ys 1 = 10-a 1-d 1； ! Funds at the beginning of the year1;

yt 1 = ys 1+ 1.06 * d 1； ! Year-end funds1;

ys2 = yt 1-a2-C2-D2； ! Funds at the beginning of the second year;

yt2 = ys2+ 1. 15 * a 1+ 1.06 * D2； ! Funds at the end of the second year;

ys3 = yt2-a3-B3-C3-D3； ! Funds at the beginning of the third year;

yt3 = ys3+ 1. 15 * a2+ 1.06 * D3； ! Funds at the end of the third year;

ys4 = yt3-a4-B4-C4-D4； ! Funds at the beginning of the fourth year;

yt4 = ys4+ 1. 15 * a3+ 1.06 * D4； ! Funds at the end of the fourth year;

ys5 = yt4-b5-C5-D5； ! Funds at the beginning of the fifth year;

yt5 = ys5+ 1. 15 * a4+ 1.25 *(B3+B4+b5)+ 1.40 *(C2+C3+C4+C5)+ 1.06 * D5； ! Funds at the end of the fifth year;

C2+C3+C4+C5 & lt； 3;

max = yt5

end

Running results:

Finding the global optimal solution in iteration: 10

Objective value: 16.98 125.

Variable value reduces cost

ys 1 0.000000 0.93573 1 1E-0 1

a 1 1.00000 0.000000

d 1 0.00000 0.000000

yt 1 0.00000 0.000000

YS2 0.000000 0. 12205 19

A2 0.000000 0.00000

C2 0.000000.3055 19

D2 0.000000 0.3580 189 e-0 1

YT2 1 1.50000

YS3 0.000000 0.8 136792 e-0 1

A3 1 1.50000 0.000000

B3 0.000000 0. 1875000

C3 0.000000 0. 1875000

D3 0.000000

YT3

YS4 0.000000 0. 106 132 1

a4 0.000000 0.206 132 1

B4 0.000000 0. 106 132 1

C4 0.000000 0. 106 132 1

D4 0.000000 0.3 1 13208 e-0 1

YT4 13.22500 0.000000

YS5 0.000000 0.2500000

B5 10.22500 0.00000

C5 3.000000

D5 0.000000 0. 1900000

YT5 16.98 125 0.000000

Line slack or excess double price

1 0.000000 1.653 125

2 0.000000 1.559552

3 0.000000 1.559552

4 0.000000 1.437500

5 0.000000 1.437500

6 0.000000 1.356 132

7 0.000000 1.356 132

8 0.000000 1.250000

9 0.000000 1.250000

10 0.000000 1.000000

1 1 0.000000 0. 1500000

12 16.98 125 1.000000

Here a 1-a4 represents the annual investment of project A, and other symbols have similar meanings.