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Decision Trees & Rollback - Finite Mathematics

  1. A consulting firm offers a new technology that would only affect the profit of the small plant in a high-demand market. What is the minimum profit the small plant with high demand would need to generate for the optimal decision to switch from 'Build Large' to 'Build Small'?
    Answer: $160,000 First, calculate the current EMV for each option. EMV(Small) = 0.5($100,000)+0.5($40,000)=$50,000+$20,000=$70,0000.5(\$100,000) + 0.5(\$40,000) = \$50,000 + \$20,000 = \$70,000. EMV(Large) = 0.5($250,000)+0.5(−$50,000)=$125,000−$25,000=$100,0000.5(\$250,000) + 0.5(-\$50,000) = \$125,000 - \$25,000 = \$100,000. Currently, the optimal decision is to build the large plant. The decision will switch when the EMV of the small plant is equal to or greater than the EMV of the large plant. Let XX be the new profit for the small plant with high demand. We set the new EMV(Small) equal to EMV(Large): 0.5(X)+0.5($40,000)=$100,0000.5(X) + 0.5(\$40,000) = \$100,000 0.5X+$20,000=$100,0000.5X + \$20,000 = \$100,000 0.5X=$80,0000.5X = \$80,000 X=$160,000X = \$160,000 Therefore, the profit for the small plant with high demand must be at least $160,000.
  2. Let the payoff for success in Project X be $100,000 and the payoff for failure be -$25,000. The decision to choose Project X over Project Y remains optimal as long as the probability of success pp is greater than or equal to what threshold value?
    Answer: p=0.520p = 0.520 When you encounter decision tree sensitivity analysis problems, you're testing how changes in probability estimates affect the optimal choice between alternatives. The key is finding the probability threshold where both projects have equal expected value. To find this threshold, you need to set up an equation where Project X's EMV equals Project Y's EMV. Project X's expected value is: p×$100,000+(1−p)×(−$25,000)p \times \$100,000 + (1-p) \times (-\$25,000). This simplifies to $125,000p−$25,000\$125,000p - \$25,000. Project Y has a fixed EMV of $40,000. Setting them equal: $125,000p−$25,000=$40,000\$125,000p - \$25,000 = \$40,000. Solving for p: $125,000p=$65,000\$125,000p = \$65,000, so p=0.520p = 0.520. This means Project X remains optimal when p≥0.520p \geq 0.520. Choice A (p=0.520p = 0.520) is correct—this is the exact breakeven point where both projects have equal expected value. Choice B (p=0.400p = 0.400) is wrong because this represents a probability below the threshold. At this level, Project Y would actually be superior to Project X. Choice C (p=0.600p = 0.600) is wrong because this is the original probability given in the problem, not the threshold value we're solving for. Choice D (p=0.680p = 0.680) is wrong because this exceeds the threshold unnecessarily. While Project X would still be optimal at this probability, it's not the minimum required threshold. Study tip: In sensitivity analysis problems, always set up equations where competing alternatives have equal expected values to find decision thresholds. The threshold represents the point where you're indifferent between choices.
  3. What is the optimal decision based on expected monetary value, and what is that value?
    Answer: Outsource; EMV = $170,000 First, calculate net payoffs (Revenue - Cost). In-house: HighQ profit = $300k - $150k = $150k; LowQ profit = $50k - $150k = -$100k. EMV(In-house) = $0.7(150k) + 0.3(-100k) = 105k - 30k = $75k. Outsource: HighQ profit = $300k - $100k = $200k; LowQ profit = $50k - $100k = -$50k. The outsource path has a two-stage chance process. If unreliable, the EMV of the outcome is $0.4(200k) + 0.6(-50k) = 80k - 30k = $50k. Now, roll back the first chance node for the outsource option: EMV(Outsource) = P(Reliable)×(Profit if reliable)+P(Unreliable)×(EMV if unreliable)P(Reliable) \times (\text{Profit if reliable}) + P(Unreliable) \times (\text{EMV if unreliable}). EMV(Outsource) = $0.8(200k) + 0.2(50k) = 160k + 10k = $170k. Comparing the two options, EMV(Outsource) = $170k is greater than EMV(In-house) = $75k. The optimal decision is to outsource.
  4. In a standard decision tree used for rollback analysis, which of the following actions is performed at a square decision node?
    Answer: The node's value is determined by selecting the maximum value among the immediate subsequent nodes. Decision tree analysis is a fundamental tool in decision-making under uncertainty, where you work backward from outcomes to determine optimal choices. The key distinction is between square nodes (decision points where you choose) and circular nodes (chance events with given probabilities). At square decision nodes, you have control over which path to take, so you naturally want to choose the option that gives you the best possible outcome. This means selecting the branch with the maximum expected value among all immediate options available from that node. You're literally picking the best choice available to you at that moment. Option A describes what happens at circular chance nodes, not square decision nodes. At chance nodes, you calculate expected value using weighted averages based on probabilities since you don't control the outcome. Option B is incorrect because decision analysis doesn't involve subtracting costs from simple averages of subsequent values—this misunderstands both the cost accounting and averaging processes. Option C confuses probability with value maximization; even if an outcome has the highest probability, it might not have the highest expected value, and probability-based selection isn't how decision nodes work anyway. The correct answer is D because decision nodes represent points where you actively choose the best available option, which means selecting the maximum value path. Study tip: Remember the shape-function relationship: squares = decisions = maximum selection, circles = chance = probability-weighted averaging. When you see a square node, always think "choose the best option available."
