATU METODE YG DIGUNAKAN UNTUK MENGATUR DISTRIBUSI DR SUMBER YG MENYEDIAKAN PRODUK KE TEMPAT TUJUAN SECARA OPTIMAL ALOKASI PRODUK KE TEMPAT TUJUAN, DG MEMPERTIMBANGKAN BIAYA PENGENGKUTAN YG BERVARIASI KARENA JARAK & KONDISI ANTAR LOKASI YG BERBEDA JADI : METODE TRANSPORTASI BERTUJUAN UNTUK MEMINIMALKAN BIAYA

2 TO FROM A B C KAPASITAS W WA 4 WB 8 WC 8 56 X XA 16 XB 24 XC 16 82 Y
YA YB YC 77 KEBUTUHAN 72 102 41 215

3 _ + + _ 36 41 TO FROM A B C KAPASITAS W WA 4 56 WB 8 WC 8 X XA 16 16
XB 66 XC 82 Y YA YB YC 77 KEBUTUHAN 72 102 41 215 _ + + _ 36 41

4 _ + + _ _ + 36 41 TO FROM A B C KAPASITAS W WA 4 56 WB 8 WC 8 X XA 16
XB 66 XC 82 Y YA YB YC 77 KEBUTUHAN 72 102 41 215 _ + + _ _ + 36 41 Q X COST / UNIT = TC ($) WA 56 4 224 XA 16 256 XB 66 24 1 584 YB 36 576 YC 41 984 3 624

5 _ + _ + _ + 36 41 A B C KAPASITAS W WA 4 56 WB 8 WC 8 X XA 16 16 XB 24
66 XC 82 Y YA YB YC 77 KEBUTUHAN 72 102 41 215 _ + _ + _ + 36 41 UN USED CLOSED PATH WB WB - WA + XA – XB = 8 – – 24 = - $ 4 WC WC - WA + XA – XB + YB – YC = 8 – – 24 = - $ 12 XC XC – XB + YB – YC = 16 – – 24 = - $ 16 YA - YA + YB – XB + XA = – = 0

6 XB 66 XC YB 36 YC 41 102 XB 66-41= XC 0+41 = YB 36+41 = YC 41-41 = 102 41 _ + 25 41 _ + 77 TO FROM A B C KAPASITAS W WA 56 WB WC X XA 16 XB 25 XC 41 82 Y YA YB 77 YC KEBUTUHAN 72 102 215

7 TO FROM A B C KAPASITAS W WA 56 WB WC X XA 16 XB 25 XC 41 82 Y YA YB 77 YC KEBUTUHAN 72 102 215 Q X COST / UNIT = TC ($) WA 56 4 224 XA 16 256 XB 25 24 600 XC 41 656 YB 77 1 232 2 968

8 _ + _ + _ + A B C KAPASITAS W WA 4 56 WB 8 WC 8 X XA 16 16 XB 24 25
XC 41 82 Y YA YB 77 YC KEBUTUHAN 72 102 215 _ + _ + _ + UN USED CLOSED PATH WB + WB – WA XA – XB = + 8 – – 24 = - 4 WC + WC – WA + XA- XC = + 8 – – 16 = + 4 YA + YA – XA + XB – YB = – 16 = 0 YC + YC – XC + XB – YB = +24 – = + 16

9 WA 56 WB XA 16 XB 25 72 102 WA 56-25 = WB 0+25 = XA 16+25 = XB 25-25 = 72 102 31 25 41 TO FROM A B C KAPASITAS W WA 31 WB WC 56 X XA 41 XB XC 82 Y YA YB 77 YC KEBUTUHAN 72 102 215 25

10 25 TO FROM A B C KAPASITAS W WA 4 31 WB 8 WC 8 56 X XA 16 41 XB 24
XC 82 Y YA YB 77 YC KEBUTUHAN 72 102 215 25 Q X COST / UNIT = TC ($) WA 31 4 124 WB 25 8 200 XA 41 16 656 XC YB 77 1 232 2 868

11 25 TO FROM A B C KAPASITAS W WA 4 31 WB 8 WC 8 56 X XA 16 41 XB 24
XC 82 Y YA YB 77 YC KEBUTUHAN 72 102 215 25 UN USED CLOSED PATH WC + WC – WA + XA – XC = + 8 – – 16 = + 4 XB + XB – WB + WA – XA = + 24 – – 16 = + 4 YA + YA – WA + WB – YB = + 8 – – 16 = - 4 YC + YC – YB + WB – WA + XA – XC = + 24 – – – 16 = + 12

12 31-31= - 56 31 46 WA 4 WB 8 + 25+31 = YA 16 0+31 = + YB 24 - 71-31= 72
31-31= WB = YA 0+31 = YB = 72 102 56 31 46 TO FROM A B C KAPASITAS W WA WB 56 WC X XA 41 XB XC 82 Y YA 31 YB 46 YC 77 KEBUTUHAN 72 102 215

