{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

3325p11hs d waiting line

# 667 note that this is equal to 1 note p0 1 33 444 296

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ystem D – 30 Single-Channel Example Probability of more than k Cars in the System Probability k 0 1 2 3 4 5 6 7 Pn > k = (2/3)k + 1 .667 ← Note that this is equal to 1 Note P0 = 1 - .33 .444 .296 .198 ← Implies that there is a 19.8% .198 Implies chance that more than 3 cars are in the system system .132 .088 .058 .039 D – 31 Single-Channel Economics Customer dissatisfaction and lost goodwill Wq Total arrivals Mechanic’s salary Total hours Total customers spend waiting per day waiting = = \$10 per hour = 2/3 hour = 16 per day = \$56 per day 2 2 (16) = 10 hours 3 3 Customer waiting-time cost = \$10 10 2 3 = \$106.67 Total expected costs = \$106.67 + \$56 = \$162.67 D – 32 Multiple-Server Example Queuing Formulas for Model B: Multiple-Server System, also Called M/M/S M = number of servers (channels) open l= average arrival rate µ= average service rate at each server (channel) The probability that there are zero people or units in the system is: 1 P0 = M M −1 n λ ∑ 1 λ ÷ + 1 ÷ M µ n= 0 n ! µ M ! µ M µ − λ for M µ > λ D – 33 Multipl...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online