Given the following table, determine the range of probability for state of nature #2, that is P (2), for which each alternative is optimal under the expected-value approach.

Question

		STATE OF NATURE
		#1	#2
	A	4	12
Alternative	B	16	2
	C	12	8

Accepted Answer

First, plot each alternative relative to P (2). To do this, plot the #1 value on the left side of the graph and the #2 value on the right side. For instance, for alternative A, plot 4 on the left side of the graph and 12 on the right side. Then connect these two points with a straight line. The tree alternatves are plotted on the graph as shown below. The graph shows the range of values of P (2) over which each alternative is optimal. Thus, for low values of P(2), [and thus high values of P (1), since P (1) + P (2) = 1.0], alternative B will have the highest expected value; for intermediate values of P (2), alternative C is best; and for higher values of P (2), alternative A is best.

To find exact values of the ranges, determine where the upper parts of the lines intersect. Not that at the intersections, the two alternatives represented by the lines would be equivalent in terms of expected value. Hence, the decision maker would be indifferent between the two at that point. To determine the intersections, you must obtain the equation of each line. This is relatively simple to do. Because these are straight lines, they hav the form y = a + bx, where a is the y -intercept value at the left axis, b is the slope of the line, and x is P (2). Slope is defined as the change in y for a one-unit change in x. In this type of problem, the distance between the two vertical axes is 1.0. Consequently, the slope of each line is equal to the righthand value minus the left-hand value. The slopes and equations are as follows:

	#1	#2	Slope	Equation
A	4	12	12 - 4 = +8	4 +8P(2)
B	16	2	2 - 16 = - 14	16 - 14P(2)
C	12	8	8 - 12 = - 4	12 - 4P(2)

From the graph, we can see that alternative B is best from P(2)= 0 to the point where that straight line intersects the straight line of alternatve C, and that begins the region where C is better. To find that point, solve for the value of P (2) at their intersection. This requires setting the two equations equal to each other and solving for P (2). Thus,

16 - 14P(2) = 12 4P(2)

Rearranging terms yields

4 =10P(2)

Solving yields P (2) .40. Thus, alternative B is best from P(2) = 0 up to P (2) = .40. B and C are equivalent at P (2) = .40.

Alternative C is best from that point until its line intersect alternatve A’s line. To find that intersection, set those two equations equal and solve for P (2). Thus,

4 + 8P(2) = 12 - 4P(2)

Rearranging terms results in

12P(2) = 8

Solving yields P (2) = .67.Thus, alternative C is best from P (2) > .40 up to P (2) = .67, where A and C are equivalent. For values of P (2) greater than .67 up to P (2) 1.0, A is best.

Question 5S.8: Given the following table, determine the range of probabilit...

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