13 UN USED NYA SUDAH BERNILAI POSITIVE SELURUHNYA ( ≥ 0 )
CLOSED PATH WA WA – XA + XB – WB = + 4 – – 8 = + 4 WC WC - XC + XA + YB – WB = – – 8 = +8 XB XB - YB + YA – XA = + 24 – – 16 = 0 YC YC – XC + XA – YA = + 24 – – 8 = + 16 UN USED NYA SUDAH BERNILAI POSITIVE SELURUHNYA ( ≥ 0 ) Q X COST / UNIT = TC ($) WA 56 8 448 WC 41 16 656 XB YC 31 248 46 736 2 744 MINIMUM COST

14 LATIHAN: 1) TO FROM A B C KAPASITAS W WA WB WC 10 X XA XB XC 20 Y YA YB YC 30 KEBUTUHAN 28 22 60 2) Kerjakan soal dalam Tjutju Tarliah no. 2 dan 5 pada halaman 155 dan 157

15 BEBERAPA METODA LAIN DALAM TRANSPORTASI

16 TO FROM A B C KAPASITAS W WA 9 WB 8 WC 5 25 X XA 6 XB 8 XC 4 35 Y YA 7
YB YC 40 KEBUTUHAN 30 45 100

17 OPTIMAL SOLUTION FOR AJAX SHIPPING TO FROM
B C KAPASITAS W WA WB WC 25 X XA XB XC 35 Y YA YB YC 40 KEBUTUHAN 30 45 100 25 15 20 15 25

18 TOTAL TRANSPORTATION COST 550
TOTAL COST OF OPTIMAL FOR AJAX SHIPPING Q X COST / UNIT = TC ($) WC 25 5 125 XA 15 6 90 XC 20 4 80 YA 7 105 YB 150 TOTAL TRANSPORTATION COST

19 NORTHWEST CORNER SOLUTION TO FROM
A B C KAPASITAS W WA WB WC 25 X XA XB XC 35 Y YA YB YC 40 KEBUTUHAN 30 45 100 25 5 25 5 40

20 TOTAL TRANSPORTATION COST 835
TOTAL COST NORTHWEST CORNER SOLUTION Q X COST / UNIT = TC ($) WA 25 9 225 XA 5 6 30 XB 8 200 XC 4 29 YC 40 360 TOTAL TRANSPORTATION COST

21 GREEDY METHOD: langkah 1) mengalokasikan barang pd cost terendah
TO FROM A B C KAPASITAS W WA WB WC 25 X XA XB XC 35 Y YA YB YC 40 KEBUTUHAN 30 45 100 35

22 Langkah 2) mengalokasikan barang pd cost terendah berikutnya, WC ($5)
TO FROM A B C KAPASITAS W WA WB WC 25 X XA XB XC 35 Y YA YB YC 40 KEBUTUHAN 30 45 100 10 35

23 Langkah 3) mengalokasikan barang pd cost terendah berikutnya, YB ($6), XA ($6) tidak dipilih karena barisnya sudah terisi penuh di XC TO FROM A B C KAPASITAS W WA WB WC 25 X XA XB XC 35 Y YA YB YC 40 KEBUTUHAN 30 45 100 15 10 35 15 25

24 TOTAL TRANSPORTATION COST 580
TOTAL COST OF THE GREEDY SOLUTION Q X COST / UNIT = TC ($) WA 15 9 135 WC 10 5 50 XC 35 4 140 YA 7 105 YB 25 6 150 TOTAL TRANSPORTATION COST

25 VAM: VOGEL APPROXIMATION METHOD
ROW/COLUMN SEC-LOWEST COST ━ LOWEST COST = OPPORT-CST ROW W 8 5 3 LARGEST ROW X 6 4 2 ROW Y 7 1 COLUMN A COLUMN B COLUMN C TO FROM A B C KAPASITAS W WA WB WC 25 X XA XB XC 35 Y YA YB YC 40 KEBUTUHAN 30 45 100 25

26 VAM: VOGEL APPROXIMATION METHOD
ROW/COLUMN SEC-LOWEST COST ━ LOWEST COST = OPPORT-CST ROW X 6 4 2 ROW Y 7 1 COLUMN A COLUMN B 8 COLUMN C 9 5 LARGEST TO FROM A B C KAPASITAS W WA WB WC 25 X XA XB XC 35 Y YA YB YC 40 KEBUTUHAN 30 45 100 25 20

27 VAM: VOGEL APPROXIMATION METHOD
ROW/COLUMN SEC-LOWEST COST ━ LOWEST COST = OPPORT-CST ROW X 8 6 2 LARGEST ROW Y 7 1 COLUMN A COLUMN B 2 LARGEST TO FROM A B C KAPASITAS W WA WB WC 25 X XA XB XC 35 Y YA YB YC 40 KEBUTUHAN 30 45 100 25 20 25

28 TOTAL TRANSPORTATION COST 540
FROM A B C KAPASITAS W WA WB WC 25 X XA XB XC 35 Y YA YB YC 40 KEBUTUHAN 30 45 100 25 15 20 15 25 Q X COST / UNIT = TC ($) WC 25 5 125 XA 15 6 90 XC 20 4 80 YA 7 105 YB 150 TOTAL TRANSPORTATION COST