TM 732 Engr. Economics for Managers Decision Analysis

GoferBroke

Prototype Ex. 2 Digger Construction is interested in purchasing 1 of 3 cranes. The cranes differ in capacity, age, and mechanical condition, but each is fully capable of performing the jobs expected. The firm anticipates a growing market and that there will be sufficient demand to justify each of the cranes. However, low, medium, and high growth estimates result in different cash flow profiles for each crane. Based on ATCF at 15%, the analyst estimates the following NPWs for each of the cranes for each of the growth market conditions.

Digger Construction

Decision Matrix EUAW

Matrix Decision Model A j = alternative strategy j under decision makers control S k = a state or possible future that can occur given A j p k = the probability state S k will occur

Matrix Decision Model V(  jk ) = the value of outcome  jk (terms of $, time, distance,.. )  jk = the outcome of choosing A j and having state S k occur

Decisions Under Certainty

Investor wishes to invest $10,000 in one of five govt. securities. Effective yields are: A 1 = 8.0% A 2 = 7.3% A 3 = 8.7% A 4 = 6.0% A 5 = 6.5% choose A 3.

Maximin Select A j : max j min k V(  jk ) e.g., Find the min payoff for each alternative.

Maximin Select A j : max j min k V(  jk ) e.g., Find the min payoff for each alternative. Find the maximum of minimums Select Crane 1 Choose best alternative when comparing worst possible outcomes for each alternative.

Maximin Select A j : max j min k V(  jk ) e.g., Find the min payoff for each alternative. Find the maximum of minimums Sell Land Choose best alternative when comparing worst possible outcomes for each alternative.

MiniMax Select A j : max j min k V(  jk ) e.g., Find the max payoff for each alternative.

MiniMax Select A j : max j min k V(  jk ) e.g., Find the max payoff for each alternative. Find the minimum of maximums Select Crane 1 Choose worst alternative when comparing best possible outcomes for each alternative.

MiniMax Select A j : max j min k V(  jk ) e.g., Find the max payoff for each alternative. Find the minimum of maximums Sell Land Choose worst alternative when comparing best possible outcomes for each alternative.

Class Problem Choose best alternative using a.Maximax criteria b.Minimin criteria

Class Problem Choose best alternative using a.Maximax criteria (best of the best) max j {15163, 16536, 18397} = 18,397 choose A 3

Class Problem Choose best alternative using a.Minimin criteria (worst of the worst) min j {11,962 10,934 10,840} = 10,840 choose A 3

Maximum Likelihood Assume S 2 a certainty

Maximum Likelihood Assume S 2 a certainty max{P A1, P A2, P A3 | p 2 =1.0} choose A 1

Most Probable Assume S 2 a certainty max{P A1, P A2, P A3 | p 2 =1.0} choose A 1

Most Probable Assume S 2 a certainty max{P A1, P A2, P A3 | p 2 =1.0} choose A 1

Most Probable Assume S 2 a certainty max{P A1, P A2, P A3 | p 2 =1.0} choose A 1

Most Probable Assume S 2 a certainty max{P A1, P A2, P A3 | p 2 =1.0} choose A 1

Most Probable Assume S 2 a certainty max{P A1, P A2, P A3 | p 2 =1.0} choose A 1

Most Probable Assume S 2 a certainty max{P A1, P A2, P A3 | p 2 =1.0} choose A 1

Most Probable Assume S 2 a certainty max{P A1, P A2, P A3 | p 2 =1.0} choose A 1

Most Probable Assume S 2 a certainty max{P A1, P A2, P A3 | p 2 =1.0} choose A 1

Assume S 2 a certainty max{P A1, P A2 | p 2 =1.0} choose A 2 Maximun Likelihood Most Probable

