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Readings. Readings. Chapter 11, Sections 1, 2, 3, 5 Waiting Line Models. Overview. Overview. Overview. Overview. Tool Summary Use analytical formulas or Management Scientist to compute performance: Probability that no units are in the system: P 0

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• Chapter 11, Sections 1, 2, 3, 5

• Waiting Line Models

• Overview

• Tool Summary

• Use analytical formulas or Management Scientist to compute performance:

• Probability that no units are in the system: P0

• Average number of units in waiting line: Lq

• Average number of units in system: L = Lq + l/m

• Average time a unit spends in waiting line: Wq =Lq/l

• Average time a unit spends in the system: W = Wq + 1/m

• Probability that an arriving unit has to wait for service: Pw

• Probability of n units in the system: Pn

• Compute total hourly cost for units in the system

• = (\$ waiting cost per hour) x (Average number of units in system)

• Note average number of units in system is the only right choice above; average time in waiting line does not count the number of units.

• Analytical Formulas for Multiple Channels

Overview

Analytical Formulas for Multiple Channels, forM/M/k under FCFS require assumptions, some of which are not 100% realistic:

• Multiple channels (with one central waiting line)

• Poisson arrival-rate distribution

• Exponential service-time distribution

• Unlimited maximum queue (waiting line) length

• Examples:

• Four-teller transaction counter in bank

• Two-clerk returns counter in retail store

• Probability that no units are in the system:

• Average number of units in waiting line:

• Average number of units in system: L = Lq +l/m

• Average time a unit spends in waiting line: Wq =Lq/l

• Average time a unit spends in the system: W = Wq + 1/m

6) Probability that an arriving unit has to wait for service: Pw = (1/k!)(l/m)k (km/(km-l)) P0

Probability of n units in the system:

[ (l/m)n /n! ] P0 for n < k

Pn =

[ (l/m)n /(k!k(n-k)) ] P0 for n > k

• M/M/2 Queuing System

Overview

M/M/2 Queuing System designates M = Markov (memoryless) arrival distribution (exponential), M = Markovservice-time distribution (Poisson), and 2 service channels, under first-come, first-served.

Question: Smith, Jones, Johnson, and Thomas, Inc. has begun a major advertising campaign which it believes will increase its business 50%. To handle the increased volume, the company has hired an additional floor trader, Fred Hanson, who works at the same speed as Joe Ferris.

Note that the new arrival rate of orders, l, is 50% higher than that of Example 1 in Lesson 2.5. Thus, l= 1.5(20) = 30 per hour.

M/M/1:

P0 = 1-l/m

Lq = l2/(m(m-l))

L = Lq +l/m

Wq =Lq/l

W = 1/(m-l)

Pw = l/m

Pn = (l/m)nP0

• Sufficient Service Rate

• Will Joe Ferris alone be able to handle the increase in orders?

• Answer: Since Joe Ferris processes orders at a mean rate of µ = 30 per hour, then  = µ = 30 and the average time a unit spends in the system is W = 1/(m-l) = 1/0 = infinity.

That implies the queue of orders will grow infinitely large. Hence, Joe alone cannot handle that increase in demand.

• Probability of n Units in System

• What is the probability that neither Joe nor Fred will be working on an order at any point in time?

• Answer: This is an M/M/k queue with  = 30 per hour,  = 30 per hour, and k = 2. The probability that neither Joe nor Fred will be working = the probability of no units in the system. Analytical Formula #1 says that is:

= 1/[(1 + (1/1!)(30/30)1] + [(1/2!)(1)2][2(30)/(2(30)-30)]

= 1/(1 + 1 + 1) = 1/3 = .333

• Average Time in System

• What is the average turnaround time for an order with both Joe and Fred working?

• Answer: The average turnaround time = the average time a unit spends in the system, W. Analytical Formula #2 and 3 say

and L = Lq + ( /µ) = 1/3 + (30/30) = 4/3. Finally,

W = L/(4/3)/30 = 4/90 hr. = 2.67 min.

• Average length of queue

• What is the average number of orders waiting to be filled with both Joe and Fred working?

• Answer: The average number of orders waiting to be filled = the average number of units in the waiting line, Lq. That was calculated earlier as 1/3.

• Economic Analysis

Overview

Economic Analysis of waiting lines maximizes profit for a firm by maximizing value for customers. Maximizing customer value trades off quick service with low purchase prices (resulting from the lower costs of having fewer or less qualified employees). Wealthy customers (like at Malibu Yogurt) prefer quick service, even if that means higher purchase prices to pay for more or better employees, but poor customers (like at Popeye's Chicken in Oxnard) prefer lower purchase prices, even if that means slower service from fewer or incompetent employees.

