ASSIGNMENT 7 1. The professor starts with the award for maintaining excellent quality so that the students know the importance of six sigma standard which is needed to achieve this award. Moreover, ISO 9000 makes students aware on what’s the minimum level of standard necessary to ensure organizations meet the needs of their customers in terms of quality.

2. Service Encounter Triad captures the relationships between the three parties in a service encounter and suggests possible sources of conflict.

Three parties in the triad are: * Service Organization * Contact Personnel * Customer

3. Different methods companies are resorting to, to shorten waits are given below. A. Animate: Disneyland Distractions; Elevator Mirror; Recorded Music B. Discriminate: Avis Frequent Renter Treatment (out of sight) C. Automate: Use computer scripts to address 75% of questions D. Obfuscate: Disneyland staged waits (e.g. House of Horrors)

4. Yield management is the process of understanding, anticipating and influencing consumer behavior in order to maximize yield or profits from a fixed, perishable resource.
The typical characteristics of service firms that could gainfully use yield management are given below. a. Relatively Fixed Capacity b. Ability to Segment Markets c. Perishable Inventory d. Product Sold in Advance e. Fluctuating Demand f. Low Marginal Sales Costs & High Marginal Capacity Change Costs
Its effectiveness in generating incremental revenues from an existing operation and customer base has made it particularly attractive to business leaders that prefer to generate return from revenue growth and enhanced capability rather than downsizing and cost cutting.

5. The production-line approach suggests that the total job be broken down into groups of simple tasks. Task grouping permits the specialization of labor skills (e.g., not everyone at a restaurant needs to be a cook). Further, the division of labor allows one to pay only for the skill that is required to perform the task.
Features of production-line approach: * Limited Discretionary Action of Personnel * Division of Labor * Substitution of Technology for People * Service Standardization

Customer as Co-producer differs from the above approach because instead of being a passive bystander, the customer represents productive labor just at the moment it is needed, and opportunities exist for increasing productivity by shifting some of the service activities onto the customer (i.e., making the customer a co-producer). Further, customer participation can increase the degree of customization.

6. The Capacity Planning decision involves tradeoff between the cost of providing a service and the cost of inconvenience of customer waiting.

7. M/M/1 Queuing System λ = 15/hr. µ = 60/3 = 20/hr. A. Probability of the system being idle so as to serve an arriving passenger immediately without making him / her wait = 1 – ρ
= 1 – (λ/µ)
= 1 – 15/20
= 0.25 B. The waiting area would be inadequate when there are more than 3 passengers in the line waiting to be served. Hence, the required probability is given by,
P (n>=4) = ρ4 = (λ/µ) 4 = (15/20)4 = 0.3164
So, the waiting area is inadequate 31.64% of time.

8. The alternatives have been evaluated below. A. In this case, both lines are independent M/M/1 queues. The total cost of the system would be double the cost of a single line and the average arrival rate would be half the total i.e., 10/hr. Lq = (ρ * λ / (µ - λ)) = 10*10 / (20*(20-10)) = 0.5
The total cost of the system is given by the sum of hourly cost of service and the hourly cost of waiting.
Thus, TC = Cs*C + Cw*λ*Ws = Cs*C + Cw*Ls = 2*(10 + 15*0.5) = $35/hr. B. In this case, first line is same as the one discussed in the last alternative with Lq=0.5. The second line is an M/G/1 queuing system.
Variance, V(t)=0 Lq = (ρ2 + λ2 * V(t)) / (2*(1 - ρ)) = (0.52 + 102 * 0) / (2 * (1 - 0.5)) = 0.25 TC = Cost of Single Clerk + Cost of ATM
= (10 + 15*0.5) + (0 + 15*0.25)
= $21.25/hr.
Hence, the least expensive arrangement is that of a single clerk and one self-serve ATM with constant service time of 3 minutes

9. M/M/1 Queuing System λ = 2/hr. µ = 60/20 = 3/hr. ρ = λ/µ = 2/3 = 0.67 E. The mean number of customers in the system is given by, Lq = λ / (µ - λ)
= 2 / (3 - 2)
= 2 customers F. The mean time in the system is given by,
Ws = 1 / (µ - λ)
The arrival rate at which an additional mechanic would be needed is given by equating the expression for mean time in the system to 90 minutes, the average time in the system for a customer above which new mechanic would be needed.
Thus, 90/60 = 1 / (3 - λ) λ = 7/3 = 2.33/hr.
So, at an arrival rate of 2.33/hr., a new mechanic would be needed.