Solving an online problem without maximizing or minimizing an objective You have a budget $B$ and a satisfaction level $L$. An indivisible product is available for sell and you are interested in buying. The product is divided into $n$ parts. Part $i$ of the product has positive value $v_i$.

On day $i$, you get to know $v_i$ and you have to decide how much price $p_i\geq0$ you would pay for this part of the product.

After $j$ days, given the values $v_1, v_2, \ldots, v_j$ and the prices you paid $p_1, p_2, \ldots, p_j$ for each value, you know that you can reach a satisfaction level of $(1+v_1p_1)(1+v_2p_2)\cdots(1+v_jp_j)$.

Your objective is to buy the product after $n$ days while you reach the satisfaction level of $L$ and pay no more than $B$.
@orlp's intuition is correct. 

lemma. No online algorithm solves the problem in the worst case.

Proof.  Consider the following instance:

$$f(p) = (1+p_1)(1+p_2)(1+p_3) \ge 8 \text{ with } B=3.$$

This instance has just one solution (one $p$ that satisfies it), namely $p=(1,1,1)$.  So, given $B=3$ and $L=8$ and just the first factor $1+p_1$, the algorithm has to choose $p_1 = 1$.

But then the adversary instead continues by defining the following instance:

$$g(p) = (1+p_1)(1+ 1.49 p_2)(1+0\cdot p_3) \ge 8 \text{ with } B=3.$$

This instance is feasible (e.g. for $p_1=p_2=3/2$ we have $g(p)\approx 8.1$). However, the algorithm committed to $p_1 = 1$, so the residual problem it is faced with is now $2(1+1.49 p_2) \ge 8$ with $p_2 \le 2$.  Since $2(1+1.49\cdot 2) &lt; 2\cdot 4 = 8$, this residual problem is not feasible.  

So the online algorithm cannot guarantee a solution, even if it knows in advance that it will face just one of the two instances above. $~~~\Box$