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NEW QUESTION: 1
Which of the following is a method for requirements change control?
A. Scope analysis
B. Requirements management
C. Communication management
D. Baselining
Answer: D
NEW QUESTION: 2
A. Option B
B. Option C
C. Option D
D. Option A
Answer: A
Explanation:
Explanation
The round-robin keyword enables round-robin address allocation for a PAT pool. Without round robin, by default all ports for a PAT address will be allocated before the next PAT address is used. The round-robin method assigns an address/port from each PAT address in the pool before returning to use the first address again, and then the second address, and so on.
Source:
http://www.cisco.com/c/en/us/td/docs/security/asa/asa90/configuration/guide/asa_90_cli_config/ nat_objects.html#61711
NEW QUESTION: 3
Heniser Pet Foods manufactures two products, X. and Y. The unit contribution margins for
Products X. and Y are US $30 and US $50, respectively. Each product uses Materials A and
B. Product uses 6 pounds of Material A and 12 pounds of Material B. Product Y uses 12 pounds of Material A and 8 pounds of Material B. The company can purchase only 1,200 pounds of Material A and 1,760 pounds of Material B. The optimal mix of products to manufacture is:
A. 146 units of and 0 units of Y.
B. 40 units of X. and 120 units of Y.
C. 120 units of and 40 units of Y.
D. 0 units of X and 100 units of Y.
Answer: C
Explanation:
Linear programming is a technique used to maximize a contribution margin function or to minimize a cost function, subject to constraints such as scarce resources or minimum/maximum levels of production. Thus, linear programming is often used for planning resource allocations. In this problem, the equation to be maximized, called the objective function, is: U3 $30)K+ $50Y. This equation is to be maximized subject to the constraints on materials. The two constraint functions are:
Material A: 6X + 12Y < 1,200 Material B: I 2X + 8Y < 1,760
One way to solve this problem is to graph the constraint lines and determine the feasible area. The optimal production level is at an extreme point within the feasible area. The graph shows that a production level of 120 units of and 40 units of Y is a feasible production level that maximizes the contribution margin.
