I have the following data from chemical kinetics research to fit the parameters of ordinary differential equations:

$$ \left[ \begin{array}{ccccccc} \text{No.}& t & y_1(t)&y_2(t) & y_3(t) & y_4(t) & y_5(t)\\ 1&30.0000 & 9.1300 & 0.0931 & 0.0899 & 0.1000 & 0.0000 \\ 2&60.0000 & 8.9300 & 0.1270 & 0.1230 & 0.2270 & 0.0049 \\ 3&90.0000 & 8.6000 & 0.1510 & 0.1390 & 0.4920 & 0.0153 \\ 4&120.0000 & 8.2800 & 0.1540 & 0.1490 & 0.7780 & 0.0249 \\ 5&150.0000 & 7.9700 & 0.1540 & 0.1570 & 1.0700 & 0.0329 \\ 6&180.0000 & 7.8600 & 0.1540 & 0.1600 & 1.1700 & 0.0348 \\ 7&210.0000 & 7.8100 & 0.1530 & 0.1530 & 1.2100 & 0.0404 \\ 8&240.0000 & 7.7700 & 0.1400 & 0.1420 & 1.2800 & 0.0432 \\ \end{array} \right] $$

The ordinary differential equations to fit have $k_1,k_2,k_3,k_4,k_5,k_6$ to be determined. $$ \left\{ \begin{array}{l} {y_1}'(t)=-{k_1} {y_1}(t)-{k_2} {y_1}(t),\\ {y_2}'(t)={k_2} {y_1}(t)-{k_3} {y_2}(t),\\ {y_3}'(t)={k_1} {y_1}(t)+{k_3} {y_2}(t)-{k_4} {y_3}(t),\\ {y_4}'(t)={k_4} {y_3}(t)-{k_5} {y_2}(t) {y_4}(t)+{k_6} {y_5}(t),\\ {y_5}'(t)={k_5} {y_2}(t) {y_4}(t)-{k_6} {y_5}(t)\\ \end{array} \right.$$

In order to solve it from conventional numerical optimization methods, my original thoughts are: first convert it into least square problems, then apply numerical optimization to it, but this requires symbolically solve a nonlinear system of ordinary differential equations into explicit solutions first, which seems difficult.

My questions are:

(1)Is it possible to determine the global (least square or similarly converted) solution to all the parameters $k_i(i=1,\cdots, 6)$ ?

(2)Is the solution unique?

(3) are there general approaches to solve such problems (globally if possible)?

Update: Further data from replicates:

$$ \begin{array}{ccccccc} \text{No}&t& y_1(t) &y_2(t) &y_3(t) & y_4(t) &y_5(t)\\ 9&30.0000 & 9.0400 & 0.1190 & 0.1040 & 0.1390 & 0.0044 \\ 10& 60.0000 & 8.8000 & 0.1640 & 0.1120 & 0.3210 & 0.0097 \\ 11&90.0000 & 8.5300 & 0.1640 & 0.1140 & 0.5630 & 0.0219 \\ 12&120.0000 & 8.1800 & 0.1600 & 0.1250 & 0.8730 & 0.0369 \\ 13&150.0000 & 7.9700 & 0.1550 & 0.1380 & 1.0600 & 0.0459 \\ 14&180.0000 & 7.7900 & 0.1580 & 0.1510 & 1.2000 & 0.0545 \\ 15&210.0000 & 7.4900 & 0.1480 & 0.1430 & 1.5000 & 0.0636 \\ 16&240.0000 & 7.0800 & 0.1390 & 0.1380 & 1.9100 & 0.0756 \\ \end{array} $$

Update 2:

I am still seeking for a working approach to obtaining solutions to the problem.