LOGO
MM – OM
Session 2a
Henry Yuliando
Process Flow
Measures
The essence of process flow including :
1. Flow time ; indicates the time needed to
convert inputs into outputs and includes
any time spent by a flow unit waiting for
processing activities to be performed.
2. Flow rate ; the number of flow units that
flow through a specific point in the process
per unit of time. Ri(t) and Ro(t)
3. Inventory ; is the total number of flow
units present within process boundaries.
Ri,1(t)
Ri,2
(t)
Process
Ro(t)
Input and Output Flow Rates for a Process with Two Entry Points
The inflow rate of passengers at a security checkpoint at
Vancouver Airport changes over time. Recall that for each of
the three flights, about half the 200 passengers for each flight
arrive between 50 and 30 minutes before flight departure. So
far for each of the three flights departing at 10 a.m.
About 100 passengers arrive between 9:10 and 9:30 a.m., a 20minute interval. This means that a total of 300 passengers
arrive during this time period for three flights, giving an inflow
rate of roughly 15 passengers per minute. The remaining 300
passengers for the three flights arrive between 8:40 and 9:10
a.m. (about 80 to 50 minute before departure) and between 9.30
and 9:40 a.m. (about 30 to 20 minute before departure).
That is the remaining 300 passengers arrive during a
total time period of 40 minutes, giving an inflow rate
of 7.5 passengers per minute, which is half the inflow
rate during the peak time period from 9:10 to 9:30
a.m. The outflow rate of the checkpoint, however, is
limited by the X-ray scanner, which cannot process
more than 18 bags per minute or, with an average of
1.5 bags per passenger, 12 passenger per minute.
Time
Inflow rate Ri
Outflow rate Ro
Buildup Rate R
Ending Inventory
(number of passengers
in line)
8:40 a.m.
8:40 – 9:10
a.m.
9:10 – 9:30
a.m.
9:30 –
9:43:20 a.m.
7.5/min.
7.5/min.
0
0
15/min.
12/min.
3/min.
7.5/min.
12/min.
4.5/min.
0
9:43:20 –
10:10 a.m.
7.5/min.
7.5/min.
0
0
• Outflow rate capacity 12 passengers/min.
• At peak period 9:10 to 9:30 a.m., the inflow rate 15 passengers/min. so that a queue (inventory)
starts building at R = 3 passengers/min.
• At 9:30, the inventory has grown = R x (9:30-9:10) = 60 passengers.
• After 9:30 a.m. X-ray scanner keeps processing at full rate 12 passengers/min., while the inflow
rate 7.5 passengers/min., so that the passenger queue is being depleted at a rate of 4.5
passengers/min.
• Thus, 60-passenger queue is eliminated after 60/4.5 = 13.33 minutes, or at 9:43:20 the queue is
empty again.
•.Example video
Inflow rates with staggered departures for Vancouver Airport
security checkpoint….
(unit / minute)
Time
(start of period)
8:40
8:50
9:00
9:10
9:20
9:30
9:40
9:50
10:00
10:10
10:20
10:30
Inflow rate for
flights arriving on
the hour, Ri,1
2.5
2.5
2.5
5
5
2.5
2.5
2.5
2.5
5
5
2.5
Inflow rate for
flights arriving 20
minutes past the
hour, Ri,2
5
2.5
2.5
2.5
2.5
5
5
2.5
2.5
2.5
2.5
5
Inflow rate for
flights arriving 40
minutes past the
hour, Ri,3
2.5
5
5
2.5
2.5
2.5
2.5
5
5
2.5
2.5
2.5
Total inflow rate
Ri
10
10
10
10
10
10
10
10
10
10
10
10
 The inventory includes all flow units within process boundaries (in –
on – off process) valid for manufacturing, services, financial, or
even information process.
 The inventory dynamics are driven by the difference between inflow
and outflow rates.
 The instantaneous inventory accumulation (buildup) rate, R(t)
as the difference between instantaneous inflow rate and outflow rate.
R(t) = Ri(t) – Ro(t)
 Within a time interval (t1, t2), inventory change = buildup rate x
length of time interval.
 For R constant = I(t2) – I(t1) = R(t) x (t2-t1)
 A stable process is one, in which in the long run, the average
inflow rate is the same as the average outflow rate.
(ex : inflow 600 passenger per hour; outflow 12 passengers per
minute or equal 720 passengers per hour; means that in the
long rung the average outflow rate is 600 passengers per hour)
 Average flow rate or throughput is the average number of
flow units that flow through (into and out of) the process per
unit of time. Denoted as R.
