1. Effect of Buffer Capacity
Process Data
Ri = 20/hour, Tp = 2.5 mins, c = 1, K = # Lines – c
Performance Measures K 4 5 6 Ii
1.23
1.52
1.79 Ti
4.10
4.94
5.72 Pb
0.1004
0.0771
0.0603 R
17.99
18.46
18.79 r
0.749
0.768
0.782

2. Economics of Capacity Decisions
Cost of Lost Business Cb
$ / customer
Increases with competition
Cost of Buffer Capacity Ck
$/unit/unit time
Cost of Waiting Cw
$ /customer/unit time
Cost of Processing Cs
$ /server/unit time
Increases with 1/ Tp
Tradeoff: Choose c, Tp, K
Minimize Total Cost/unit time
= Cb Ri Pb + Ck K + Cw I (or Ii) + c Cs

3. Optimal Buffer Capacity
Cost Data
Cost of telephone line = $5/hour, Cost of server = $20/hour, Margin lost = $100/call, Waiting cost = $2/customer/minute
Effect of Buffer Capacity on Total Cost K
$5(K + c)
$20 c
$100 Ri Pb
$120 Ii
TC ($/hr) 4 25 20
200.8
147.6
393.4 5 30
154.2
182.6
386.4 6 35
120.6
214.8
390.4

4. Optimal Processing Capacity
K = 6 – c Pb Ii
TC ($/hr) = $20c + $5(K+c) + $100Ri Pb+ $120 Ii 1 5
0.0771
1.542
$386.6 2 4
0.0043
0.158
$97.8 3
0.0009
0.021
$94.2
0.0004
0.003
$110.8

5. Performance Variability
Effect of Variability
Average versus Actual Flow time
Time Guarantee
Promise
Service Level
P(Actual Time  Time Guarantee)
Safety Time
Time Guarantee – Average Time
Probability Distribution of Actual Flow Time
P(Actual Time  t) = 1 – EXP(- t / T)

6. Effect of Blocking and Abandonment
Blocking: the buffer is full = new arrivals are turned away
Abandonment: the customers may leave the process before being served
Proportion blocked Pb
Proportion abandoning Pa

7. Net Rate: Ri(1- Pb)(1- Pa) Throughput Rate: R=min[Ri(1- Pb)(1- Pa),Rp]
Effect of Blocking and Abandonment
Net Rate: Ri(1- Pb)(1- Pa) Throughput Rate: R=min[Ri(1- Pb)(1- Pa),Rp]

8. Example 8.8 - DesiCom Call Center
Arrival Rate Ri= 20 per hour=0.33 per min
Processing time Tp =2.5 minutes (24/hr)
Number of servers c=1
Buffer capacity K=5
Probability of blocking Pb=0.0771
Average number of calls on hold Ii=1.52
Average waiting time in queue Ti=4.94 min
Average total time in the system T=7.44 min
Average total number of customers in the system I=2.29

9. Example 8.8 - DesiCom Call Center
Throughput Rate
R=min[Ri(1- Pb),Rp]= min[20*( ),24]
R=18.46 calls/hour
Server utilization:
R/ Rp=18.46/24=0.769

10. Example 8.8 - DesiCom Call Center
Number of lines 5 6 7 8 9 10
Number of servers c 1
Buffer Capacity K 4
Average number of calls in queue
1.23
1.52
1.79
2.04
2.27
2.47
Average wait in queue Ti (min)
4.10
4.94
5.72
6.43
7.08
7.67
Blocking Probability Pb (%)
10.04
7.71
6.03
4.78
3.83
3.09
Throughput R (units/hour)
17.99
18.46
18.79
19.04
19.23
19.38
Resource utilization
.749
.769
.782
.793
.801
.807

11. Capacity Investment Decisions
The Economics of Buffer Capacity
Cost of servers wages =$20/hour
Cost of leasing a telephone line=$5 per line per hour
Cost of lost contribution margin =$100 per blocked call
Cost of waiting by callers on hold =$2 per minute per customer
Total Operating Cost is $386.6/hour