Question: The advertising campaign of Smith, Jones, Johnson and Thomas, Inc. was so successful that business doubled. The mean rate of stock orders arriving at the exchange is now 40 per hour and the company must decide how many floor traders to employ. Each floor trader hired can process an order in an average time of 2 minutes. (So far,  = 40/hr. and m = 30/hr.)

The brokerage firm has determined the average waiting cost per minute for an order to be \$.50. (So, you can charge \$.50 more per order if you can process it an average of 1 minute faster.) Floor traders hired will earn \$20 per hour in wages and benefits. Hence, compare the total hourly cost of hiring 2 traders with that of hiring 3 traders.

= (Total salary cost per hour)

+ (Total hourly cost for orders in the system)

+ (\$30 waiting cost per hour) x (Average number of orders in system)

= 20k + 30L.

 = 40/hr.

m = 30/hr.

Cost = 20k + 30L

• This is an M/M/2 queue with  = 40 per hour and  = 30 per hour. Analytical Formulae #1, 2, 3

P0= 1 / [1+(1/1!)(40/30)]+[(1/2!)(40/30)2(60/(60-40))]

= 1 / [1 + (4/3) + (8/3)] = 1/5

say the average number of units in the system is:

L = Lq + ( /µ) = 16/15 + 4/3 = 2.40

Hence, total cost = (20)(2) + 30(2.40) = \$112.00 per hour

 = 40/hr.

m = 30/hr.

Cost = 20k + 30L

• This is an M/M/3 queue with  = 40 per hour and  = 30 per hour. Analytical Formulae #1, 2, 3

P0= 1/[[1+(1/1!)(40/30)+(1/2!)(40/30)2]+

[(1/3!)(40/30)3(90/(90-40))] ]

= 1 / [1 + 4/3 + 8/9 + 32/45] = 15/59

say the average number of units in the system is:

L = .1446 + 40/30 = 1.4780

Hence, total cost = (20)(3) + 30(1.4780) = \$104.35 per hour

• System cost comparison

Wage Waiting Total

Cost/HrCost/HrCost/Hr

Thus, the total cost of having 3 traders is less than that of 2 traders.

• Economic Analysis with Teamwork

Overview

Economic Analysis with Teamwork maximizes firm’s profits and customer’s value by trading off having more service channels (like registers in a grocery) with teams of workers (like a cashier and bagger working at the same register in a grocery) providing faster service in each channel.

• A fast-food franchise is considering adding a drive-up window to a particular location.

• Assume customer arrivals follow a Poisson probability distribution, with an arrival rate of l = 24 cars per hour.

• Assume customer service times follow an exponential distribution.

• Arriving customers place orders at an intercom station at the back of the parking lot and then drive to the service window to pay for and receive their orders.

• The following three service alternatives are being considered.

• System A. One employee fills the order and takes the money from the customer. The average service time for this alternative is 2 minutes (30 customers per hour). All together, (k, l, m) = (1, 24, 30).

• System B. One employee fills the order while a second employee takes the money from the customer. The average service time for this alternative is 1.25 minutes (48 customers per hour). All together, (k, l, m) = (1, 24, 48).

• System C. Two service windows, each with an employee that fills the order and takes the money from the customer. The average service time for this alternative is 2 minutes (30 customers per hour) for each channel. All together, (k, l, m) = (2, 24, 30).

• Customer waiting time is valued at \$25 per hour.

• So, you can charge \$25/60 more per order if you can process orders an average of 1 minute (1/60 hour) faster.

• The cost of each employee is \$6.50 per hour.

• Each channel costs \$20 for equipment and space.

• Which system is most profitable?

• The labor plus equipment-space cost for each channel of each system is:

• The waiting plus channelcost for each each system is:

• System B is thus most profitable (it costs the minimum, \$58.00). That is, one employee fills the order while a second employee takes the money from the customer.

• If, instead, the drive-up window were for a location in a poor part of town, where customer waiting time is valued at \$2 per hour, which system is most profitable?

• Answer: Change the waiting plus channel cost for each system from to reduce time value from \$25 to \$2. System A is now most profitable.

• Finally, if the drive-up window were for outside the colony in Malibu, where customer waiting time is valued at \$200 per hour, which system is most profitable?

• Answer: System B is most profitable, but if the value of time were high enough, then System C would be most profitable.

BA 452 Quantitative Analysis

End of Lesson B.8