BE1401 Business Operations and Processes
Answer Key for Exercises on Little’s Law
1. Responding to emails:
The flow unit is an email message, and the process boundary is from the arrival of
message into your inbox until the message is responded to and archived in
another folder or deleted. The average flow rate is 50 messages per day and the
average WIP is 150 messages. Hence, the average flow time is 150/50 = 3 days.
2. Selling your flat:
The flow unit is a flat for sale on the market, and the process boundary is from the
listing of the flat for sale until it is sold to its next owner. The average WIP is 25
flats and the average flow rate is 75 flats per year. Hence, the average flow time is
25/75 = 1/3 year or about 122 days, which means the agent’s average flow time
of 100 days is understated.
3. Accounts receivable:
The flow unit is one dollar of accounts receivable, and the process boundary is
from the moment the customer is billed for an equipment sale until the payment
is received. The average flow rate is $300 million per year and the average WIP is
$45 million. Hence, the average flow time is 45/300 = 0.15 year or about 55 days.
4. Lone Star Insurance Co:
The flow unit is an insurance claim, and the process boundary is from the filing of
the claim until it is completely processed and a final decision has been made on
the claim.
a. The average flow rate is 10,000 claims per quarter and the average flow time is
2 quarters. Hence, the average WIP is 10,000 x 2 = 20,000 pending claims.
b. The total workload is 10,000 claims per quarter multiplied by 4 hours per
claim, which gives 40,000 hours per quarter. Each rep can work 400 hours per
quarter. Hence, the required number of reps is 40,000/400 = 100 reps.
Furthermore, the average caseload is 20,000 pending claims to be distributed
among the 100 reps, which results in 200 claims per rep.
5. Public transport bus company:
a. The flow unit is a dedicated bus, and the process boundary is from the start of one
loop until the rest period is over. Here, the average flow rate is the reciprocal of the
frequency, i.e. 1/frequency. The average flow time would be the journey time plus
the rest time. The table below summarizes the computations.
Service No. Buses required during peak
Buses required during non-peak
201
45/10 = 5
60/15 = 4
202
35/10 = 4
50/15 = 4
203
25/10 = 3
45/15 = 3
204
55/10 = 6
70/15 = 5
Total buses
required
18
16
b. The flow unit here is one generic bus and its process boundary is loop 201,
then rest, followed by loop 202, then rest, and so on until the rest after loop
204. For peak period, the average flow time is 45 + 35 + 25 + 55 = 160 minutes
while the average flow rate is still 1 bus per 10 minutes. This gives a WIP of
160/10 = 16 buses. For non-peak period, the average flow time is 60 + 50 + 45
+ 70 = 225 minutes, while the average flow rate is still 1 bus per 15 minutes.
This gives a WIP of 225/15 = 15 buses. In the dedicated scenario, the fleet size
is 18 while here the fleet is 16, giving us a savings of 2 buses. This is one of the
benefits of using a “pooling strategy”.