Question

A city in Ohio is considering replacing its fleet of gasoline powered cars with electric cars. The manufacturer of the electric cars claims that this municipality will experience significant cost savings over the life of the fleet if it chooses to pursue this conversion. If the manufacturer is correct, the city will save about $1.5 million. If the new technology employed within the electric cars is faulty as some critics suggest, it will cost the city $675,000. A third possibility is that less serious problems will arise and the city will break even in the conversion. A consultant hired by the city estimates the probabilities of these three outcomes are 0.30, 0.30, and 0.40, respectively. The city has an opportunity to implement a pilot program that would indicate the potential cost or savings resulting from the switch to electric cars. The pilot program involves renting a small number of electric cars for three months and running them under typical conditions. This program would cost the city $75,000. The city’s consultant believes that the results of the pilot program would be significant but not conclusive. She submits the following compilation of probabilities based on the experience of other cities to support her contention.

Savings	Loss		Breakeven
Indicates Saving	0.6	0.1	0.3
Indicates Loss	0.1	0.4	0.5
Indicates Breakeven	0.4	0.2	0.4

For example, the first column of her table indicates that given that the conversion to electric cars actually results in a savings, the conditional probabilities that the pilot program will indicate that the city saves money, loses money, and breaks even are 0.6, 0.1, and 0.3. What actions should the city take to maximize its expected savings?

Solve with Decision Tree clearly in Excel

Answer 1

We have defined the following abbreviations:

IS: Indicates Saving
IL: Indicates Loss
IB: Indicates Breakeven

S: Actual Savings
L: Actual Loss
B: Actual Breakeven

IS
Priors		Conditionals		Joint		Posteriors
P(S)	0.3	P(IS | S)	0.60	P(S) x P(IS | S)	0.18	P(S | IS)	0.49
P(L)	0.3	P(IS | L)	0.10	P(L) x P(IS | L)	0.03	P(L | IS)	0.08
P(B)	0.4	P(IS | B)	0.40	P(B) x P(IS | B)	0.16	P(B | IS)	0.43
Total	1			P(IS) =	0.37		1.00
IL
Priors		Conditionals		Joint		Posteriors
P(H)	0.3	P(IL | S)	0.10	P(S) x P(IL | S)	0.03	P(S | IL)	0.13
P(M)	0.3	P(IL | L)	0.40	P(L) x P(IL | L)	0.12	P(L | IL)	0.52
P(L)	0.4	P(IL | B)	0.20	P(B) x P(IL | B)	0.08	P(B | IL)	0.35
Total	1			P(IL) =	0.23		1.00
IB
Priors		Conditionals		Joint		Posteriors
P(H)	0.3	P(IB | S)	0.30	P(S) x P(IB | S)	0.09	P(S | IB)	0.23
P(M)	0.3	P(IB | L)	0.50	P(L) x P(IB | L)	0.15	P(L | IB)	0.37
P(L)	0.4	P(IB | B)	0.40	P(B) x P(IB | B)	0.16	P(B | IB)	0.40
Total	1			P(IB) =	0.40		1.00

Decision Tree

Based on the decision tree output, it is obvious that the pilot is pretty much useless and it is better to go with the conversion without the pilot.