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```					                                                      3. Process Flow Measures

Auto-Moto Financial Services- The Old Process
1

Auto-Moto receives 1,000 applications per month. In the old
process, each application is handled in a single activity, with
20% of applications being approved. 500 were in the process at
any time. Average flow time T = ?
200/month
1000/month   Process
Ip=500
800/month
RT = I
T = I/R = 500/1,000 months = 0.5 month or 15 days.
The firm recently implemented a new loan application process.
In the new process, applicants go through an initial review and
are divided into three categories.
Discussion: How operational power destroys the walls of
poverty.
3. Process Flow Measures

New Process: The Same R, But smaller I
2

Subprocess A    70%                200/month
Review                  Accepted
IA = 25          30%

25%
10%
1000/month   Initial             Subprocess B
25%    Review
Review
IR = 200            IB = 150         90%
50%

800/month
R = 1000                                                Rejected
I = IIR + IA + IB = 200 + 25 + 150 = 375
Inventory reduced to 375 from 500 in the old process.
Since R is constant, therefore T has reduced.
T = I/R = 375/1000 = 0.375 month or 0.375(30) = 11.25 days
The new process has decreased the processing time from 15 days to 11.25 days.
3. Process Flow Measures

Flow Time at Each Sub-process (or activity)
3
Average Flow Time for sub-process IR.
Throughput RIR = 1,000 applications/month
Average Inventory IIR = 200 applications
TIR = 200/1,000 = 0.2 months = 6 days in the IR sub-process

Average Flow Time for sub-process A.
Throughput RA = 250 applications/month
Average Inventory IA = 25 applications
TA = 25/250 months = 0.1 months = 3 days in sub-process A.

Average Flow Time for sub-process B.
Throughput RB = 250 applications/month
Average Inventory IB = 150 applications
TB = 150/250 months = 0.6 months = 18 days in sub-proces B
3. Process Flow Measures

Routing, Flow Time, and Percentage of Each Flow units
4

One flow unit at very macro level:           Application
1000 flow units/month at very micro level:   Each specific application
Two flow units:                              Accepted and rejected
Five flow units:                             Accepted-A, Accepted-B
Rejected-IR, Rejected-A, Rejected-B
Accepted-A: IR, A
Accepted-B: IR, B
Rejected-IR: IR
Rejected-A: IR, A
Rejected-B: IR, B

TIR = 6 days
TA = 3 days
TB = 18 days
We also need percentages of each of the five flow units
3. Process Flow Measures

New Process: Intermediate Probabilities
5

Subprocess A   70%                       20%
Review                 Accepted
T=3             30%

25%
10%
100%   Initial           Subprocess B
25%   Review
Review
T=6               T = 18           90%
50%

80%
Rejected
3. Process Flow Measures

New Process: Intermediate Probabilities
6

Subprocess A    17.5%                         20%
Review                      Accepted
T=3              7.5%

25%
2.5%
100%   Initial           Subprocess B
25%   Review
Review
T=6               T = 18            22.5%
50%

50%                       80%
Rejected
3. Process Flow Measures

Flow Time of the Accepted Applications
7

Under the Original Process – the average time spent by an application in the
process is 15 days (approved or rejected).

In the new process: 15 days reduced to 11.25 days.
On average, how long does it take to approve an applicant?
On average, how long does it take to reject an applicant?

Accepted-A: IR, A  Accepted-A(T) = 6 + 3 = 9         Accepted-A = 17.5 %
Accepted-B: IR, B  Accepted-B(T) = 6 + 18 = 24       Accepted-B = 2.5 %

Average Flow time of an accepted application =
[0.175(9)+0.025(24)] / (0.175+.025) = 10.875

The average flow time has reduced from 15 to 11.25. In addition, the flow
time of accepted applications has reduced to 10.875. That is what the firm
really cares about, the flow time of the accepted applications.
3. Process Flow Measures

Flow Time of Rejected Applications
8

Rejected-IR: IR     Rejected-IR(T) = 6  Rejected-IR(%) = 50%
Rejected-A: IR, A  Rejected-A(T) = 6+3 = 9  Rejected-A(%) = 7.5%
Rejected-B: IR, B  Rejected-B(T) = 6+18 = 24  Rejected-B(%) = 22.5%

Average Flow time of a rejected application =
0.5       0.075       0.225
(6)        (9)        (24)  = 11.343
0.8        0.8         0.8

