code: int getPagePosition(AbstractIndexPage indexPage) {
  		//We know that the element exists, so we iterate to list.length instead of nEntires 
  		//(which is not available in this context at the moment.
  		for (int i = 0; i < subPages.length; i++) {
  			if (subPages[i] == indexPage) {
  				return i;
  			}
  		}
  		throw DBLogger.newFatalInternal("Leaf page not found in parent page: " + 
  				indexPage.pageId + "   " + Arrays.toString(subPageIds));
  	}
docs: This method will fail if called on the first page in the tree. However this should not
  happen, because when called, we already have a reference to a previous page.
  @param oidIndexPage
  @return The position of the given page in the subPage-array with 0 <= pos <= nEntries.