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n the previous two US presidential elections, very long wait times have been witnessed at precincts...

n the previous two US presidential elections, very long wait times have been witnessed at precincts (voting stations) in states that ultimately decided the election (Florida in 2000 and Ohio in 2004). In Iowa City as well, some voters complained about the long lines in some precincts, with most complaints coming from precinct A. In 2018, the average number of voters arriving at Precinct A was 35 per hour and the arrivals of voters was random with inter-arrival times that had a coefficient of variation of 1 (CVa=1). Iowa City had deployed one voting machine in Precinct A. Suppose that each voter spent on average 100 seconds in the voting booth (this is the time needed to cast a vote using a voting machine), with a standard deviation of 120 seconds.

A. Given the long wait times for Precinct A this 2018 election, City is thinking of alternative solutions to improve voting conditions. In preparation for the 2020 election, one of the proposed solutions is to replace the old voting machine in Precinct A two new high tech. voting machines. (With two machines, the precinct will have two voting booths. ) Assume that voter turnout is expected to have similar characteristics in 2020 as in the previous election, except casting a vote using a high tech machine will take exactly 100 seconds. Under this proposal, how long on average would a voter have to wait in line in Precinct A before casting a vote.

a. Less than half a minute
b. More than a minute, but less than 2 minutes c. Exactly 6 minutes
d. Exactly 10 min
e. Exactly 11 min

B. Under the proposed solution, what would be the ratio of the average number of people casting their votes over the average number of people waiting in line?

a. Below 2
b. Between 2 and 3

c. Between 3 and 5

d. Between 5 and 6

e. Above 6

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Answer #1

Issues:

delay, long wait times;

Arrival rate = Lambda = 35 per hour

Service time = 100 seconds (Standard Deviation = 120 seconds)

Servce rate = Miu = 3600/100 = 36 customers per hour

Utilization factor = Chi = Lambda / Miu = 35 / 36 = 0.97

( Chi < 1 )

Ls = Chi / (1-Chi) = number of customers in the system (both in Queue and being served)

Lq = Ls – Chi = number of customers waiting in the Queue

Ws = Ls / Lambda = Average time spent by a customer in the system (both Queue and being served)

Wq = Lq / Lambda = Average time spent by a customer just in the Queue

Ls = Chi / (1-Chi) = 0.97/(1-0.97)= 32.33

Lq = Ls – Chi = 32.33 – 0.97 = 31.36

Ws = Ls / Lambda = 32.33 / 35 = 0.923714 ; 0.923714 * 60 minutes = 55.42 minutes

Wq = Lq / Lambda = (31.36 / 35 ) * 60 = 53.76 minutes

with an extra machine, C = 2 ( there are 2 server) the model of the ques vecomes Ma/Ms/C:GD/Nc/Ncs

Where

Ma = Arrival rate

Ms = Service Rate

C = number of servers

GD = General discipline

Nc = Number of Customers allowed in the system

Ncs = Number of Calling Source?

The only change from the previous part of the question is C has become 2 from 1.

Hence C = 2

There is a small change in the formula:

Lq = chi ^ C+1 / ((c-1)!*(c-chi)^2

Lq = 0.97 ^ 3 / ( ( 1!) * (2-0.97)^2 = 0.86

Ls = Lq + chi = 0.86 + 0.97 = 1.83

Wq = Lq / Lambda = 0.86/35 = 0.02457; in min = 0.02457*60 = 1.4742 minutes = 1.5 minutes

Ws = Wq + 1 / Miu = 1.5 + 1/36 = 1.53 minutes = 2 minutes approximately

Huge improvement of just 2 minutes of waiting time instead of almost an hour at 53.76 minutes

Hence during the next president election they will wait just for 2 minutes

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