1. Establish a model and set the series c(n) as the total deposit at the beginning of the nth year. Obviously, the problem is to minimize c( 1), that is, to minimize the total amount in 1 year to meet the requirement of paying the most. The total deposit consists of three parts, namely, c (n) = x(n)+0.98y(n)+0.965z(n) (1-1), where x (n), y (n) and z (n) are short-term deposits and six-year deposits respectively. In order to unify the expression form, the number of copies of X can be decimal, and one copy is 1 ten thousand yuan; The other shares y and z are integers. X, Y and Z series are basic independent variables, which determine the investment mode and proportion. The income s(n) at the end of each year is related to the investment of the previous year, that is, s (n) =1.04x (n-1)+1.04y (n-6)+1.03z (n-/kloc). Therefore, c (n+1) = s (n)-F(n) (1-3), where f (n) is the bonus F = [10,1,. Simplification of the problem, because in 15, x (n > 16- 1)= 0y(n & gt； 16-6)= 0(2- 1)z(n & gt； 16- 13)=0, that is, the sequence is finite. The intermediate variables c and s satisfy equations (1- 1) to (1-3), where the values of n are all in the range of 1 to 15. Obviously, c( 16)=0, and c( 1) is what you want. Since F is known, there are five groups of independent variables ***x, Y, Z, C and S, the number of which is15+10+3+14+15 = 57, which satisfies15. The dependent variable is 1, that is, c( 1). The constraints to be satisfied are (1-4) and *** 15 inequality. 3. Specific solution As can be seen from the previous analysis, this is a typical linear programming problem with constraints. Therefore, it is suggested to use lp function in matlab to realize it. You can check the relevant data specifically, so I won't go into details here.

Hope to adopt