Quantum Mathematics

Running head: QUANTUM MATHEMATICS 1
Quantum Mathematics
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QUANTUM MATHEMATICS 2
Question 1
Let
The arrival rate be λ
Service rate be μ
Lqis the average number of customers in waiting line
W is the average time a customer will spend in the queue system
W q is the average waiting time in the line per customer
We are given the following data:
λ = 6 customers per hour
μ = 10 customers per hour
Average number of customers in waiting line
Average length, Lq = λ2
μ (μ− λ) = 62
10(10−6) = 0.9
On average, there are 0.9 cars in the queue
Average time a car spend in total queue system
W = 1
( μ−λ) = 1
(10−6) = 0.25 hours
Therefore, the average waiting time in the entire system per car is 0.25 hours
Individual car waiting time in line
Individual car waiting time, W q = λ
μ (μ− λ) = 6
10(10−6) = 0.15 hours
Therefore, the individual car waiting time in line is 0.15 hours
Comment:
If the arrival rate increases to 12 cars per hour an infinite line would be formed which is higher
than the normal service rate. If the rate of competing is a task increases, the corresponding time

QUANTUM MATHEMATICS 3
for completing the task decreases significantly (Ocampo, Paycha & Reyes, 2001; Wrede, Spiegel
& Arangno, 2002).
Question 2
λ = 12 per hour
μ = 60
4 = 15 per hour
For one window;
W q = λ
μ (μ− λ) = 12
15(3−60) = 0.26 hours ≅ 16 minutes
For two windows,
μ remain constant (μ=15 per hour ) but the arrival time for each window is now split
λ = 6 per hour
W q = λ
μ (μ− λ) = 6
15(9) = 0.044 hours ≅ 2.67 minutes
The difference in waiting time for λ = 6 and λ = 15 is calculated as:
Difference = 16-2.67 = 13.33 minutes hence a reduction of waiting time to 13.33 minutes from
16 minutes after changing arrival time.
At t=13.33 minutes, the estimated bank revenue is approximated as:
Bank revenue = 13.33 * $2000 = $26660.
The cost of window = $20,000
Since $26660> $20000, then a second drive in would increase banks revenue therefore a second
drive-in window should be installed.
Question 3
λ = 200 per day
μ = 220 per day
a)