Mathematical modeling of transportation problem

Guo Dunqing replied:

It is getting late. Give a formal answer tomorrow.

20 14-8-27 answer:

Principles of traffic planning (F)

(1) Total oil transportation of two vehicles U= total oil supply ∑ U =14000+3000+6000+16000+15000+59000,

=u 1+u2, and u 1, u2≤39000, u 1, u2 are the oil throughput of two vehicles respectively;

(2) The sum of the two-vehicle transportation distances and the shortest minL=l 1+l2, and l 1+l2 are the two-vehicle transportation distances respectively.

(3) The principle of arranging transportation routes by rows

(1) The priority is that the initial replenishment point and the last replenishment point are close, namely S2=55, S3=32, S4=70 and S6 = 73.

(2) Give priority to the two replenishment points with short distances, namely: S(A 1, A5)= 12, S(A2, A4)=20, S(A6, A2)=28, S(A6, A4)=49, …

The distance between the two feeding points is equivalent, such as S(A5, A 1)=S(A 1, A5)= 12.

(3) The comprehensive distance is the shortest, and the distance between the initial supply point and the final supply point is the shortest. If the distance between two supply points is short, the supply point with shorter distance between the initial supply point and the final supply point can be adopted.

(4) Type of transportation plan (F)

(1) three+three models, that is, each vehicle has three supply points.

(2) Type II+IV, that is, each vehicle is supplied with two supply points and each vehicle is supplied with four supply points.

Transportation Scheme (F)-the combination of transportation schemes (F), the fuel quantity and transportation distance of each vehicle.

F 1= f 1+ f2，f 1=(A2，A 1，A3)，f2=(A4，A5，A6)

f 1u 1 = 3000+6000+ 14000 = 23000 < 39000，

f 1 U2 = 16000+ 15000+5000 = 36000 < 39000，23000+36000 = 59000；

f 1 l 1 = 55+93+ 180+32 = 360，f 1 L2 = 70+85+73+73 = 30 1

f 1L = f 1 l 1+f 1L 2 = 360+30 1 = 66 1 .

F2= F2f 1+ F2f2，F2f 1=(A2，A5，A3)，F2f2=(A4，A 1，A6)

f2u 1 = 3000+ 15000+ 14000 = 32000 < 39000，

F2 U2 = 16000+6000+5000 = 27000 < 39000，32000+27000 = 59000；

F2 l 1 = 55+83+ 174+32 = 344，F2l2=70+99+72+73=3 14

F2L = F2 l 1+F2 L2 = 344+3 14 = 658 .

What is given above is the calculation method of the content of the transportation scheme, which does not fully respect the principle of arranging the transportation path, so it is not optimal.

F3= F3f 1+ F3f2，F3f 1=(A5，A 1，A6)，F3f2=(A3，A2，A4)

f3u 1 = 15000+ 14000+5000 = 34000 < 39000，

F3 U2 = 6000+3000+ 16000 = 25000 < 39000，34000+25000 = 59000；

F3 l 1 = 140+ 12+72+73 = 297，F2l2=32+85+20+70=207

F3L = F2 l 1+F2 L2 = 297+207 = 504 .

F4= F4f 1+ F4f2，F4f 1=(A4，A6，)，F4f2=(A2，A5，A 1，A3，)

f4u 1 = 16000+5000 = 2 1000 < 39000，

F4 U2 = 3000+ 15000+ 14000+6000 = 38000 < 39000，2 1000+38000 = 59000；

F4 l 1=70+49+73= 192，F2 L2 = 55+83+ 12+ 180+32 = 362

F4L = F2 l 1+F2 L2 = 192+362 = 554 .

F5= F5f 1+ F5f2，F5f 1=(A3，A4，)F5f2=(A2，A 1，A5，A6)

f5u 1 = 6000+ 16000 = 22000 < 39000，