140520200849
The work speed of three information processing centers carrying out banking transactions was analyzed. Each of the centers has several independent servers processing information. It was found that there is a difference in the work of individual bank servers.
The purpose of the exercise is to indicate the general performance characteristics of the banking information processing system.
The following formulas come from the study: Hillier F.S., Lieberman G.J. Instruction of Operation Research. New York: McGrew Hill 1990 oraz Z. Jędrzejczyk, K.Kukuła, J.Skrzypek, A. Walkosz Badania operacyjne w przykładach i zadaniach, Wydawnictwo Naukowe PWN
Arrival rate: λ
These are transactions arriving within a minute. It is a stream of transaction orders sent from bank customers sent online.
In [1]:
import numpy as np
np.random.seed(148)
Λ1 = (np.random.random_sample(2354)*4690)
Λ2 = (np.random.random_sample(4899)*2319)
Λ3 = (np.random.random_sample(3511)*5491)
a1= sum(Λ1)
p1= len(Λ1)
λ1 = 1/(p1/a1)
a2= sum(Λ2)
p2= len(Λ2)
λ2 = 1/(p2/a2)
a3= sum(Λ3)
p3= len(Λ3)
λ3 = 1/(p3/a3)
print('Number of transactions: ', p1)
print('Sume of transactions time: print('Arrival_λ1: print('-----------------------------------------------')
print('Number of transactions: ', p2)
print('Sume of transactions time: print('Arrival_λ2: print('-----------------------------------------------')
print('Number of transactions: ', p3)
print('Sume of transactions time: print('Arrival_λ3: print('-----------------------------------------------')
Service rate: μ
This parameter represents the execution time of individual banking transactions by the banking transaction system.Each of the three data centers has a number of servers. A single server is treated as an individual information flow channel.
In [2]:
r1 = 5
r2 = 6
r3 = 8
In [3]:
np.random.seed(148)
M1 = (np.random.random_sample(2354)*959)
M2 = (np.random.random_sample(4899)*790)
M3 = (np.random.random_sample(3511)*903)
a1= sum(M1)
p1= len(M1)
μ1 = 1/(p1/a1)
a2= sum(M2)
p2= len(M2)
μ2 = 1/(p2/a2)
a3= sum(M3)
p3= len(M3)
μ3 = 1/(p3/a3)
print('Number of transactions: ', p1)
print('Sume of service time: print('Service_μ: print('-----------------------------------------------')
print('Number of transactions: ', p2)
print('Sume of service time: print('Service_μ: print('-----------------------------------------------')
print('Number of transactions: ', p3)
print('Sume of service time: print('Service_μ: print('-----------------------------------------------')
Traffic intensity: ϱ
Traffic intensity is the ratio of arrival rate to service rate.
$$\text{Traffic intensity equation:} \\
\bbox[#f3f3f3,12px,border:1px solid black] {
{ϱ}=\frac{λ}{{μ}*r}
} \qquad (1)
$$
\bbox[#f3f3f3,12px,border:1px solid black] {
{ϱ}=\frac{λ}{{μ}*r}
} \qquad (1)
$$
In [4]:
ϱ1 = λ1/(μ1*r1)
if ϱ1 >= 1:
print("The system is unstable, we do not carry out further calculations!")
print('Arrival rate λ1 > service rate μ1, ϱ1= else:
print("The system is stable, you can calculate the probabilities!")
print('Arrival rate λ1 > service rate μ1')
print('Arrival rate λ1: print('Service rate μ1: print('Traffic intensity ϱ1: ϱ2 = λ2/(μ2*r2)
if ϱ2 >= 1:
print("The system is unstable, we do not carry out further calculations!")
print('Arrival rate λ2 > service rate μ2 ,ϱ2= else:
print("The system is stable, you can calculate the probabilities!")
print('Arrival rate λ2 > service rate μ2')
print('Arrival rate λ2: print('Service rate μ2: print('Traffic intensity ϱ2: ϱ3 = λ3/(μ3*r3)
if ϱ3 >= 1:
print("The system is unstable, we do not carry out further calculations!")
print('Arrival rate λ3 > service rate μ3, ϱ3= else:
print("The system is stable, you can calculate the probabilities!")
print('Arrival rate λ3 > service rate μ3')
print('Arrival rate λ3: print('Service rate μ3: print('Traffic intensity ϱ3:
Multiple service channel
We assume in this algorithm that the arrival of transactions (arrival rate) has a normal distribution. Transactions are directed to several servers with the same level of service. To maintain system balance, inequality must be maintained:
$\begin{align}
λ < rμ &&\text{where r is the number of service channels} \tag 5\\
\end{align}$
$$\text{The probability that there are no queues at r service stations:} \\
\bbox[#f3f3f3,12px,border:1px solid black] {
p(n=0) = \displaystyle\frac{1}{sum_{n=1}^\infty}
} \qquad (6)
$$
\bbox[#f3f3f3,12px,border:1px solid black] {
p(n=0) = \displaystyle\frac{1}{sum_{n=1}^\infty}
} \qquad (6)
$$
The probability that there are no queues at r service stations.
