Question

In past presidential elections in the United States, 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 NYC as well, some voters complained about the long lines in some precincts, with most complaints coming from precinct A. In 2016, the average number of voters arriving at Precinct A was 27 per hour and the arrival of voters was random. Suppose service times were also random and that each voter spent on average 5 minutes in the voting booth (this is the time needed to cast her/his vote using a voting machine). what would be the utilization rate with the minimum number of voting machines to barely satisfy the demand

Answer 1

Answer:

The utilization rate with the minimum number of voters = 0.75

Step-by-step explanation:

Rate of arrival of voters = 27 voters per hour

If each voter spends 5 minutes in the voting booth,

In 1 hour, number of voters = 60/5 = 12. i.e. service rate = 12 voters per hour

Utilization rate, UR= $\frac{Arrival rate}{service rate * number of voters}$...............(1)

Maximum value for UR = 1. i.e UR ≤ 1

$\frac{Arrival rate}{service rate * number of voters} \leq 1\\\frac{27}{12* number of voters} \leq 1\\27 \leq 12* number of voters\\2.25 \leq number of voters\\number of voters \geq 2.25$

number of voters cannot take a decimal value, therefore,

The minimum number of voters = 3

Substituting number of voters into equation (1)

$UR = \frac{27}{12 * 3} \\UR = 0.75$