Bayes’ Decision Rule E[A 1 ] > E[A 2 ] > E[A 3 ] choose A 1

Bayes’ Decision Rule E[A 1 ] > E[A 2 ] choose A 1

Expectation E[A 1 ] > E[A 2 ] > E[A 3 ] choose A 1

Expectation E[A 1 ] > E[A 2 ] > E[A 3 ] choose A 1

Expectation E[A 1 ] > E[A 2 ] > E[A 3 ] choose A 1

Expectation E[A 1 ] > E[A 2 ] > E[A 3 ] choose A 1

Expectation E[A 1 ] > E[A 2 ] > E[A 3 ] choose A 1

Expectation E[A 1 ] > E[A 2 ] > E[A 3 ] choose A 1

Laplace Principle If one can not assign probabilities, assume each state equally probable. Max E[P Ai ] choose A 1

Expectation-Variance If E[A 1 ] = E[A 2 ] = E[A 3 ] choose A j with min. variance

Sensitivity Suppose probability of finding oil (p) is somewhere between 15 and 35 percent.

Sensitivity Suppose probability of finding oil (p) is somewhere between 15 and 35 percent.

Sensitivity Suppose probability of finding oil (p) is somewhere between 15 and 35 percent.

Sensitivity

Sensitivity Plot Prob. of Oil Expected Value Drill Sell

Sensitivity We know E[payoff] = 700(p) -100(1-p) = 800p - 100

Sensitivity Sensitivity Plot Prob. of Oil Expected Value Drill Sell

Aspiration-Level Aspiration: max probability that payoff > 60,000 P{P A1 > 60,000} = 0.8 P{P A2 > 60,000} = 0.3 P{P A3 > 60,000} = 0.3 Choose A 2 or A 3

Aspiration-Level Aspiration: max probability that payoff > 60,000 P{P A1 > 60,000} = 0.8 P{P A2 > 60,000} = 0.3 P{P A3 > 60,000} = 0.3 Choose A 2 or A 3

Class Problem Determine alternative A j if aspiration level is NPW > $14,000.

Class Problem Determine alternative A j if aspiration level is Payoff > $14,000.

Class Problem Determine alternative A j if aspiration level is Payoff > $14,000. P{P A1 > 14,000} = 0.1 P{P A2 > 14,000} = 0.1 P{P A3 > 14,000} = 0.4Choose A 3

Hurwicz Principle  = 0.3 Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk ) max{12,922 12,615 13,107} = 13,107 choose A 3

Hurwicz Principle  = 0.3 Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk ) max{12,922 12,615 13,107} = 13,107 choose A 3

Hurwicz Principle  = 0.3 Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk ) max{12,922 12,615 13,107} = 13,107 choose A 3

Hurwicz Principle  = 0.3 Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk ) max{12,922 12,615 13,107} = 13,107 choose A 3

Hurwicz Principle  = 0.3 Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk ) max{12,922 12,615 13,107} = 13,107 choose A 3

Hurwicz Principle  = 0.3 Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk ) max{12,922 12,615 13,107} = 13,107 choose A 3

Hurwicz Principle  = 0.3 Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk ) max{12,922 12,615 13,107} = 13,107 choose A 3

Hurwicz Principle  = 0.3 Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk ) Note:   = 1.0 MaxiMax  = 0.0 MaxiMin

Hurwicz Principle  = 1.0 MaxiMax = best of the best = max{max k V(  jk )} max{15,163 16,536 18,397} = 18,397 choose A 3

Hurwicz Principle  = 0.0 MaxiMin = best of the worst = max{min k V(  jk )} max{11,962 10,934 10,840} = 11,962 choose A 1

Class Problem You personally assess your boss’s risk level  to be approximately.3. Use Hurwicz’s principle to analyze the value matrix and determine the appropriate alternative.

Hurwicz Principle Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk )

Hurwicz Principle Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk ) max{50500, 48400, 45000} = 50,500

Hurwicz Principle Select j: max j {H j =  max k [V(  jk )]+(1-  )min k (V(  jk ) max{50500, 48400, 45000} = 50,500 choose A 1

Savage Principle (Minimax Regret) Build table of regrets: R jk = max j [V(  jk )] - V(  jk ) (max in each column less cell value)

Savage Principle (Minimax Regret)

Minimize the maximum regret Min {3,234 1,861 1,122} = 1,122 choose A 3

Class Problem Being somewhat of a pessimist, you constantly worry about lost opportunities. Compute a regret matrix and determine an alternative which minimizes the maximum regret.