$$\text{The probability that there is no queue n = 0 in the system:} \\
\bbox[#f3f3f3,12px,border:1px solid black] {
p(n=0) =\frac{1}{\displaystyle\sum_{i=0}^{r-1} \frac{?^i}{i!}+\frac{?^r}{{\left(r-{?}\right)}{\left(r-1\right)!}}}
} \qquad (7)
$$
\bbox[#f3f3f3,12px,border:1px solid black] {
p(n=0) =\frac{1}{\displaystyle\sum_{i=0}^{r-1} \frac{?^i}{i!}+\frac{?^r}{{\left(r-{?}\right)}{\left(r-1\right)!}}}
} \qquad (7)
$$
$$\text{Average number of items waiting in queue:} \\
\bbox[#f3f3f3,12px,border:1px solid black] {
ψ =\frac{{?^{r+1}{p(n=0)}}}{{(r-?)}^2{(r-1)!}}
} \qquad (8)
$$
\bbox[#f3f3f3,12px,border:1px solid black] {
ψ =\frac{{?^{r+1}{p(n=0)}}}{{(r-?)}^2{(r-1)!}}
} \qquad (8)
$$
$$\text{The probability that there is no queue n = 0 in the system:} \\
\bbox[#f3f3f3,12px,border:1px solid black] {p(n) =\begin{cases}
\frac{{ψ^n}{p(n=0)}}{n!}, & \text{if $n\le r$}
\frac{{r^{r-n}}{ψ^n}{p(n=0)}}{r!}, & \text{if $n\gt r$}
\end{cases}
} \qquad (9)
$$
\bbox[#f3f3f3,12px,border:1px solid black] {p(n) =\begin{cases}
\frac{{ψ^n}{p(n=0)}}{n!}, & \text{if $n\le r$}
\frac{{r^{r-n}}{ψ^n}{p(n=0)}}{r!}, & \text{if $n\gt r$}
\end{cases}
} \qquad (9)
$$
The probability that the system has more than zero elements and elements is more than r channels.
$$\text{The probability that there is no queue n = 0 in the system:} \\
\bbox[#f3f3f3,12px,border:1px solid black] {
p(n>n_0) =\frac{{r^{r-n_{0}}}{?^{n_{0}+1}}{p(n=0)}}{(r-{?})r!}
} \qquad (10)
$$
\bbox[#f3f3f3,12px,border:1px solid black] {
p(n>n_0) =\frac{{r^{r-n_{0}}}{?^{n_{0}+1}}{p(n=0)}}{(r-{?})r!}
} \qquad (10)
$$
$$\text{The probability that the waiting time of an item in the queue is longer n that t.:} \\
\bbox[#f3f3f3,12px,border:1px solid black] {
p(t>t_0) ={{p{(n>r-1)}}{e^{μt(r-ψ)}}}
} \qquad (11)
$$
\bbox[#f3f3f3,12px,border:1px solid black] {
p(t>t_0) ={{p{(n>r-1)}}{e^{μt(r-ψ)}}}
} \qquad (11)
$$
In [5]:
λ=λ1
μ=μ1
r=r1
We check for inequality:
λ<rμ
In [6]:
print('Arrival rate λ: print('Service rate μ: ϱ = λ/(r*μ)
print('Traffic intensity ϱ:
In [7]:
if ϱ > 1:
print("The system is unstable, we do not carry out further calculations!")
print('Arrival rate λ > split service rate μ')
else:
print("The system is stable, you can calculate the probabilities!")
print('Arrival rate λ > split service rate μ')
In [8]:
def factorial(n):
return 1 if (n==1 or n==0) else n * factorial(n - 1);
Since there are two airports, the time to accept the plane is 2*μ
Probability of no queues, pattern no. 7
$${
p(n=0) =\frac{1}{\displaystyle\sum_{i=0}^{r-1} \frac{?^i}{i!}+\frac{?^r}{{\left(r-{?}\right)}{\left(r-1\right)!}}}
}\qquad (7)$$
In [9]:
p7=1/((1+ϱ)+(((ϱ**r)/((r-ϱ)*factorial(r-1)))))
print('Probability of no queues ?(?=0)≈
Probability that you will have to wait in queue, template no. 10¶
$${
p(n>n_0) =\frac{{r^{r-n_{0}}}{?^{n_{0}+1}}{p(n=0)}}{(r-{?})r!}
} \qquad (10)$$
In [10]:
if ϱ > 1:
print("The system is unstable, we do not carry out further calculations!")
