Is it possible for breadth first search to have a larger operating time (O(n)) than IDDFS

I have an exam in an hour, and theres something in the lecture slides that I disagree with. Theres a nice little table saying that the time complexity of BFS is O(b^(d+1)), and the time complexity of IDDFS O(b^d), where b is a branching factor and d is the depth of the solution. I have no i开发者_运维问答dea where he got the +1 for the BFS time complexity, and further more, implementation efficiency asside, with my understanding of IDDFS I have no idea why BFS would expand more nodes. Am I insane?