nurulramah

fN(i) = min {Cij + f(N+1)j}

Tahap =N = 4

i f4(i) = min{Cij +f(4+1)j} f(4)i*

j = 10
8 C8,10 +f(5)10
1090 + 0 = 1090 1090
9 C9,10 +f(5)10
1390 + 0 = 1390 1390

Tahap =N = 3

i f3(i) = min{Cij +f(4)j} f(3)i j*
J = 8 J= 9
5 C5,8+f(4)8
610+1090=1700 C5,9+f(4)9
790+1390=2180 1700 8
6 C6,8+f(4)8
540 + 1090=1630 C6,9+f(4)9
540+1390=1930 1630 8
7 C7,8+f(4)8
790 +1090=1880 C7,9+f(4)9
270+1390=1660 1660 9









Tahap =N = 2

i f2(i) = min{Cij +f(3)j} f2(i) j*
J=5 J=6 j=7
2 C2,5+f(3)5
680+1700=2380 C2,6+f(3)6
740+1630=2370 C2,7+f(3)7
1050+1660=2710
2370 6
3 C3,5+f(3)5
580+1700=2280 C3,6+f(3)6
760+1630=2390 C3,7+f(3)7
660+1660=2320 2280 5
4 C4,5+f(3)5
510+1700=2210 C4,6+f(3)6
400+1630=2030 C4,7+f(3)7
830+1660=2490 2030 6


Tahap =N = 1
i f1(i) = min{Cij +f(2)j} f1(i) j*
J=2 J=3 j=4
1 C1,2+f(2)2
350+2370=2720
C1,3+f(2)3
900+2280=3180
C1,4+f(2)4
770+2030 =2800
2720 2

Route terpendek : 1 – 2 – 6 – 8 – 10
Jarak tempuh : 2720