TA B L E in Software

Create European Article Number 13 in Software TA B L E

Activity System State Backup Generation Backup Verification Clusters: Cluster State Verification Clusters: Print Queue Status Verification User Management User Password Reset Directory Service Log Event Verification Account Management Security Group Management DNS Event Log Verification Available Free Space Verification Data Backup Management Shared Folder Management File Replication Service Event Log Verification Run As Shortcuts General Service Status Verification System Event Log Verification Security Event Log Verification Service and Admin Account Management Activity Log Maintenance Print Queue Management

6.22. Give an O(nt) algorithm for the following task. Input: A list of n positive integers a1 , a2 , . . . , an ; a positive integer t. Question: Does some subset of the ai s add up to t (You can use each ai at most once.) 6.23. A mission-critical production system has n stages that have to be performed sequentially; stage i is performed by machine Mi . Each machine Mi has a probability ri of functioning reliably and a probability 1 ri of failing (and the failures are independent). Therefore, if we implement each stage with a single machine, the probability that the whole system works is r1 r2 rn . To improve this probability we add redundancy, by having mi copies of the machine Mi that performs stage i. The probability that all mi copies fail simultaneously is only (1 ri )mi , so the probability that stage i is completed correctly is 1 (1 ri )mi and the probability that the whole n system works is i=1 (1 (1 ri )mi ). Each machine Mi has a cost ci , and there is a total budget B to buy machines. (Assume that B and ci are positive integers.) Given the probabilities r1 , . . . , rn , the costs c1 , . . . , cn , and the budget B, nd the redundancies m1 , . . . , mn that are within the available budget and that maximize the probability that the system works correctly. 6.24. Time and space complexity of dynamic programming. Our dynamic programming algorithm for computing the edit distance between strings of length m and n creates a table of size n m and therefore needs O(mn) time and space. In practice, it will run out of space long before it runs out of time. How can this space requirement be reduced (a) Show that if we just want to compute the value of the edit distance (rather than the optimal sequence of edits), then only O(n) space is needed, because only a small portion of the table needs to be maintained at any given time. (b) Now suppose that we also want the optimal sequence of edits. As we saw earlier, this problem can be recast in terms of a corresponding grid-shaped dag, in which the goal is to nd the optimal path from node (0, 0) to node (n, m). It will be convenient to work with this formulation, and while we re talking about convenience, we might as well also assume that m is a power of 2. Let s start with a small addition to the edit distance algorithm that will turn out to be very useful. The optimal path in the dag must pass through an intermediate node (k, m/2) for some k; show how the algorithm can be modi ed to also return this value k. (c) Now consider a recursive scheme: procedure find-path((0, 0) (n, m)) compute the value k above find-path((0, 0) (k, m/2)) find-path((k, m/2) (n, m)) concatenate these two paths, with k in the middle Show that this scheme can be made to run in O(mn) time and O(n) space. 6.25. Consider the following 3- PARTITION problem. Given integers a1 , . . . , an , we want to determine whether it is possible to partition of {1, . . . , n} into three disjoint subsets I, J, K such that ai =

costs and buy the October PG 52.50 call for 2.90, we achieve a long delta with the call option of 46.9 percent. Outright this would be a limited risk position with a breakeven at 55.40 on the underlying. If we sell the October PG 42.50 put for a premium of 2.25, we have at total net premium debit of .65 per option spread and we have an additional long delta of 20.6 percent. The total combined delta on the position is 67.5 percent long the underlying stock. Now let s look at the model for this trade as shown in Figure 6.14. The risk reversal has a similar angle to the underlying with the pro t building sharply as the market rises and losing sharply as the market falls. The important point to the risk reversal is the breakeven points on each side of the trade. There is a at area on the model where both options expire as worthless and the total loss is the premium paid for the spread. Unlike the outright purchase of the underlying market, however, the risk reversal does not start losing signi cantly at expiration until it reaches $42.50, while the underlying would start losing immediately. Calculating breakeven for the risk reversal depends on whether the spread is applied for a debit or credit. If it s a debit spread,

ARCHITECT REGISTRATION EXAMINATION (ARE)

The state space representation converts each n-th order di erential equation into n coupled rst order di erential equations. Such a representation is often desirable for ease of implementation and analysis, as the state space representation allows the use of vector and matrix techniques. State space techniques relevant to both continuous and discrete-time systems are discussed in Section 3.2 3.5.

Consider the following quadratic equation in polynomial standard form: 4x 2 + 64 = 0 How can this equation be rewritten in binomial factor form