The Hindu Business Line : Estimating the cost of processing claims

The analysis suggested that more appropriate cost drivers and behaviour for automobile accident claims are: 0.2 per cent of Astha Insurance policyholders' property claims; + 0.6 per cent of other parties' property claims; and + 0.8 per cent of total personal injury claims

Data from two recent automobile accident claims are presented in Table 1.

Required: Estimate the cost of processing each claim using data from account analysis and then the activity analysis.

How would you recommend that Astha estimate the cost of processing claims?

Answer: i) The cost of processing of insurance claims is presented in Table 2.

ii) When there is no personal injury claim, activity analysis will result into lower cost of processing and, hence, activity analysis is recommended.

When there is personal injury claim, account analysis will result in lower cost of processing and, hence, account analysis is recommended.

Queuing theory

VISWAKARMA Enterprises conducted a study on its maintenance shop and found that the inter arrival times at tool crib were exponential with an average time of 10 minutes.

The length of the service time (amount of the time taken by the tool-crib operator to meet the needs of the maintenance man) is assumed to be exponentially distributed, with mean of six minutes.

Find:

i) The probability that a person arriving at the booth of the shop will have to wait.

ii) The average length of the queue that forms and the average time that an operator spends in the queue system.

iii) The manager of the shop will install a second booth when an arrival would have to wait 10 minutes or more for the service.

By how much must the rate of arrival be increased in order to justify a second booth?

iv) Estimate the fraction of the day that tool crib operator will be idle.

v) The probability that there will be six or more operators waiting for the service.

Answer: The Average arrival rate A, 60/10 = 6

Average service rate S, 60/6 =10

Traffic intensity = R = A/S = 0.6 or 60 per cent

i) The probability that a person arriving will have to wait, that is, the probability that the system is busy = R = 60 per cent

ii) The average length of the queue Lq = R2/1-R = 0.6 x 0.6/1-0.6 = 0.36/0.4 = 0.9

Average time spent in the queue = Lq/A = 0.9/6 x 60 = 9 minutes

iii) Let A be the required rate of arrival for the waiting time in queue to be 10 minutes

Wq = Lq/A' = R2/A'(1-R) = A'/ S(S- A')

Considering the second formula, that is, A'/S (S-A') = 1/6 and simplifying, we get 6A'= S2-SA'

We know service rate S = 10

Hence 6A' = 100 - 10A'

16A'=100

A'=100/16= 6.25

Check: Wq = A'/S (S- A') = 6.25 x 60 / 10 x (10 - 6.25) = 10 minutes

iv) Fraction of the day that the tool crib operator will be idle is nothing but the probability of the system being idle = 1-R = 1- 0.60 = 0.40 or 2/5 of the day

v) Probability that there are six or more persons in the system:

P (n = 6) =R{circ}N = 0.6 {circ} 6 = 4.67 per cent

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