 Solution

With initial threshold zero and unit step costs, the thresholds are $0,1,\dots ,d$, so there are $d+1$ passes. Starting the first pass at one instead omits the zero-threshold pass when the root is not a goal.

At depth $i$, at most $b^i$ search nodes are admitted in each pass with threshold at least $i$. Across all passes, the full-to-cutoff visit bound isThis is attained when the only depth-$d$ goal is visited last. If a goal test occurs before expansion, the goal is visited but not expanded, giving the corresponding worst-case expansion countFor $b\ne 1$, the visit bound is $V(b,d)=[b^{d+2}-(d+2)b+d+1]/(b-1{)}^2$, hence the worst-case number expanded is $\Theta (b^d)$ for fixed $b>1$. For $b=1$, it is $\Theta (d^2)$. An earlier goal reduces the final pass, so its exact count cannot be inferred from $b,d$ alone. Children generated and immediately rejected above a threshold are not counted as expanded; counting those generated nodes adds the corresponding branching overhead.