Consider a street hot dog vendor that has a demand that follows a Poisson distribution with mean of 0.75 customer/minute while the time to serve customer follows a negative exponential distribution with the mean of 1 min. 1. Utilization of the vendor 2. The probability that a new customer arrival has to wait 3. The probability that there is no customer in the system 4. What is the probability that the system will have 5 or more customers? 5. Average number of customers in the system 6. Average number of customers in the waiting line 7. Average time a customer spends in system 8. Average waiting time of a typical customer in line 9. If service time becomes an constant of 1 min, what is the new “Average waiting time of a typical customer in line”? (M/D/1)
Consider the street hot dog vendor (M/M1 example in slide 31)
•
= 0.75 customer/min
•
= 1 customer/min
•
Customer arrival rate now has increased from 0.75 customer/min to 2.2
customers/min. Answer below questions:
A.
What will be the new system utilization
B.
What is the minimum number of new helpers he should hire to manage this new
demand?
C.
On average, how much time a customer spends in the queue if the vendor hires 2 extra
help?
I ONLY NEED HELP WITH PART 2 A,B,C
street hot dog vendor 

Mean arrival rate, ? 
2.2 
customers/min 
Mean service rate, ? 
1 
customers/min 
A. 

The new system utilization, ?=?/? 

220% 

B. 

the minimum number of new helpers needed will make the utilization as 100% (or 1) 

1 = ?/(m?) 

1 = 2.2/(m*1) 

Solve for m: 

m 
2.20 
helpers 
= 
3 
(rounding off to the next whole number) 
C. time a customer spends in the queue if the vendor hires 2 extra help : 

Time in the queue, Wq=?/(?*(??)) 

here, ? = 2* 1 = 2 customers/hour 

i.e. 
?<? 

hence, this form of queue is not feasible, thus the average time customer spends in the queue cannot be calculated. 
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