quiz 2: problem (a)
DESCRIPTION
Quiz 2: Problem (a). Recommendation: Drill for oil. Quiz 2: Problem (b). Let p be the probability of oil. EMV(Drill) = 700 p – 100 (1-p) EMV(Sell) = 90 Sell when 700 p – 100 (1-p) < 90 Sell when p < 0.2375 Drill when p 0.2375. Quiz 2: Problems (c), (d), and (e). Random Events - PowerPoint PPT PresentationTRANSCRIPT
Quiz 2: Problem (a)
Alternative Oil Dry EMVDrill for oil $700,000 -$100,000 $100,000Sell the land $90,000 $90,000 $90,000Probability 0.25 0.75
Payoff
Recommendation: Drill for oil.
Quiz 2: Problem (b)
• Let p be the probability of oil.
• EMV(Drill) = 700 p – 100 (1-p)
• EMV(Sell) = 90
• Sell when 700 p – 100 (1-p) < 90Sell when p < 0.2375
Drill when p 0.2375
Quiz 2: Problems (c), (d), and (e)
• Random Events– Oil = the land contains oil
– Dry = the land is dry
– FS = favorable sounding
– US = unfavorable sounding
– P(FS|Oil) = 0.6, P(US|Oil) = 0.4
– P(US|Dry) = 0.8, P(FS|Dry) = 0.2
Quiz 2: Problems (c), (d) and (e)
0.25
Dry
0.8
0.2
0.6
P(Oil and FS) = 0.15
P(Oil and US) = 0.10
P(Dry and FS) = 0.15
P(Dry and US) = 0.60
0.75
FS
0.4US
FS
US
Oil
P(FS) = 0.15 + 0.15 = 0.3P(US) = 0.10 + 0.60 = 0.7
P(Oil|FS) = 0.15/0.3 = 0.5P(Dry|US) = 0.6/0.7 = 0.86
Decision Tree for Problem f
1
FS (0.3)
US (0.7)
2Oil (0.5)
Dry (0.5)Drill
Sell
800-100-30=670
-130
60
3Oil (0.14)
Dry (0.86)Drill
Sell
670
-130
EMV(2) = (0.5)(670-130) = 270
EMV(3) = (0.14)(670) + (0.86)(-130) = -18
60
EMV(1) = (0.7)(60) + (0.3)(270) = 123
Optimal Policy
• EMV(No Sounding) = $100,000
• EMV(w/Sounding) = $123,000
• Conclusion:– Pay for the sounding
• If favorable drill
• If unfavorable sell
• EMV = $123,000