(20 pts) In a single time-step binomial model, suppose A(0) = 50, A(1) = 55, S(0) = 100, and 140, with probability p = 0.3; S(1) = 90, with probability 1 - p = 0.7. (9a) (10 pts) Find the one time-step return r on the risk-free asset, the one time-step return K on the stock, and the risk-neutral probability p*. (9b) (10 pts) Find the time zero price P(0) of a put option with strike price K = 120 and exercise time T = 1. 2. Carsales.com Limited (CAR) is an Australian company that (unsurprisingly) owns the online marketplace www.carsales.com.au. At the time of writing, the share price of CAR is $15.92. Assume this is time zero. You wish to compare premium prices of a European call calculated with the Black- Scholes model, with premium prices calculated with a binomial model. The call has strike price K = $17.00, and expires in 90 days so T = 90/365 years. The yearly volatility of CAR shares is estimated to be = 4.395. Assume the continuously compounding interest rate is r=4% pa. (a) Calculate the call premium using the Black-Scholes model. (b) Consider a three-step binomial CRR model. (i) Assuming interest rates are constant over the life of the call, calculate the return R over one time step. (ii) Calculate the up and down factors u and d in this three-step model. (iii) Calculate the risk neutral probability in this three-step model. (iv) Construct a three-step binomial pricing tree for the call and calculate its premium. This is the most complicated one. Inorder successor Search for the node with a key value larger than than the key of the given node. This is the most complicated algorithm. That says: Method 2 (Search from root) Parent pointer is NOT needed in this algorithm. The Algorithm is divided into two cases on the basis of right subtree of the input node being empty or not. Input: node, root // node is the node whose Inorder successor is needed. Output: succ // succ is Inorder successor of node. 1. If right subtree of node is not NULL (that means, if there is a right subtree of that node), then succ lies in right subtree. Do the following. Go to right subtree and return the node with minimum key value in the right subtree. 2. If right subtree of node is NULL (that means, there is no right subtree), then start from the root and use search like technique. Do the following. Travel down the tree, if a node's data is greater than root's data then go right side, otherwise, go to left side. 3. [not in text] last left is the inorder successor 4. For all the questions, list the traversal nodes land why you go left or right. a. Find the inorder successor of node 4 b. Find the inorder successor of node 2 c. Find the inorder success of node 5. 5. Note: algorithm 1 is slightly easier, but assumes that every node has a pointer back to its parent. Therefore in that algorithm, you can back up. That is not the case in the data structure that we have used for a binary search tree (BST). That one has only two pointers, one to the left child (the smaller one) and one to the right child (the larger one). So there is a trade-off between space, using an extra pointer to point back to the parent, and convenience. An extra pointer costs 50% more space (3 pointers instead of 2 per node). That is quite a lot. 6. Note: lease follow the question and answer the follwing question, thanks in advance