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` .
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.