print('Arrival rate λ > split service rate μ')
else:
n0 = 1
p10 = ((r**(r-n0))*(?**(n0+1))*p7)/((r-?)*factorial(r))
print('Probability to wait in queue ?(?>?0)≈
Probability of two items in the queue: $p(n>2)$ where $n_0 =2$¶
In [11]:
n0 = 2
In [12]:
if ϱ > 1:
print("The system is unstable, we do not carry out further calculations!")
print('Arrival rate λ > split service rate μ')
else:
n0 = 2
p11 = ((r**(r-n0))*(?**(n0+1))*p7)/((r-?)*factorial(r))
print('Probability of two items in the queue: ?(?>2)≈
The probability that the item will have to queue over k minutes: $p(t>?_0)$¶
$${
p(n>n_0) =frac{{r^{r-n_{0}}}{?^{n_{0}+1}}{p(n=0)}}{(r-{?})r!}
} qquad (10)$$$${
p(t>t_0) ={{p{(n>r-1)}}{e^{μt(r-ψ)}}}
} qquad (11)
$$
In [13]:
## time in minutes
t0=0.05
t0=t0/60
print('n - unspecified number of units in the queue')
print('number of channels r =', r)
$ p(n>r-1) = p(n>1)$
equation ⟹ 10 for $p(n>1)$
In [14]:
n0=1
p12 = ((r**(r-n0))*(?**(n0+1))*p7)/((r-?)*factorial(r))
p12
Out[14]:
In [15]:
# wzór 11
if ϱ > 1:
print("The system is unstable, we do not carry out further calculations!")
print('Arrival rate λ > split service rate μ')
else:
p13 = p12*(np.e**(-μ*t0*(r-?)))
print('The probability that the item will have to queue over n minutes: ?(?>?0)≈
What is the average number of transactions waiting?!¶
${
ψ =\frac{{?^{r+1}{p(n=0)}}}{{(r-?)}^2{(r-1)!}}
} \qquad (8)$
In [16]:
if ϱ > 1:
print("The system is unstable, we do not carry out further calculations!")
print('Arrival rate λ > split service rate μ')
else:
p14 = ((?**(r+1))*p7)/(((r-?)**2)*factorial(r-1))
print('The average number of items waiting ψ ≈
In [17]:
def queueing_test(λ,μ,r):
def factorial(n):
return 1 if (n==1 or n==0) else n * factorial(n - 1);
print("-----------------------------------------------------------------------")
print('Arrival rate λ: print('Service rate μ: ϱ = λ/(μ*r)
print('Traffic intensity ϱ: print("-----------------------------------------------------------------------")
if ϱ > 1:
print("The system is unstable, we do not carry out further calculations!")
print('Arrival rate λ > split service rate μ')
else:
print("The system is stable, you can calculate the probabilities!")
print('Arrival rate λ > split service rate μ')
print("-----------------------------------------------------------------------")
p7=1/((1+ϱ)+(((ϱ**r)/((r-ϱ)*factorial(r-1)))))
print('P7 Probability of no queues ?(?=0)≈ print("-----------------------------------------------------------------------")
n0 = 2
p11 = ((r**(r-n0))*(?**(n0+1))*p7)/((r-?)*factorial(r))
print('P11 Probability of two items in the queue: ?(?>2)≈ print("-----------------------------------------------------------------------")
n0=1
p12 = ((r**(r-n0))*(?**(n0+1))*p7)/((r-?)*factorial(r))
## time in minutes
t1=0.0165 ## 1 SEKUNDA PRZESTÓJ
t1=t1/60
p13A = p12*(np.e**(-μ*t1*(r-?)))
print('P13 The probability that the item will have to queue over 1.0 SEC: ?(?>?0)≈ ## time in minutes
t0=0.05 ## 3 SEKUNDY PRZESTÓJ
t0=t0/60
p13B = p12*(np.e**(-μ*t0*(r-?)))
print('P13 The probability that the item will have to queue over 3.0 SEC: ?(?>?0)≈ ## time in minutes
t2=0.0834 ## 5 SEKUNDA PRZESTÓJ
t2=t2/60
p13C = p12*(np.e**(-μ*t2*(r-?)))
print('P13 The probability that the item will have to queue over 5.0 SEC: ?(?>?0)≈ print("-----------------------------------------------------------------------")
p14 = ((?**(r+1))*p7)/(((r-?)**2)*factorial(r-1))
print('P14 The average number of items waiting ψ ≈ print("-----------------------------------------------------------------------")
First server room
In [18]:
queueing_test(λ1,μ1,r1)
Second server room
In [19]:
queueing_test(λ2,μ2,r2)
Third server room
In [20]:
queueing_test(λ3,μ3,r3)
As you can see above, we tested the performance of three bank information processing centers. The probability of no queue in the first center is p7 = 50% even though the traffic intensity ϱ indicator is there as much as 98%, in the second p = 67% and in the third p7 = 57% even though the traffic intensity ϱ there is less than 0.76 there. The difference is even better between the first and third IT center in the category of probability of waiting time for a transaction for 1 second, 3 and 5 seconds.
