Here is a test problem without integer restriction:

>> n=6; C=7.4 ;c = 5*rand(n,1); d = 2*rand(n,1); >> disp(c) 2.107028387658593 0.400055793339331 0.396769045866593 0.361881245993311 4.501705030405445 4.846541494332056 >> disp(d) 0.782613917623122 0.627170209734511 1.106627159874890 1.584096124994781 1.596628166160420 1.726517081119803 

and here is the output from running the above program:

Calling SDPT3 4.0: 25 variables, 12 equality constraints For improved efficiency, SDPT3 is solving the dual problem. ------------------------------------------------------------ num. of constraints = 12 dim. of sdp var = 12, num. of sdp blk = 6 dim. of linear var = 6 dim. of free var = 1 *** convert ublk to lblk ******************************************************************* SDPT3: Infeasible path-following algorithms ******************************************************************* version predcorr gam expon scale_data HKM 1 0.000 1 0 it pstep dstep pinfeas dinfeas gap prim-obj dual-obj cputime ------------------------------------------------------------------- 0|0.000|0.000|1.7e+01|3.4e+00|1.9e+03| 1.261398e+02 0.000000e+00| 0:0:00| chol 1 1 1|1.000|0.598|3.9e-06|1.4e+00|5.6e+02| 1.429249e+02 -3.878636e+01| 0:0:00| chol 1 1 2|1.000|0.966|1.0e-06|4.9e-02|7.2e+01| 5.719567e+01 -6.592350e+00| 0:0:00| chol 1 1 3|1.000|0.540|2.8e-08|2.3e-02|1.0e+01| 3.460381e+00 -5.835158e+00| 0:0:00| chol 1 1 4|1.000|0.442|1.6e-07|1.3e-02|4.8e+00|-2.227469e-01 -4.589011e+00| 0:0:00| chol 1 1 5|0.921|0.925|1.7e-08|9.6e-04|4.7e-01|-1.759148e+00 -2.201532e+00| 0:0:00| chol 1 1 6|1.000|0.575|1.1e-09|4.1e-04|1.7e-01|-1.906104e+00 -2.063560e+00| 0:0:00| chol 1 1 7|0.981|0.919|3.2e-09|3.3e-05|1.3e-02|-1.939996e+00 -1.951918e+00| 0:0:00| chol 1 1 8|0.961|0.971|2.6e-10|9.6e-07|3.9e-04|-1.941595e+00 -1.941964e+00| 0:0:00| chol 1 1 9|0.956|0.918|7.3e-11|9.8e-07|3.3e-05|-1.941656e+00 -1.941686e+00| 0:0:00| chol 1 1 10|0.980|0.964|1.7e-11|8.4e-08|1.6e-06|-1.941660e+00 -1.941661e+00| 0:0:00| chol 1 1 11|1.000|0.966|3.2e-13|3.9e-09|1.3e-07|-1.941660e+00 -1.941660e+00| 0:0:00| chol 1 1 12|1.000|0.984|4.6e-14|3.2e-10|5.5e-09|-1.941660e+00 -1.941660e+00| 0:0:00| stop: max(relative gap, infeasibilities) < 1.49e-08 ------------------------------------------------------------------- number of iterations = 12 primal objective value = -1.94166032e+00 dual objective value = -1.94166032e+00 gap := trace(XZ) = 5.46e-09 relative gap = 1.12e-09 actual relative gap = 1.02e-09 rel. primal infeas (scaled problem) = 4.58e-14 rel. dual " " " = 3.23e-10 rel. primal infeas (unscaled problem) = 0.00e+00 rel. dual " " " = 0.00e+00 norm(X), norm(y), norm(Z) = 2.7e+00, 4.2e+00, 1.0e+01 norm(A), norm(b), norm(C) = 6.9e+00, 3.4e+00, 1.4e+01 Total CPU time (secs) = 0.20 CPU time per iteration = 0.02 termination code = 0 DIMACS: 7.9e-14 0.0e+00 5.4e-10 0.0e+00 1.0e-09 1.1e-09 ------------------------------------------------------------------- ------------------------------------------------------------ Status: Solved Optimal value (cvx_optval): +1.94166 >> disp(x) % this is the optimal value of x 0.399054349910791 2.815385122487830 2.241149560765849 1.944410952569123 0.000000007535945 0.000000006728979 

The last 2 elements of x would be exactly zero if the optimization problem were solved exactly. The slight difference from zero is due to solver optimality tolerance.

Because the denominator must be positive, the objective function, and hence the optimization problem is convex, and can be readily formulated and solved using CVX or a similar convex optimization tool.

cvx_begin variable x(n) minimize(sum(inv_pos(c + d.* x))) x >= 0 sum(x) == C cvx_end 

Because the denominator must be positive, the objective function, and hence the optimization problem is convex, and can be readily formulated and solved using CVX or a similar convex optimization tool.

cvx_begin variable x(n) minimize(sum(inv_pos(c + d.* x))) x >= 0 sum(x) == C cvx_end 

You change x >= 0 to x >= small_number if you prefer. Strict inequalities are treated by the solvers as if they are non-strict inequalities.

It does not help to restrict x to integer values. That would make the problem more computationally difficult to solve. Nevertheless, to do so, merely change variable x(n) to variable x(n) integer .