a) In a supermarket, the average arrival rate of customers is 10 every 30 minutes, following Poisson process. The average time taken by a cashier to list and calculate the customer’s purchase is two and a half minutes following exponential distribution. What is the probability that the queue length exceeds six? What is the expected time spent by a customer in the system?
b) A project schedule has the following characteristics:
Activity Time (Weeks) Activity Time (Weeks)
1-2 4 5-6 4
1-3 1 5-7 8
2-4 1 6-8 1
3-4 1 7-8 2
3-5 6 8-10 5
4-9 5 9-10 7
1)Compute earliest start time (Es) and latest finish time (Lf) for each event.
2)Compute Total float and critical path.
a) To calculate the probability that the queue length exceeds six, we can use the Erlang-C formula.
The formula for the probability that the queue length exceeds a certain number (n) in an M/M/1 queue system is:
P(n) = (ρ^n / n!) * ((1 - ρ) / (1 - ρ^(c+1)))
where ρ is the traffic intensity or the utilization factor (λ/μ), λ is the arrival rate, and μ is the service rate.
In this case, the arrival rate (λ) is 10 customers every 30 minutes, which can be converted to an arrival rate of 20 customers every hour. The service rate (μ) is 1 customer every 2.5 minutes, which can be converted to a service rate of 24 customers every hour.
ρ = λ / μ = 20/24 = 5/6
Using the formula above with n = 6 and c = 1 (since there is one server in the cashier system), we can calculate the probability that the queue length exceeds six:
P(6) = ((5/6)^6 / 6!) * ((1 - 5/6) / (1 - (5/6)^2))
= (0.3349 / 720) * (1/6)
= 0.0000787
Therefore, the probability that the queue length exceeds six is approximately 0.0000787 or 0.00787%.
To calculate the expected time spent by a customer in the system, we can use Little's Law.
The formula for Little's Law is:
L = λ * W
where L is the long-term average number of customers in the system, λ is the arrival rate, and W is the long-term average time spent by a customer in the system.
In this case, the arrival rate (λ) is 10 customers every 30 minutes, which can be converted to an arrival rate of 20 customers every hour.
L = 20 * W
We need to find the value of W. To calculate W, we can use the formula for the average waiting time in an M/M/1 queue system:
W = 1 / (μ - λ)
In this case, the service rate (μ) is 1 customer every 2.5 minutes, which can be converted to a service rate of 24 customers every hour.
W = 1 / (24 - 20) = 1 / 4 = 0.25 hour
Therefore, the expected time spent by a customer in the system (W) is 0.25 hour or 15 minutes.
b) To calculate the earliest start time (Es) and latest finish time (Lf) for each event in the project schedule, we can use the forward and backward pass method in the Critical Path Method (CPM).
1) Forward Pass:
Starting with activity 1, the earliest start time (Es) is 0 since it is the first activity. The earliest finish time (Ef) for activity 1 is therefore 4 (4 weeks after the project starts).
For activity 2, the earliest start time (Es) is the same as the earliest finish time (Ef) of activity 1 (4), so Es = Ef = 4. The earliest finish time (Ef) for activity 2 is 8.
For activity 3, the earliest start time (Es) is the same as the earliest finish time (Ef) of activity 1 (4), so Es = Ef = 4. The earliest finish time (Ef) for activity 3 is 5.
For activity 4, the earliest start time (Es) is the same as the earliest finish time (Ef) of activity 2 (8), so Es = Ef = 8. The earliest finish time (Ef) for activity 4 is 9.
For activity 5, the earliest start time (Es) is the same as the earliest finish time (Ef) of activity 2 (8), so Es = Ef = 8. The earliest finish time (Ef) for activity 5 is 14.
For activity 6, the earliest start time (Es) is the same as the earliest finish time (Ef) of activity 5 (14), so Es = Ef = 14. The earliest finish time (Ef) for activity 6 is 18.
For activity 7, the earliest start time (Es) is the same as the earliest finish time (Ef) of activity 5 (14), so Es = Ef = 14. The earliest finish time (Ef) for activity 7 is 22.
For activity 8, the earliest start time (Es) is the same as the earliest finish time (Ef) of activity 6 (18), so Es = Ef = 18. The earliest finish time (Ef) for activity 8 is 19.
For activity 9, the earliest start time (Es) is the same as the earliest finish time (Ef) of activity 4 (9), so Es = Ef = 9. The earliest finish time (Ef) for activity 9 is 14.
For activity 10, the earliest start time (Es) is the same as the earliest finish time (Ef) of activity 9 (14), so Es = Ef = 14. The earliest finish time (Ef) for activity 10 is 21.
2) Backward Pass:
Starting with activity 10, the latest finish time (Lf) is the same as the earliest finish time (Ef) of activity 10 (21), so Lf = Ef = 21. The latest start time (Ls) for activity 10 is therefore 14.
For activity 9, the latest finish time (Lf) is the same as the earliest start time (Es) of activity 10 (14), so Lf = Es = 14. The latest start time (Ls) for activity 9 is 7.
For activity 8, the latest finish time (Lf) is the same as the earliest start time (Es) of activity 10 (14), so Lf = Es = 14. The latest start time (Ls) for activity 8 is 13.
For activity 7, the latest finish time (Lf) is the same as the earliest start time (Es) of activity 8 (13), so Lf = Es = 13. The latest start time (Ls) for activity 7 is 5.
For activity 6, the latest finish time (Lf) is the same as the earliest start time (Es) of activity 10 (14), so Lf = Es = 14. The latest start time (Ls) for activity 6 is 7.
For activity 5, the latest finish time (Lf) is the same as the earliest start time (Es) of activity 6 (7), so Lf = Es = 7. The latest start time (Ls) for activity 5 is 2.
For activity 4, the latest finish time (Lf) is the same as the earliest start time (Es) of activity 9 (7), so Lf = Es = 7. The latest start time (Ls) for activity 4 is 4.
For activity 3, the latest finish time (Lf) is the same as the earliest start time (Es) of activity 4 (4), so Lf = Es = 4. The latest start time (Ls) for activity 3 is 3.
For activity 2, the latest finish time (Lf) is the same as the earliest start time (Es) of activities 4 and 5 (4), so Lf = Es = 4. The latest start time (Ls) for activity 2 is 0.
For activity 1, the latest finish time (Lf) is the same as the earliest start time (Es) of activities 2 and 3 (4), so Lf = Es = 4. The latest start time (Ls) for activity 1 is 0.
To compute the total float and critical path, we can subtract the earliest start time (Es) from the latest start time (Ls) for each activity:
Activity | Es | Ls | Total Float | Critical Path
--------------------------------------------------
1-2 | 4 | 0 | 4 | No
1-3 | 4 | 0 | 4 | No
2-4 | 8 | 4 | 4 | No
3-4 | 4 | 3 | 1 | No
3-5 | 4 | 2 | 2 | No
4-9 | 9 | 4 | 5 | No
5-6 | 14 | 7 | 7 | Yes
5-7 | 14 | 5 | 9 | Yes
6-8 | 18 | 7 | 11 | Yes
7-8 | 22 | 13 | 9 | Yes
8-10 | 19 | 14 | 5 | No
9-10 | 14 | 7 | 7 | Yes
From the table, we can see that the critical path is 5-7-8-10, with a total length of 24 weeks. The activities on the critical path are 5-7, 7-8, 8-10.
Activities with a total float of zero are on the critical path.
Therefore, the critical path is 5-7-8-10 and the total length of the critical path is 24 weeks.
W