Recitation 2

Service Engineering
Question 1
Recitation 2
Little's Law and Capacity Analysis
Workforce Management: A certain Japanese company has 36,000 employees, 20% of
whom are women. The average term of employment for a woman is 37 months, whereas
for men it is 200 months. Assume that the total employment level and the mix of men and
women are relatively stable over time.
Question: How many new employees per year does the company hire?
Solution:
L W = average number of women in system = 36, 000 × 0.2 = 7, 200 women.
λ 7200
W = = 194.
6 women / month is the average number of new women employees
37
hired per month.
L M = 36, 000 × 0.8 = 28, 800 men, and λ 28,
800
M = = 144 men/month.
200
Thus, the company hires an average of 194.6 + 144 = 338.6 new employees per month, or
equivalently, 338.6 × 12 = 4063.2 new employees per year.
Discussion:
Pay attention that it is not correct to think like W = 0.2*37 + 0.8*200 !!!
The above expression is meaningless!
Instead of this, we have:
λM
λW
LM LW
L
W = 200* + 37* = + = = 106.3
λ + λ λ + λ λ + λ λ + λ λ + λ
M W M W M W M W M W
λM
Note: = 0.43 ≠0.8
λ + λ
M W
λW
And = 0.57 ≠0.2
.
λ + λ
M W
1

Question 2
MBPF Finance Inc. makes loans to qualified buyers of prefabricated garages from its
parent company, MBPF Inc.
Having just re-engineered its application-Processing operations, MBPF Finance is now
evaluating the effect of its changes on service performance. The subsidiary receives about
1000 loan applications per 30-day working month and makes accept/reject decisions based
on an extensive review of each application.
Prior to January 1998 (under “Process 1”) MBPF Finance Processed each application
individually. On Average, 20% of all applications received approval. An Internal audit
showed that, on average, MBPF Finance had about 500 applications in Process at various
stages of the approval procedure, but on which no decisions had yet been made.
In response to customer complaints about the time taken to Process each application,
MBPF hired a consultants company that suggested the following changed to the Process,
thereby creating “Process 2”.
1. Because the percentage of approved applications is fairly low. An initial Review
Team should be set up to preprocess all applications according to strict but fairly
mechanical guidelines.
2. Each application would fall into one of three categories: (A) looks excellent (B)
needs more details evaluation (C) reject. Type A and B applications would be
forwarded to different specialist subgroups.
3. Each subgroup would then evaluate the applications in its domain and make accept /
reject decisions.
Process 2 was implemented on an experimental basis. The company found that, on average,
25% of all applications were of type A, 25% were of B and 50% were of C. Typically,
about 70% of type A and 10% of type B were approved on review. Internal-audit checks
further revealed that, on average, 200 applications were with the Initial Review Team
undergoing pre-Processing. Only 25 were with the subgroup A Team undergoing the next
stage of Processing and approximately 150 were with the subgroup B Team.
Graphically Process 2 can be described by the following flow chart:
2

1000/month
Initial
Review
LIR=200
Questions:
a) What was the service level under Process 1 – i.e. what was the average Processing
time for a request?
b) What is the average Processing time of a request under Process 2 ?
c) Did the service level for approved requests improve in the transition from Process 1
to Process 2 ?
25%
50%
25%
Subgroup
A Review
LA =25
Subgroup
B Review
LB =150
d) Assume now that 5 Insurance Agents are working in subgroup A (and they share the
load evenly). Each agent can handle, on average, 75 requests per month. What is the
average utilization level of the agents in subgroup A? Similarly, assume also that
there are 8 agents in subgroup B, each capable of handling 40 requests per-months.
What is the average utilization level of the agents in subgroup B?
e) Assume now that, on average, 20% of the requests sent initially to subgroup A need
in fact more detailed processing. These requests are sent to B after they complete the
processing at A. What is now the utilization level of B agents?
f) What is the utilization profile of agents in B, That is, what are the fractions of time,
in the long run, that they allocate to requests from Initial Review, to requests from
A, and what is the fraction of idle time?
3
70%
30%
10%
90%
Accepted
Rejected
200/month
800/month

Solution:
a) According to Little’s law the average Processing time would be 500/1000=0.5
month or 0.5*30=15 days.
b) Under Process 2 the average requests undergoing Processing is LIR+LA+LB=375. So
the average Processing time is 375/1000=11.25 days. Or, if we look at the different
“Processing stations” we have that
W (IR) = 200/1000 = 0.2= 6 days.
W (A) = 25/ (1000*0.25) =0.1= 3 days.
W (B) = 150/ (1000*0.25) =0.6 = 18 days.
And therefore the average Processing would be
0 . 5*
0.
2 + 0.
25*
( 0.
2 + 0.
1)
+ 0.
25*
( 0.
2 + 0.
6)
= 0.
375 = (* 30)
= 11.
25 days
c) Note that for approved requests we have λ = 200 and all requests spend 0.2 in IR.
25*
0.
7 + 150*
0.
1 17.
5 + 15
Hence W = 0 . 2 +
= 0.
2 + = 0.
3625 = (* 30)
= 10.
875 days .
200
200
Or, we can calculate
200*
0.
25*
0.
7 + 200*
0.
25*
0.
1+
25*
0.
7 + 150*
0.
1
W =
= 0.
3625 = 10.
875 days
200
So we see that the Processing time for approved requests did improve. However it is
important to note that the averages do not tell the whole story and still all the
approved requests going through B “enjoy” a 24 days processing time. That’s
because
15
λ B = 1000 * 0.
25*
0.
1 = 25 AndWB = 0 . 2 + = 0.
8 = (* 30)
= 24 days .
25
d) The input rate to A is 250 and the Processing capacity is 5*75 so the utilization level
is 250/375=2/3.
In the same way for subgroup B we have Utilization level = 250/ (8*40) = 250/320
= 0.78125.
e) Now we have the input rate for B is 250+50=300, so the utilization level of B would
be 300/320=0.9375
f) 250/320=78% of the time is dedicated to requests from Initial Review and
50/320=16% of the time is dedicated to requests coming from A. Finally,
20/320=6% of the time agents are idle.
4