6-Multiple Targets

6- Understanding Multiple Target issue

Previously we had one target which needed to be achieved. This could be done with multiple possible values of w (weight) and b(bias) however, as you may know once 2 points are there they form a line and only one equation solves it.

Physical Interpretation

we have a ice cream seller who sold 12 icecreams on 1 degree temprature so we can have x=1 and y =12 and w & b can have infinite values however if we put the condition that he sold 14 ice creams on 2 degree temprature, suddenly the equation can be solved and we get a single unique solution of y = 2x + 10

Now we are basically adjusting w and b through calculating gradient from loss of one target. but if there are multiple targets to achieve, we will get multiple gradients.

Multiple gradient problem

these are losses essentially for our current setup of w and b for each of targets eg. example 1 → gradient1 = -2 example 2 → gradient2 = -8 example 3 → gradient3 = -18 example 4 → gradient4 = -32 Now how do we resolve this, simple, calculate average. average of gradients = (-2 - 8 - 18 - 32) / 4 = -15 And rest of steps are same. we minimize loss for this -15 value. None of the targets will get exactly 0 loss. but it will be the closest we can get.