Set full-time waiter:
9 ~ 12+ 13 ~ 17:x 1
9 ~ 13+ 14 ~ 17:x2
Part-time waiter:
9~ 13: x3
10~ 14: x4
1 1~ 15: x5
12~ 16: x6
13~ 17: x7
Objective function: min {100 (x1+x2)+40 (x3+x4+X5+X6+X7)}
Constraints:
9~ 10 segment is not less than 4:
x 1+x2+x3 & gt; =4;
10~ 1 1 period shall not be less than 3:
x 1+x2+x3+x4 & gt; =3;
Similarly, you can always write down:
x 1+x2+x3+x4+X5 & gt; =4;
x2+x3+x4+X5+X6 & gt; =6;
x 1+x4+X5+X6+x7 & gt; =5;
x 1+x2+X5+X6+x7 & gt; =6;
x 1+x2+X6+x7 & gt; =8;
x 1+x2+x7 & gt; =8;
Otherwise, the total number of part-time waiters will be limited:
x3+x4+X5+X6+x7 & lt; =3.
Again, note that this is an integer programming, using mathematica to run the following statement:
Linear programming [{ 100, 100, 40, 40, 40,
40}, {{ 1, 1, 1, 0, 0, 0, 0}, { 1, 1, 1, 1, 0, 0, 0}, { 1, 1, 1, 1, 1,
0, 0}, {0, 1, 1, 1, 1, 1, 0}, { 1, 0, 0, 1, 1, 1, 1}, { 1, 1, 0, 0,
1, 1, 1}, { 1, 1, 0, 0, 0, 1, 1}, { 1, 1, 0, 0, 0, 0, 1}, {0,
0, - 1, - 1, - 1, - 1, - 1}}, {4, 3, 4, 6, 5, 6, 8,
8, -3}, automatic, integer]
The results are as follows:
{3, 4, 0, 2, 0, 0, 1}
Corresponding to the values of x 1 to x7, respectively.