  5. To maximize expected utility, which strategy should the student choose, and what is the expected utility of that choice?
    Answer: Study Group; Expected Utility = 90 The expected utility (EU) for each strategy is calculated by summing the products of each outcome's utility and its probability. For 'Cram': EU(Cram) = 0.6(100)+0.4(50)=60+20=800.6(100) + 0.4(50) = 60 + 20 = 80 utils. For 'Study Group': EU(Study Group) = 0.5(100)+0.5(80)=50+40=900.5(100) + 0.5(80) = 50 + 40 = 90 utils. Comparing the two expected utilities, EU(Study Group) = 90 is greater than EU(Cram) = 80. Therefore, the optimal strategy is to join the Study Group, which has an expected utility of 90.
  6. Based on a rollback analysis using expected monetary value (EMV), what is the optimal initial decision and the overall expected value of the project?
    Answer: Conduct the survey; the project's EMV is $80,000. The problem is solved using rollback analysis. First, evaluate the 'Launch Immediately' branch: EMV(Launch Immediately) = 0.5($180,000)+0.5(−$60,000)=$90,000−$30,000=$60,0000.5(\$180,000) + 0.5(-\$60,000) = \$90,000 - \$30,000 = \$60,000. Since $60,000 > $0 (for not launching), the value of this path is $60,000. Next, evaluate the 'Conduct Survey' branch. If the survey is positive, EMV(Launch) = 0.8($200,000)+0.2(−$50,000)=$160,000−$10,000=$150,0000.8(\$200,000) + 0.2(-\$50,000) = \$160,000 - \$10,000 = \$150,000. This is greater than $0, so the decision is to launch. If the survey is negative, EMV(Launch) = 0.3($150,000)+0.7(−$80,000)=$45,000−$56,000=−$11,0000.3(\$150,000) + 0.7(-\$80,000) = \$45,000 - \$56,000 = -\$11,000. This is less than $0, so the decision is not to launch (value is $0). The expected value at the survey's chance node is 0.6($150,000)+0.4($0)=$90,0000.6(\$150,000) + 0.4(\$0) = \$90,000. Finally, subtract the survey cost: $90,000−$10,000=$80,000\$90,000 - \$10,000 = \$80,000. Comparing the initial options, EMV(Survey) = $80,000 is greater than EMV(Launch Immediately) = $60,000. Thus, the optimal decision is to conduct the survey, and the project's EMV is $80,000.
  7. Given that the overall EMV of the project is $150, what must be the value of the payoff XX?
    Answer: $50 The overall EMV of the project is the value at the initial decision node. This value is the maximum of the values of the branches, so EMV(Project) = max(Value(A), Value(B)). We are given Value(A) = $140 and EMV(Project) = $150. Since $150 > $140, the optimal decision must have been to choose Alternative B, and the value of Alternative B must be $150. The value of Alternative B is its expected monetary value: EMV(B) = 0.5($250)+0.5(X)0.5(\$250) + 0.5(X). We set this equal to $150: $150=0.5($250)+0.5X\$150 = 0.5(\$250) + 0.5X $150=$125+0.5X\$150 = \$125 + 0.5X $25=0.5X\$25 = 0.5X X=$50X = \$50
  8. Based on the expected monetary value, which crop should the farmer plant, and what is the expected profit?
    Answer: Plant Corn; Expected Profit = $58,000 First, calculate the expected profit (EMV) for each option by considering revenues and costs. For Corn: The profit in a wet season is $50,000−$10,000=$40,000\$50,000 - \$10,000 = \$40,000. The profit in a dry season is $80,000−$10,000=$70,000\$80,000 - \$10,000 = \$70,000. EMV(Corn) = 0.4($40,000)+0.6($70,000)=$16,000+$42,000=$58,0000.4(\$40,000) + 0.6(\$70,000) = \$16,000 + \$42,000 = \$58,000. For Soybeans: The profit in a wet season is $70,000−$15,000=$55,000\$70,000 - \$15,000 = \$55,000. The profit in a dry season is $60,000−$15,000=$45,000\$60,000 - \$15,000 = \$45,000. EMV(Soybeans) = 0.4($55,000)+0.6($45,000)=$22,000+$27,000=$49,0000.4(\$55,000) + 0.6(\$45,000) = \$22,000 + \$27,000 = \$49,000. Comparing the two, EMV(Corn) = $58,000 is greater than EMV(Soybeans) = $49,000. Therefore, the farmer should plant Corn for an expected profit of $58,000.

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