12. Example 8.9 - Effect of Buffer Capacity on Total Cost
Number of lines n 5 6 7 8 9
Number of CSR’s c 1
Buffer capacity K=n-c 4
Cost of servers ($/hr)=20c 20
Cost of tel.lines ($/hr)=5n 25 30 35 40 45
Blocking Probability Pb (%)
10.04
7.71
6.03
4.78
3.83
Lost margin = $100RiPb
200.8
154.2
120.6
95.6
76.6
Average number of calls in queue Ii
1.23
1.52
1.79
2.04
2.27
Hourly cost of waiting=120Ii
147.6
182.4
214.8
244.8
272.4
Total cost of service, blocking and waiting ($/hr)
393.4
386.6
390.4
400.4
414

13. Example 8.10 - The Economics of Processing Capacity
The number of line is fixed: n=6
The buffer capacity K=6-c c K
Blocking Pb(%)
Lost Calls RiPb
(number/hr)
Queue length Ii
Total Cost ($/hour) 1 5
7.71%
1.542
1.52
30+20+(1.542x100)+(1.52x120)=386.6 2 4
0.43%
0.086
0.16
30+40+(0.086x100)+(0.16x120)=97.8 3
0.09%
0.018
0.02
30+60+(0.018x100)+(0.02x120)=94.2
0.04%
0.008
0.00
30+80+(0.008x100)+(0.00x120)110.8

14. Variability in Process Performance
Why considering the average queue length and waiting time as performance measures may not be sufficient?
Average waiting time includes both customers with very long wait and customers with short or no wait.
We would like to look at the entire probability distribution of the waiting time across all customers.
Thus we need to focus on the upper tail of the probability distribution of the waiting time, not just its average value.

15. Average of 20 customers per hour
Example WalCo Drugs
One pharmacist, Dave
Average of 20 customers per hour
Dave takes Average of 2.5 min to fill prescription
Process rate 24 per hour
Assume exponentially distributed interarrival and processing time; we have single phase, single server exponential model
Average total process is;
T = 1/(Rp – Ri) = 1/(24 -20) = 0.25 or 15 min

16. Example Probability distribution of the actual time customer spends in process (obtained by simulation)

17. Example 8.11 - Probability Distribution Analysis
65% of customers will spend 15 min or less in process
95% of customers are served within 40 min
5% of customers are the ones who will bitterly complain. Imagine if they new that the average customer spends 15 min in the system.
35% may experience delays longer than Average T,15min

18. Service Promise: Tduedate , Service Level & Safety Time
SL; The probability of fulfilling the stated promise. The Firm will set the SL and calculate the Tduedate from the probability distribution of the total time in process (T).
Safety time is the time margin that we should allow over and above the expected time to deliver service in order to ensure that we will be able to meet the required date with high probability
Tduedate = T + Tsafety
Prob(Total time in process <= Tduedate) = SL
Larger SL results in grater probability of fulfilling the promise.

19. Due Date Quotation
Due Date Quotation is the practice of promising a time frame within which the product will be delivered.
We know that in single-phase single server service process; the Actual total time a customer spends in the process is exponentially distributed with mean T.
SL = Prob(Total time in process <= Tduedate) = 1 – EXP( - Tduedate /T)
Which is the fraction of customers who will no longer be delayed more than promised.
Tduedate = -T ln(1 – SL)

20. 95% of customers will get served within 45 min
Example WalCo Drug
WalCo has set SL = 0.95
Assuming total time for customers is exponential
Tduedate = -T ln(1 – SL)
Tduedate = -T ln(0.05) = 3T
Flow time for 95 percentile of exponential distribution is three times the average T
Tduedate = 3 * 15 = 45
95% of customers will get served within 45 min
Tduedate = T + Tsafety
Tsafety = 45 – 15 = 30 min
30 min is the extra margin that WalCo should allow as protection against variability

21. Higher the utilization; Longer the promised time and Safety time
Relating Utilization and Safety Time: Safety Time Vs. Capacity Utilization
Capacity utilization ρ % % % %
Waiting time Ti Tp Tp Tp Tp
Total flow time T= Ti + Tp Tp Tp Tp Tp
Promised time Tduedate Tp Tp Tp Tp
Safety time Tsafety = Tduedate – T Tp Tp Tp Tp
Higher the utilization; Longer the promised time and Safety time
Safety Capacity decreases when capacity utilization increases
Larger safety capacity, the smaller safety time and therefore we can promise a shorter wait

22. Managing Customer Perceptions and Expectations
Uncertainty about the length of wait (Blind waits) makes customers more impatient.
Solution is Behavioral Strategies
Making the waiting customers comfortable
Creating distractions
Offering entertainment

23. Thank you
Questions?