 For the case of Vancouver Airport, it is known that from 8:40
to 9:10 a.m., the inventory or queue is zero. From 9:10 through
9:43 a.m., the inventory builds up linearly to a maximum size
of 60. Thus the average inventory during that period is 60/2 =
30.
 Finally, the inventory is zero again is from 9:43 to 10:10 a.m.,
when the cycle repeats with the next inventory buildup. Thus,
the average queue size is the time-weighted average:
I = (33 min x 30 passengers + 27 min x 0 passengers)/60 min
= 17.5 passengers
 Further, average inventory over time is denoted by I.
 Average flow time (T) is as the average (of the flow times) across
all flow units that exit the process during a specific span of time.
 Little`s Law : average inventory equals throughput times average
flow time, or I = R x T.
 To determine the average time spent by a passenger in a checkpoint
queue, is that
T = I/R = 17.5 passengers/10 passengers per minute =
= 1.75 minutes pax/minute
where R = 600 passengers per hour or 10 passengers per minute.
 Process I : HOM financial services provides financing to
qualified creditors. It receives about 1,000 applications per
month and makes accept/reject decisions based on an
extensive review of each applications.
 Assume a 30-day working month.
 Up to present, HFS process each application individually. On
average, 20% of all applications received approval.
 Based on an audit, it revealed that HFS on average had about
500 applications in process at various stages of the
approval/rejection procedure.
 In response to customer complaints about the time taken to
process each application, HFS called for a measure to improve
its performance.
It was suggested that (Process II) :
 An Initial Review Team should be set up to
preprocess according to strict but fairly mechanical
guidelines.
 Each application would fall into one of three
categories: A (looks excellent), B (needs more
detailed evaluation), and C (reject summarily). A
and B applications would be forwarded to different
specialist subgroups.
 Each subgroup would then evaluate the applications
in its domain and make accept/reject decisions.
Process II was implemented on an experimental basis.
 HFS found that on average, 25% of all applications were A
category, 25% B, and 50% C. Typically about 70% of all A
applications and 10% of all B applications were approved on
review. (Recall that all C applications were rejected). It was also
noticed that 200 applications were with the Initial Review Team
undergoing preprocessing. Just 25, however, were with the
Subgroup A Team undergoing the next stage of processing and
about 150 with the Subgroup B Team.
 Under Process I : Throughput R = 1,000 applications/month
Average inventory I = 500 applications
Average flow time T = I/R = 0.5 months = 15 days.
Subgroup A
Review
II = 25
70%
Initial
Review
IIR = 200
25%
50%
200/month
10%
25%
1000/month
Accepted
Subgroup B
Review
II = 150
30%
90%
Rejected
800/month
 Throughput R = 1000 applications/month
 Average inventory I = 200 + 150 + 25 = 375 applications.
 Average flow time T = I/R = 0.375 months = 11.25 days.
 Detailed analysis and calculate the average flow time of each type group under
process II can be done. As seen on the above figure, there are 200 applications
with the Initial Review Team (average inventory). The performance is that
Throughput = 1000 applications/month
and
Average inventory = 200 applications
then
Average flow time TIR = IIR/RIR = 200/1000 months
= 0.2 months = 6 days
 For subgroup A process, there are 25 applications. Since 25% of all incoming
applications are categorized as type A, on average, 250 of the 1000 applications
received per month are categorized as type A. Therefore,
Throughput RA
= 250 applications/month
Average inventory IA = 25 applications
then
Average flow time TA = IA/RA = 25/250 months = 0.1 months = 3 days
 Similarly, for subgroup B process, there are 25% of incoming applications are
250, and 150 applications with subgroup B. That is,
Throughput R B
= 250 applications/month
Average inventory IB = 150 applications
then
Average flow time TA = IA/RA = 150/250 months = 0.6 months = 18 days
 And for subgroup C process, since this application type leave immediately
means that the average inventory are 0.
 Recall that both subgroup process must through Initial Review
Team process first which is taking 6 days process, thus the average
flow time of each type application under process II:
 Type A applications spend, on average, 9 days in the process.
 Type B applications spend, on average, 24 days in the process.
 Type C applications spend, on average, 6 days in the process.
 To find the average flow time across all applications under process
II it using the weighted average across three application types.
