Answer:
For 10 minutes: E[h1(x)] = $1 and E[h2(x)] = $1.125
For 15 minutes: E[h1(x)] = $1.5 and E[h2(x)] = $1.385
Explanation:
Given:
- Flat rate 1st charge = $ 0.10 / min
- Flat rate 2nd charge = $ 0.99/min 0< t < 20 mins
- Overhead rate for 2nd charge = $0.10 / min
- They follow an exponential distribution λ.
Find:
For each plan
- Which plan is better if expected call duration is 10 minutes?
- Which plan is better if expected call duration is 15 minutes?
Solution:
- The expected duration for flat plans of 10 and 15 mins are:
10 mins: E(h_1(x)) = mins * Rate
E(h_1(x)) = 10 * 0.10 = $1
15 mins: E(h_1(x)) = mins * Rate
E(h_1(x)) = 15 * 0.10 = $ 1.5
- For second plan, you need to work out the probability of a call exceeding 20 minutes. Calls under 20 minutes cost 99 cents. Calls over 20 minutes 100 cents, and then have an expected duration of 10 or 15 minutes for (a) and (b) respectively-- So calls longer than 20 minutes have an expected cost of 99+100 for part (a), and 99+150 for part (b).
- So calls longer than 20 minutes have an expected cost of 99+100 for part (a), and 99+150 for part (b).
- The CDF is the probability that the time will be LESS THAN t, and we will call that probability P.
So for part (a):
P = p(time < 20) = CDF(20) = 1 - e^(-20/10) = 1 - e^(-2) = 0.865
E(h2(x)) = P * .99 + (1-P)(.99 + 1) = 0.865*0.99 + (1 - .865)*(1.99) = $1.125
for part (b):
P = p(time < 20) = CDF(20) = 1 - e^(-20/15) = 1 - e^(-4/3) = 0.736
E(h2(x)) = P * .99 + (1-P)(.99 + 1.5) = 0.736*0.99 + (1 - .736)*(2.49) = $1.385
- Hence, for 10 mins call Plan 1 is better. However, for 15 mins call its plan 2