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MGMT 361
Operations Management
(OM)
Process Analysis – part 3 (minimum throughput, examples, practice)
Prof. Julia A. Kalish
Example 2.7
CT for Stage 1:
CT for Stage 2:
Bottleneck:
Process CT :
Process Capacity:
Min TPT:
= 3.75 min / unit (1 / 16 * 60)
= 4 min / unit
Stage 2
= 4 min / unit (based on the bottleneck)
= ¼ [u / min] = 15 units / hour (1/4 * 60)
= 6 + 4 = 10 min (M2 + M3)
Stage 1
Stage 2
M3 4 min
M2 6 min
M1 10 min
Capacity of M1: 6 [units/hr]
((1/10) * 60))
Capacity of M2: 10 [units/hr]
((1/6) * 60))
Total for Stage 1: 16 [units/hr]
(6 + 10 )
Design parameters
Mgmt 361 Module 2 - Process Analysis 2
Mgmt 361 Module 2 - Process Analysis 3
M1
M2
M3
Buffer (2 min)
6 10 14 18 22 26 30 34 38 42
Idle time ignored in WIP calculations
5
4
4
5
6
6
6
2
1
1
2
3
3
3
8
7
7
8
9
9
9
Schedule: Start a job every 12 minutes on M1 (0, 12, 24, ..).
Start a job every 6 minutes on M2 (0, 6, 12, ..).
Example 2.7 – How much WIP?
CT:
Average TPT:
WIP M1:
M2:
Buffer:
M3:
Total:
4 [min/unit]
1/3*(10+14+12)=12 [min]
10 / 12 = 5/6 units (10 min of work + 2 idle mins)
1 units (6/6)
2 / 12 = 1/6 units (2 min of buffer + 10 min empty)
1 unit (4/4)
3 units (5/6 + 1 +1/6 + 1)
M1
M2
M3
Buffer
6 10 14 18 22 26 30 34 38 42
Idle time ignored in WIP calculations
5
4
4
5
6
6
6
2
1
1
2
3
3
3
8
7
7
8
9
9
9
Little’s Formula WIP = TPT / CT [Q] = [T] / [T /Q ].
3 = 12 / 4
[units] = [min] / [min / unit]
Stage 1
Stage 2
M3 4 min M2
6 min
M1 10 min
Mgmt 361
Module 2 - Process Analysis
4
Run Time Parameters
Other Examples of When Little’s Law Can be Used
Mgmt 361 Module 2 - Process Analysis 5
Parameter relationships
When a process becomes stable, there is a relationship between process parameters – Little’s formula.
WIP = Flow time * Flow rate
= FT * FR
[Q ] = [ T ] * [ Q /T ]
WIP = Flow Time / Cycle Time
= FT / CT = TPT / CT
[ Q ] = [ T ] / [T /Q ]
Inventory = Throughput rate * Flow time
Quantity [Q]
Time [ T ]
Cost [$]
Capacity, output rate, input rate: [Q/T]
Cycle time: [T/Q]
WIP [Q]
Throughput Time - TPT : [ T ] Mgmt 361 Module 2 - Process Analysis 6
1
2
3
Flow time = 45 [customers] / 100 [customers / hr]
Flow time = 0.45 hours
= 0.45 [hours] * 60 [min / hr]
= 27 minutes
200 customers arrive in a two-hour period. 45 customers inside. What is the time spent in the facility?
Ex: 2.8
Flow rate = 100 [cust. / hr]
WIP = 45 [cust.]
flow time = ?
Little’s formula for processes with variable processing times / arrival times.
WIP = Flow time* Flow rate
Mgmt 361 Module 2 - Process Analysis 7
2
Flow time = WIP/ Flow rate
Mgmt 361 Module 2 - Process Analysis 8
Ex: 2.9
6000 claims per year (50 Weeks). Processing time 2 weeks. # of applications in the process?
Flow rate =
=
Flow time =
WIP = ?
WIP =
=
2
WIP = Flow rate * Flow time [Q] = [Q / T ] * [T ]
$100M per year. Accounts receivables: $15M.
Payment days?
Ex: 2.10
Mgmt 361 Module 2 - Process Analysis 9
2
Mgmt 361 Module 2 - Process Analysis 10
Ex: 2.11
15 customers waiting. Processing 2.5 customers per minute. How long will the customer wait?
Flow rate = 3000 [$] / (1/6) [year] = 18,000 [$] / [year]
WIP = 3000 [$]
Flow time = 1/6 [year]
Average balance $3000. Money turned over 6 times per year (every 2 months). How many dollars per
year?
Ex: 2.12
Mgmt 361 Module 2 - Process Analysis 11
WIP = Flow rate * Flow time
Flow rate = WIP / Flow time
2
Assembly Operations Example 2.13: What is the Process Capacity?
M1 9 min
AA
BB
M2 5 min
X1 3 min
Y1 5 min
P1 4 min
R1 16 min
R2 18 min
R3 15 min
Stage 1
Stage 2
Stage 4
Stage 3 Stage 5
St Cycle Time
1
2
3
4
5
3.21 [min/unit]
3.00 [min/unit]
5.00 [min/unit]
4.00 [min/unit]
5.41 [min/unit]
Bottleneck
AA BB
Process Capacity = 1/5.41 [u/min] * 60 [min/h] 11 units / hour
WIP cannot be determined, different product after assembly !
Mgmt 361 Module 2 - Process Analysis 12
Assembly Operations Example 2.13 – continued
M1 9 min
AA
BB
M2 5 min
X1 3 min
Y1 5 min
P1 4 min
R1 16 min
R2 18 min
R3 15 min
Stage 1
Stage 2
Stage 4
Stage 3 Stage 5
Minimum TPT ?
From Stage 1, you can get at the end of 5 minutes
From Stages 2 and 3, you can get at the end of 8 minutes
Both are required for the assembly. So you can start assembly after 8 minutes. Add 4 minutes on P1 then 15 minutes on R3.
AA BB
Mgmt 361 Module 2 - Process Analysis 13
Some Insights and Summaries
1. Capacity of a process = maximum output rate.
2. If input rate > capacity, process will not become stable. (TPT will
become higher and higher.)
3. To increase capacity of process, add resources at the bottleneck stage.
Careful: This will shift bottleneck to another stage !
4. Two schedules can have same cycle time (output rate) but different
WIP and TPT.
5. If an important job is to be completed as fast as possible: Free up
fastest resources at each stage; time required = minimum TPT.
6. Analysis for nondeterministic cases complex. Since processing times
are variables, bottleneck operations can shift frequently.
Mgmt 361 Module 2 - Process Analysis 14