Check our computations:
Average flow time of an application
0.8(11.343)+0.2(10.875) = 11.25
3. Process Flow Measures

ER1; Problem 3.4 MBPF
9

A hospital emergency room (ER) is currently organized so that
all patients register through an initial check-in process. Each
patient is seen by a doctor and then exits the process, either
with a prescription or with admission to the hospital.
55 patients per hour arrive at the ER, 10% are admitted to the
hospital and the rest will leave with a simple prescription.
On average, 7 people are waiting to be registered and 34 are
registered and waiting to see a doctor.
The registration process takes, on average, 2 minutes per
patient. Among patients who receive prescriptions, average
time spent with a doctor is 5 minutes. Among those admitted
to the hospital, average time is 30 minutes.
Assume the process to be stable; that is, average inflow rate
equals average outflow rate.
3. Process Flow Measures

Directions
10

o) Draw the flow process chart
b) On average how long a patient spend in ER?
c) On average how many patients are in ER?
Hints:
Compute flow time in buffer 1
Compute average activity time of Doctor
Compute the average flow time in this process
Compute average flow time for a simple prescription
patient
Compute average flow time for a potential admission
patient
Compute number of patients in Doctor activity
Compute the average number of patients in the process.
3. Process Flow Measures

Flow Time
11
10%
Buffer 1   Registration       Buffer 2          Doctor
R =55/hr
I=7                          I = 34
T = 2 min                           T1 = 5min      90%
T2 = 30min
Simple Prescription
R            I               T
Buffer 1          55           7              7.6
Registeration     55                           2
Buffer 2          55           34             37.1
Doctor            55                          7.5       TD= 0.9(5)+0.1(30) = 7.5
54.2
Re-Check
TPA= 7.6+2+37.1+30 = 76.7
TSP= 7.6+2+37.1+5 = 51.7               TP= 0.1(76.7)+0.9(51.7) = 54.2
3. Process Flow Measures

Inventory
12

10%
Buffer 1      Registration        Buffer 2         Doctor
R =55/hr
I=7                              I = 34
T = 2 min                           T1 = 5min      90%
T2 = 30min
Simple Prescription

R               I          T
Buffer 1               55              7         7.6
Registeration          55              1.8        2
Buffer 2               55              34       37.1
Doctor                 55              6.9       7.5
54.2
3. Process Flow Measures

The Little’s Law
13
10%
Buffer 1   Registration          Buffer 2          Doctor
R =55/hr
I=7                             I = 34
T = 2 min                              T1 = 5min      90%
T2 = 30min
Simple Prescription
R               I                  T
Buffer 1            55              7                 7.6
Registeration       55             1.8                 2
Buffer 2            55             34                37.1
Doctor              55             6.9                7.5
49.7              54.2

Recheck             I = 49.7, R = 55. Is our T= 54.2 correct?
RT= I
R = 55/60 = 0.916667 minutes
RT = I  T=I/R = 49.7/0.916667 = 54.2
3. Process Flow Measures

ER2; Problem 3.5 MBPF
14

A triage system has been proposed for the ER described in Exercise
3.4. As mentioned earlier, 55 patients per hour arrive at the ER.
Patients will be registered as before and it takes an average of 2
minutes per patient. They will then be quickly examined by a nurse
practitioner who will classify them as Simple Prescriptions or
Potential Admits. Planners anticipate that the initial examination by
triage nurses will take 3 minutes. They expect that, on average, 20
patients will be waiting to register and 5 will be waiting to be seen
by the triage nurse. Planners expect eh Simple Prescriptions area to
have, on average, 15 patients waiting to be seen. As before, once a
patient’s turn come, each will take 5 minutes of a doctor’s time. The
hospital anticipates that, on average, the emergency area will have
only 1 patient waiting to be seen . As before, once that patient’s turn
comes, he or she will take 30 minutes of a doctor’s time. Assume
that, as before, 90% of all patients are Simple Prescriptions, assume,
too, that the triage nurse is 100% accurate in making classifications.
3. Process Flow Measures

Directions
15

o) Draw the flow process chart
a) On average how many patients are in ER?
b) On average, how long a patient spend in ER?
c) Compute average flow rate in buffer 3 and buffer 4.
d) Compute average flow time in all buffers.
e) Compute average number of patients in all activities.
f) On average, how long does a simple prescription patient
spend in the ER?
d) On average, how long does a Potential Admit patient
spend in the ER?
3. Process Flow Measures

