1. We take into consideration that we have a k-bit binary with decrement and increment operations. Using the aggregate method, we can calculate the worst case scenario. With decrement operation, the worst case is O(k) as all zeroes are flipped to ones. With increment operation, the worst case operation is O(k) as we flip all ones to zeroes. With each costing O(k) since bits are flipped, the total cost would be O(nk) since we have each operation costing O(k) and we have sequence of n operations. Thus, the sequence of n operations in a k-bit binary counter will cost O(nk) using aggregate method. 2. Using an accounting method of analysis, we denote ci as the ith operation cost. We will have ci= { i if i is truly an exact power of 2 { 1 otherwise $3 will be charged for each operation. When i is not an exact power of 2, we will pay $1 and have $2 as credit. When i is the exact power of 2, then we will pay $i with credit. With this information, we will have: Operation Cost Actual Cost Remaining Credit 1 3 1 2 2 3 2 3 3 3 1 5