An insurance company has one adjuster in the branch office. People with claims against the company are found to arrive in a Poisson fashion during an 8 to 5 workday. Determine the hourly service and arrival rates using an 8-hour workday. The amount of time that the adjuster spends with a claimant is exponentially distributed. Claimants are processed in the order of their arrival.
You are the manager of this branch and you wanted to investigate service provided by your branch to claimants using waiting lines performance measures. You asked your adjuster to collect data on (1) time when clients arrived, and (2) how much time he (she) spent with each client for one typical 8-hour workday. Below are the data collected:
Customer Arrival Time |
8:00 |
8:26 |
9:31 |
10:07 |
10:26 |
10:29 |
10:41 |
10:47 |
11:21 |
1:24 |
1:25 |
1:48 |
2:00 |
2:05 |
2:21 |
3:10 |
3:30 |
3:50 |
4:05 |
4:30 |
Length of Time with Claimant (Minutes) |
13 |
13 |
12 |
10 |
12 |
13 |
18 |
24 |
20 |
12 |
14 |
21 |
3 |
14 |
15 |
10 |
24 |
13 |
14 |
15 |
Answer the following questions, assuming this observation day is very typical of the adjuster’s 8 to 5 workday. Assume also that the adjuster takes an hour-long lunch break when (s)he is idle so that there is no interruption on the service time of any customer arriving.
Dear student, according to our guidelines, we can answer only first four subparts at a time. Kindly, please post the remaining questions separately. Thank you for understanding.
1. Number of customers arriving in a 8 hours day = 20
average arrival rate lambda = 20/8 =2.5 per hour
Average service rate mu = AVERAGE of given length of time= 14.5 min that is 60/14.5 per hour =4.14 per hour
Utilisation rho = lambda /mu = 2.5 /4.14 =0.6039 = 60.39%
2. Idle time for adjuster in minutes = 480 x ( 1-2.5 /4.04) = 190.13 minutes
3. Probability of having no customer in system P(0) = 1-rho =1-0.6039 = 0.3961
4. Lq ( number of customers waiting = (lambda)2 / mu ( mu-lambda)
= 2.5x2.5 / 4.14 x(4.14-2.5)
= 0.921
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