You can try f= (1/n)loge(loss1)+(loss2), for a suitable 'n' value that scales down the first term in the sum to the range of the loss2 function values. However this will work only if loss1 is a positive variable.

Thank you Debopam Ghosh , I'll try it right now since actually, loss1 gives only positive values. Thank you again.

If both loss functions are bounded i.e. their values may change from min1 to max1 and from min2 to max2 respectively, then you can use linear scaling which transforms each function into the [0,1] interval. The transformation can be done in two versions:

1) minimum transforms to 0, maximum to 1, all other values proportionally in the ascending order (the larger original value, the transformed value closer to 1).

2) minimum transforms to 1, maximum to 0, all other values proportionally in the descending order (the larger original value, the transformed value closer to 0).

This scaling has no restrictions on negatives or zero values.

Thank you, Mr. Przemysław Kowalik . It is a good idea, however, I wonder if this transformation affects the optimization performance of the algorithm?

Mebarka Allaoui , I suppose there is nothing to worry about. If you want to minimize the sum of losses, the transformation will be

(loss1(x)-min1)/(max1-min1)+(loss2(x)-min2)/(max2-min2)

Here x stands for a single variable or a vector x1, x2,....., xn.

We can transform it further

loss1(x)/(max1-min1)-min1/(max1-min1)+loss2(x)/(max2-min2)-min2/(max2-min2).

Because min1/(max1-min1) and min2/(max2-min2) are both fixed values (independent of x) then real minimization is performed on the function loss1(x)/(max1-min1)+loss2(x)/(max2-min2). It is almost an original, minimized function but with both components divided by fixed numbers. I don't think if it affects the algoritm seriously.

Alright, Thank you Przemysław Kowalik for your detailed clarification.