Process Flow and TR=I Table
16

5.5/hr
Buffer 1   Registration   Buffer 2   Triage Nurse
55/hr

49.5/h
r      Buffer 4          Simple Prescription

R         R/min                   I                    T
Buffer 1                              55        0.916667                20
Registeration                         55        0.916667                                      2
Buffer 2                              55        0.916667                 5
Triage Nurse                          55        0.916667                                      3
Buffer 3                              5.5       0.091667                 1
Buffer 4                             49.5         0.825                 15
Simple Prescription                  49.5         0.825                                       5
3. Process Flow Measures

Inventory and Flow Time; Macro Method
17

R       R/min           I              T
Buffer 1                     55      0.916667        20            21.8
Registeration                55      0.916667        1.8             2
Buffer 2                     55      0.916667         5             5.5
Triage Nurse                 55      0.916667        2.8             3
Buffer 3                     5.5     0.091667         1            10.9
Potential Admission          5.5     0.091667        2.8            30
Buffer 4                    49.5       0.825         15            18.2
Simple Prescription         49.5       0.825         4.1             5
a) On average, how many patients are in ER?      IER= 52.5
b) On average, how long will a patient spend in ER?
Method 1: Macro Method, a single flow unit
RER= 55/60 flow units / min , IER= 52.5

TER = 52.5/(55/60) = 57.2 minutes
3. Process Flow Measures

Flow Time; Micro Method
18

R         R/min            I            T
Buffer 1                         55        0.916667         20           21.8
Registeration                    55        0.916667         1.8          2.0
Buffer 2                         55        0.916667          5           5.5
Triage Nurse                     55        0.916667         2.8          3.0
Buffer 3                         5.5       0.091667          1           10.9
Potential Admission              5.5       0.091667         2.8          30.0
Buffer 4                        49.5         0.825          15           18.2
Simple Prescription             49.5         0.825          4.1          5.0

Method 2: Micro Method, two flow units, Potential Admission and Simple Prescription
TPA= 21.8+2+5.5+3+10.9 +30 = 73.2
TSP= 21.8+2+5.5+3+18.2 +5 = 55.5
TER= .1(73.2)+.9(55.5) = 57.3
c) On average, how long will a potential admission patient spend in ER?        73.2
3. Process Flow Measures

ER3; Problem 3.6 MBPF
19

Refer again to Exercise 3.5. Once the triage system is put in
place, it performs quite close to expectations. All data conform
to planners’ expectations except for one set-the classifications
made by the nurse practitioner. Assume that the triage nurse
has been sending 91% of all patients to the Simple Prescription
area when in fact only 90% should have been so classified. The
remaining 1% is discovered when transferred to the emergency
area by a doctor. Assume all other information from Exercise 3.5
to be valid.
3. Process Flow Measures

Directions
20

o) Draw the flow process chart
a) On average how many patients are in ER?
b) On average, how long a patient spend in ER?
c) Compute average flow rate in buffer 3 and buffer 4.
d) Compute average flow time in all buffers.
e) Compute average number of patients in all activities.
f) On average, how long does a simple prescription patient
spend in the ER?
d) On average, how long does a Potential Admit patient spend
in the ER?
3. Process Flow Measures
Problem 3.6
Process Flow and Throughputs
21
4.95/hr
Buffer 1   Registration   Buffer 2   Triage Nurse
55/hr
0.55/hr

50.05/h
r       Buffer 4         Simple Prescription

R          R/min                    I                      T
Buffer 1                             55         0.916667                 20
Registeration                        55         0.916667                                         2
Buffer 2                             55         0.916667                  5
Triage Nurse                         55         0.916667                                         3
Buffer 3                             5.5        0.091667                  1
Buffer 4                            50.05       0.834167                 15
Simple Prescription                 50.05       0.834167                                         5
3. Process Flow Measures

Inventory and Flow Time; Macro Method
22

R      R/min          I              T
Buffer 1                  55     0.916667       20            21.8
Registeration             55     0.916667       1.8             2
Buffer 2                  55     0.916667        5             5.5
Triage Nurse              55     0.916667       2.8             3
Buffer 3                  5.5    0.091667        1            10.9
Potential Admission       5.5    0.091667       2.8            30
Buffer 4                 50.05   0.834167       15            18.0
Simple Prescription      50.05   0.834167       4.2             5
Macro: Average number of patients in the system =
20+1.8+5+2.8+1+2.8+15+4.2 = 52.6
Average flow time = I/R = 52.6/(55/60) = 57.3
3. Process Flow Measures

