How to make Maven copy the JAR that the project depends on to the web-inF/LIB directory and solve it (GOALS “Copy-DEPENDENCIES”, “UNPACK”))

2022-12-28   ES  

Question:

A certain airline operation
A

B

C
Three cities, these routes take off and reach the following table every day.
Set aircraft to stay at the airport
loss costs are roughly and stay
The square of the time is proportional.
Each aircraft landing from landing
At least needed to take off from get off work
2
hour preparation time,
Try a decision to stop
cost loss is the smallest
allocate flight scheme.
Flight

15th

Take off
City
Take off
time
Arrival
City

Arrival

time

101
A
9:00
B
12:00
102
A
10:00
B
13:00
103
A
15:00
B
18:00
104
A
20:00
C
24:00
105
A
22:00
C
2:00
(the next day)
106
B
4:00
A
7:00
107
B
11:00
A
14:00
108
B
15:00
A
18:00
109
C
7:00
A
11:00

110

C
15:00
A
19:00
111
B
13:00
C
18:00
112
B
18:00
C
23:00
113
C
15:00
B
20:00
114 C 7:00 B 12:00
A city loss fee:

city lling costs:

C city loss fee:

problem solving ideas:

A

B

C
City takes off as the “task” to be completed,
The aircraft that arrives can be regarded as a pending distribution
“People” to complete the task.
As long as the plane reaches two hours, it can be distributed to complete the take -off task.
can be different in the city
A

B

C
List the efficiency matrix of the distribution problem,
Among them, the numbers are stopped by the aircraft
Losses.
MATLAB code is as follows:
%applicable to any N -order coefficient matrix 
 Clear all; 
 C = [4 9 64 169 225 
 361 400 625 36 64 
 225 256 441 4 16 
 484 529 16 81 121 
 196 225 400 625 9];%efficiency matrix C 
 n = SIZE (C, 1);%Calculate the number of ran c n 
 C = c (:);%Calculate the target function coefficient, and arrange the matrix C according to the column into a column vector. 
 A = []; b = [];%without unlimited constraints 
 Ae = zeros (2*n, n^2);%Calculating the restricted coefficient matrix A 
 for i = 1: n 
 for j = (i-1)*n+1: n*i 
 Ae (i, j) = 1; 
 end 
 for k = i: n: n^2 
 Ae (n+i, k) = 1; 
 end 
 end 
 Be = Ones (2*n, 1);%equivalent to restrain the right end b 
 Xm = zeros (n^2,1);%decision variable lower boundary XM 
 XM = Ones (n^2,1);%decision variable upper boundary XM 
 [x, z1] = linprog (c, a, b, ae, be, xm, xm);%use linprog to solve 
 x = reshape (x, n, n);%row the column vector x into a n -order square array by column 
 DISP ('The optimal solution matrix is:');%output assignment scheme and the best value 
 Assignment = Round (X)%uses Round for four houses and five 
 Disp ('The optimal interpretation as:'); 
 Z1 
 %% 
 %Suitable for any N -order coefficient matrix 
 Clear all; 
 C = [256 529 9 625 36 
 225 484 4 576 25 
 100 289 441 361 576 
 64 225 361 289 484 
 256 529 9 625 36];%efficiency matrix C 
 n = SIZE (C, 1);%Calculate the number of ran c n 
 C = c (:);%Calculate the target function coefficient, and arrange the matrix C according to the column into a column vector. 
 A = []; b = [];%without unlimited constraints 
 Ae = zeros (2*n, n^2);%Calculating the constraint coefficient matrix A 
 for i = 1: n 
 for j = (i-1)*n+1: n*i 
 Ae (i, j) = 1; 
 end 
 for k = i: n: n^2 
 Ae (n+i, k) = 1; 
 end 
 end 
 Be = Ones (2*n, 1);%equivalent to restrain the right end b 
 Xm = zeros (n^2,1);%decision variable lower boundary XM 
 XM = Ones (n^2,1);%decision variable upper boundary XM 
 [x, z2] = linprog (c, a, b, ae, be, xm, xm);%use linprog to solve 
 x = reshape (x, n, n);%row the column vector x into a n -order square array by column 
 DISP ('The optimal solution matrix is:');%output assignment scheme and the best value 
 Assignment = Round (X)%uses Round for four houses and five 
 Disp ('The optimal interpretation as:'); 
 Z2 
 %% 
 %Suitable for any N -order coefficient matrix 
 Clear all; 
 C = [49 225 225 49 
 25 169 169 25 
 169 441 441 169 
 64 256 256 64];%efficiency matrix C 
 n = SIZE (C, 1);%Calculate the number of ran c n 
 C = c (:);%Calculate the target function coefficient, and arrange the matrix C according to the column into a column vector. 
 A = []; b = [];%without unlimited constraints 
 Ae = zeros (2*n, n^2);%Calculating the restricted coefficient matrix A 
 for i = 1: n 
 for j = (i-1)*n+1: n*i 
 Ae (i, j) = 1; 
 end 
 for k = i: n: n^2 
 Ae (n+i, k) = 1; 
 end 
 end 
 Be = Ones (2*n, 1);%equivalent to restrain the right end b 
 Xm = zeros (n^2,1);%decision variable lower boundary XM 
 XM = Ones (n^2,1);%decision variable upper boundary XM 
 [x, z3] = linprog (c, a, b, ae, be, xm, xm);%use linprog to solve 
 x = reshape (x, n, n);%row the column vector x into a n -order square array by column 
 DISP ('The optimal solution matrix is:');%output assignment scheme and the best value 
 Assignment = Round (X)%uses Round for four houses and five 
 Disp ('The optimal interpretation as:'); 
 Z3 
 %% 
 Z = Z1+Z2+Z3

result is as follows:

OPTImal Solution Found. 

 The optimal solution matrix is: 

 Assignment = 

      0 1 0 0 0 
      0 0 0 1 0 
      0 0 0 0 1 
      0 0 1 0 0 
      1 0 0 0 0 

 The optimal solution is: 

 Z1 = 

    273 


 Optimal solution found. 

 The optimal solution matrix is: 

 Assignment = 

      0 0 0 0 1 
      1 0 0 0 0 
      0 1 0 0 0 
      0 0 0 1 0 
      0 0 1 0 0 

 The optimal solution is: 

 Z2 = 

    848 


 Optimal solution found. 

 The optimal solution matrix is: 

 Assignment = 

      0 1 0 0 
      0 0 1 0 
      0 0 0 1 
      1 0 0 0 

 The optimal solution is: 

 Z3 = 

    627 


 z = 

         1748

z1 results:

z2 result:

 

z3 Result:

So the solution that minimizes the cost of staying is:

101(A)→110(C) →104(A) →107(B) →103(A) →109(C) →112(B) →101(A)

102(A) →106(B) →102(A)

105(A) →108(B) →114(C) →111(B) →113(C) →105(A)

Losses: 273K+848K+627K = 1748K

source

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