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Free IIA IIA-CIA-Part3 Exam Questions (page: 26)

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PDF with 1555 Questions

A post office serves customers in a single line at one service window. During peak periods, the rate of arrivals has a Poisson distribution with an average of 100 customers per hour and service times that are exponentially distributed with an average of 60 seconds per customer. From this, one can conclude that the:

  1. Queue will expand to infinity.
  2. Server will be idle one-sixth of the time.
  3. Average rate is 100 customers per hour.
  4. Average customer waiting time is 2.5 minutes.

Answer(s): A

Explanation:

One hundred customers arrive in line per hour and only 60 are serviced per hour. Accordingly, the queue will expand to infinity during peak periods.



The arrival times in a waiting-line (queuing) model follow which probability distribution?

  1. Binomial.
  2. Chi-square.
  3. Poisson.
  4. Exponential.

Answer(s): C

Explanation:

Queuing models assume that arrivals follow a Poisson process: The events (arrivals) are independent, any number of events must be possible in the interval of time, the probability of an event is proportional to the length of the interval, and the probability of more than one event is negligible if the interval is sufficiently small.



The feasible solution region is bounded by the lines connecting points:

  1. 3, 4, 6, and 7
  2. 1, 5, 6, and 8.
  3. 2, 4, 5 6, and 8.
  4. 3, 5, 6, and 7.

Answer(s): A

Explanation:

A model consisting of a system of functions may be used to optimize an objective function. If the functions in the model are all linear, the model is a linear programming model. Linear programming is a technique to determine optimal resource allocation. Several solution methods are available to solve linear programming problems. The graphical method, the easiest technique, is limited to simple problems. Here, the graph consists of three lines, each representing a production constraint. The lines connecting points 3, 4, 6, and 7 bound the feasible solution region. Product mixes of X and Y that lie outside this boundary cannot be produced and/or sold because the demand constraint (line 3, 4), the labor constraint (line 4, 6), and the material constraint (line 6, 7) are binding.



If a series of profit lines for X and Y are drawn on the graph, the mix of X and Y that will result in the maximum profit can be determined from?

  1. The last point in the feasible solution region touched by a profit line.
  2. Any point on the boundary of the feasible solution region touched by a profit line.
  3. The first point on the feasible solution region boundary that intersects a profit line.
  4. Any point on the demand constraint that intersects a profit line.

Answer(s): A

Explanation:

A profit line has negative slope because the profit from sales of one product increases as the profit from sales of the other product declines. Moving the profit line rightward (while maintaining its slope) to the last point in the feasible region determines the solution.



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 X 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:

  1. 146 units of X and 0 units of Y.
  2. 0 units of X and 100 units of Y.
  3. 120 units of X and 40 units of Y.
  4. 40 units of X and 120 units of Y.

Answer(s): 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: US $30X + $50Y. This equation is to be maximized subject to the constraints on materials. The two constraint functions are:
Material A: 6X+12Y1.200
Material B: 12X + 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 X and 40 units of Y is a feasible production level that maximizes the contribution margin.



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