T
RC
RA
RB
(TIR  TA ) 
(TIR  TB ) 
(TIR  TC )
RA  RB  RC
RA  RB  RC
RA  RB  RC
so
250
250
500
T
(6  3) 
(6  18) 
(6)
250 250 500
250 250 500
250 250 500
 Under process II also, 70% of type A applications (175 out of 250 per month,
on average) are approved, as are 10% of type B applications (25 out of 250 per
month, on average). Thus the aggregate rate at which all applications are
approved equals 175+25 = 200 applications per month. Hence, average flow
time for approved applications can be calculated by:
Tapproved
175
25

(TIR  TA ) 
(TIR  TB )
175 25
175 25
175
25

(6  3) 
(6  18)  10.875days
200
200
 Similarly, average flow time for rejected applications can
be calculated by:
75
225
Trejected 
(TIR  TA ) 
(TIR  TB ) 
75  225 500
75  225 500
500
(TIR )
75  225 500
75
225
500

(6  3) 
(6  18) 
(6)  11.343days
800
800
800
Analyzing Financial Flows Through Financial Statements
Table 1
MBPF Consolidated Statements of Income and Retained Earnings 2010
Net sale
Costs and expenses
Cost of goods sold
Selling, general and administration expenses
Interest expense
Depreciation
Other (income) expenses
Total costs and expenses
Income before taxes
Provision for income taxes
Net income
Retained earnings, beginning of year
Less cash dividends declared
Retained earnings at end of year
Net income per common share
Dividend per common share
250.0
175.8
47.2
4.0
5.6
2.1
234.7
15.3
7.0
8.3
31.0
2.1
37.2
0.83
0.21
Table 2
MBPF Consolidated Balance Sheet as of December 2010
Current assets
Cash
Short-term investments at cost (approximate market)
Receivables, less allowances of $0.7 million
Inventories
Other current assets
Total current assets
Property, plant, and equipment (at cost)
Land
Building
Machinery and equipment
Construction in progress
Subtotal
Less accumulated depreciation
Net property, plant, and equipment
Investments
Prepaind expenses and other deferred charges
Other assets
Total assets
(Selected) current liabilities
Payables
2.1
3.0
27.9
50.6
4.1
87.7
2.1
15.3
50.1
6.7
74.2
25.0
49.2
4.1
1.9
4.0
146.9
11.9
Table 3
MBPF Inventories and Cost of Goods Details
Cost of goods sold
Raw materials
Fabrication (L&OH)
Purchased parts
Assembly (L&OH)
Total
Inventory
Raw materials (roof)
Fabrication WIP (roof)
Purchased parts (base)
Assembly WIP
Finsihed goods
Total
50.1
60.2
40.2
25.3
175.8
6.5
15.1
8.6
10.6
9.8
50.6
...flow analysis
 Throughput R = $175.8 million/year (Cost of goods sold- table 1)
 Average inventory I = $50.6 million (Inventories – table 2)
 Average flow time T = I/R = 50.6/175.8 year = 0.288 years = 14.97 
15 weeks
 For the accounts-receivable (AR)
Throughput RAR = $250 million/year (Net sales – table 1)
Average inventory IAR = $27.9 million (Receivables – table 2)
Average flow time TAR = IAR / RAR = 0.112 years = 5.80  6 weeks
 For the accounts-payable (AP) or purchasing = $50.1 + $40.2 =
$90.3 million, average inventory in purchasing = $11.9 million  TAP
= IAP / RAP = 0.13 years= 6.9 weeks
Cash to cash cycle performance
Average lag = 15 + 6 =21 weeks is the
cost-to-cash cycle (ROE)
Investment in the form of purchased parts
and raw materials after 6.9 weeks, cashto-cash cycle (Net Trade Cycle) = 21 – 6.9
= 14.1 weeks
Targeting improvement
The throughput rate through fabrication :
$50.1 million/year in raw materials
+ $60.2 million in labor and overhead
= $110.3 million/year
The throughput through assembly area :
$110.3 million/years in roofs
+ $ 40.2 million/year in bases
+ $ 25.3 million/year in labor & overhead
= $175.8 million
$60.2/yr
$50.1/yr
$6.5
Raw
materials
(roofs)
$25.3/yr
$15.1
Fabrication
(roofs)
$110.3/yr
$10.6
Assembly
$40.2/yr
$40.2/yr
$8.6
Purchased
parts
(bases)
$175.8/yr
$9.8
Finished
goods
$175.8/yr
Flow times through MBPF
Raw
Materials
Fabrication
Purchase
d Parts
Assembly
Finished
Goods
50.1
0.96
6.5
6.75
110.3
2.12
15.1
7.12
40.2
0.77
8.6
11.12
175.8
3.38
10.6
3.14
175.8
3.38
9.8
2.90
Throughput R
$/year
$/week*
Inventory I ($)
Flow time T=I/R (weeks)*
*=rounding to 52 weeks per year
Flow Rate R
(million $/week)
5.0
Finished goods
Accounts
receiveable
Fabrication
Assembly
3.38
7.12
3.14
2.90
5.80
2.12
0.96
0.77
Purchased parts
11.12
Raw
materials
6.75
BASUTA case