Flow Time; Simple Prescription
23

R      R/min       I               T
Buffer 1               55     0.916667    20              21.8
Registeration          55     0.916667    1.8               2
Buffer 2               55     0.916667     5              5.5
Triage Nurse           55     0.916667    2.8               3
Buffer 3               5.5    0.091667     1              10.9
Potential Admission    5.5    0.091667    2.8              30
Buffer 4              50.05   0.834167    15              18.0
Simple Prescription   50.05   0.834167    4.2               5
Micro Method Compute SP and PA first TSP= 32.3+18+5 = 55.3
Common =
21.8 +2+5.5+3 = 32.3
3. Process Flow Measures

Flow Time; Potential Admission and Overall
24

R      R/min      I         T
Buffer 1               55     0.916667   20       21.8
Registeration          55     0.916667   1.8        2
Buffer 2               55     0.916667    5        5.5
Triage Nurse           55     0.916667   2.8        3
Buffer 3               5.5    0.091667    1       10.9
Potential Admission    5.5    0.091667   2.8       30
Buffer 4              50.05   0.834167   15        18
Simple Prescription   50.05   0.834167   4.2        5

0.09

0.01

0.9
3. Process Flow Measures

Flow Time; Potential Admission and Overall
25
R      R/min      I      T
Buffer 1               55     0.916667   20    21.8
Registeration          55     0.916667   1.8     2
Buffer 2               55     0.916667    5     5.5
Triage Nurse           55     0.916667   2.8     3
Buffer 3               5.5    0.091667    1    10.9
Potential Admission    5.5    0.091667   2.8    30
Buffer 4              50.05   0.834167   15     18
Simple Prescription   50.05   0.834167   4.2     5
TPA1= 32.3 +10.9+30 = 73.2 ……(4.95 PA patients out of 5.5 PA patient: 90%)
TPA2= 73.2+18+5 = 96.2 (0.55 PA patients out of 5.5 PA patient: 10%)
TPA = 73.2(.9) + 96.2(.1) =75.5
TSP= 55.3 is 90% of flow units, and TPA =75.5 is for 10% of flow units
T = 55.3 (.9) + 75.5(.1) = 57.32                                          Recheck

T= 55.3 (0.9) + 73.2 (0.09) + 96.2 (0.01)
Recheck Again
T= 57.32
3. Process Flow Measures

ER4; Problem 3.6b MBPF
26
80 patients per hour arrive at a hospital emergency room (ER).
All patients first register through an initial registration process.
On average there are 9 patients waiting in the Rg-Buffer.
The registration process takes 6 minutes.
Patients are then examined by a triage nurse practitioner;m.
On average there are 2 patients waiting in the Tr-Buffer in front of the triage process,
and the triage classification process takes 6 minutes.
On average, 91% of the patients are sent to the Simple-prescription process and the
On average, there are 6 patients waiting in the Simple-prescription buffer (Sp-Buffer)
in front of this process.
A physician spends 6 minutes on each patient in the Simple-prescription process.
In addition, on average, 2 patients per hour are sent to the Hospital-admission buffer
buffer (Hs-Buffer) after being examined for 6 minutes in the Simple prescription
process.
On average, 0.9 patients are waiting in the Hospital-admission buffer (Hs-Buffer).
A physician spends 25 minutes on each patient in the Hospital-admits process.0
3. Process Flow Measures

Process Flow and Throughputs
27
HsBuff:0.9       Hs:25
9%
RgBuff:9          Rg:5           TrBuff:2   Tr:6
80/hr
2/hr

91%
SpBuff:6          Sp:6
R            R/min         I    T
RgBuff           80       1.333333          9
Rg               80       1.333333                5
TrBuff           80       1.333333          2
Tr               80       1.333333                6
SpBuff 3         72.8     1.213333          6
Sp               72.8     1.213333                6
HsBuff           7.2           0.12        0.9
HsBuff           7.2           0.12               25
3. Process Flow Measures

Process Flow and Throughputs
28
HsBuff:0.9       Hs:25
9%
RgBuff:9          Rg:5           TrBuff:2   Tr:6
80/hr
2/hr

