Function y = myinvest (x);

y =-[0 0 0 1. 15 1.25 1.40 0 0 0 0 1.06]* x；

Then run the command line (preferably in another m file):

Clear all;

My idea is this: set A and D at the beginning of each year, so that there are four A's and five D's, and D, B, C and C are fixed.

% X =[a 1 a2 a3 a4 b c d 1 D2 D3 D4 D5]

% actually needs the largest positive value1.15a4+1.25b+1.40c+1.06d5, which is equivalent to finding the smallest negative value, so the fmincon function is used.

% max([0 0 0 1. 15 1.25 1.40 0 0 0 0 1.06]* X ')

% inequality constraint, only C.

a =[0 0 0 0 1 0 0 0 0 0]；

b = 3；

% equals one * * * Five: Cash on hand at the beginning of next year equals the balance at the end of last year.

% a 1 a2 a3 a4 b c d 1 D2 D3 D4 D5

aeq =[ 1 0 0 0 0 1 0 0 0 0 0

0 1 0 0 0 1 - 1.06 1 0 0 0

- 1. 15 0 1 0 1 0 0 - 1.06 1 0 0

0 - 1. 15 0 1 0 0 0 0 - 1.06 1 0

0 0 1. 15 0 0 0 0 0 0 1.06 - 1];

beq =[ 10 0 0 0]'；

options = optimset('LargeScale '，' off '，' MaxFunEvals '， 1e4，' MaxIter '， 1 E4)；

x = fmincon(@myinvest，zeros( 1 1， 1)，A，B，Aeq，Beq， 1e-6+zeros( 1 1， 1)，[]，[]，options

The solved x is the corresponding [a1a2a3a4bc12d3d4d5].

7. 1698% smashed a in the first year, so don't vote again.

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In the third year, 8.2453% hit B.

3.0000% got a C in the second year.

2.8302% got a D in the first year, so don't vote again.

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