91%
SpBuff:6          Sp:6
R            R/min         I    T
RgBuff           80       1.333333          9
Rg               80       1.333333                6
TrBuff           80       1.333333          2
Tr               80       1.333333                6
SpBuff 3         72.8     1.213333          6
Sp               72.8     1.213333                6
HsBuff           9.2      0.153333         0.9
HsBuff           9.2      0.153333                25
3. Process Flow Measures

Process Flow and Throughputs
29
HsBuff:0.9       Hs:25
9%
RgBuff:9          Rg:5           TrBuff:2     Tr:6
80/hr
2/hr

91%
SpBuff:6          Sp:6
R            R/min         I     T
RgBuff           80       1.333333          9     6.75
RgBuff           80       1.333333                5.00
TrBuff           80       1.333333          2     1.50
Tr               80       1.333333                5.00
SpBuff 3         72.8     1.213333          6     4.95
Sp               72.8     1.213333                6.00
HsBuff           9.2      0.153333         0.9    5.87
HsBuff           9.2      0.153333                25.00
3. Process Flow Measures

Process Flow and Throughputs
30
HsBuff:0.9       Hs:25
9%
RgBuff:9          Rg:5           TrBuff:2     Tr:6
80/hr
2/hr

91%
SpBuff:6          Sp:6
R            R/min         I     T
RgBuff           80       1.333333          9     6.75
RgBuff           80       1.333333         8.0    6.00
TrBuff           80       1.333333          2     1.50
Tr               80       1.333333         6.7    5.00
SpBuff 3         72.8     1.213333          6     4.95
Sp               72.8     1.213333         7.3    6.00
HsBuff           9.2      0.153333         0.9    5.87
HsBuff           9.2      0.153333         3.8    25.00
3. Process Flow Measures

Process Flow and Throughputs
31
HsBuff:0.9       Hs:25
9%
RgBuff:9          Rg:5           TrBuff:2      Tr:6
80/hr
2/hr

91%
R            R/min         I       T             SpBuff:6          Sp:6

RgBuff           80       1.333333          9       6.75
RgBuff           80       1.333333         8.0       6
Average flow time =
TrBuff           80       1.333333          2       1.5      T = I/R =
Tr               80       1.333333         6.7       5
SpBuff 3         72.8     1.213333          6     4.945055   43.68/(80/60) = 32.76
Sp               72.8     1.213333         7.3       6
HsBuff           9.2      0.153333         0.9    5.869565
HsBuff           9.2      0.153333         3.8       25
43.68
3. Process Flow Measures

Process Flow and Throughputs
32

R      R/min      I       T
RgBuff      80    1.333333   9.00    6.75
RgBuff      80    1.333333   6.67    5.00
TrBuff      80    1.333333   2.00    1.50
Tr          80    1.333333   8.00    6.00      19.25
SpBuff 3   72.8   1.213333   6.00    4.95
Sp         72.8   1.213333   7.28    6.00      10.95
HsBuff     9.2    0.153333   0.90    5.87
HsBuff     9.2    0.153333   3.83   25.00      30.87
Micro Method Compute SP and PA first TSP= 19.25+10.95 = 30.2
Common =
6.75+6+1.5+5 = 19.25
3. Process Flow Measures

Process Flow and Throughputs
33

R      R/min      I      T
RgBuff     80     1.333333   9.00   6.75
RgBuff     80     1.333333   8.00   6.00
TrBuff     80     1.333333   2.00   1.50
Tr         80     1.333333   6.67   5.00    19.25
SpBuff 3   72.8   1.213333   6.00   4.95
Sp         72.8   1.213333   7.28   6.00    10.95
HsBuff     9.2    0.153333   0.90   5.87
HsBuff     9.2    0.153333   3.83   25.00   30.87
TPA1= 19.25 +30.87 = 50.12 ……(7.2 PA patients out of 9.2 PA patients)
TPA2= 19.25 +10.95 +30.87 = 61.07 (2 PA patients out of 9.2 PA patients)
TPA = 50.12(7.2/9.2) + 61.07(2/9.2) =
TPA = 50.12(0.782609) + 61.07(0.217391) =52.5

T= 30.20(70.8/80) + 50.12 (7.2/80) + 61.02 (2/80)
T = 32.76                                                              Recheck

```
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