Throughput Part 2

Transcription

Throughput Part 2
Throughput Part 2
Effective Capacity
&
Throughput Leverages
Based on the book: Managing Business Process Flows.
Capacity Waste and Theoretical Capacity



Effective capacity of a resource unit is 1/Tp. Unit load Tp , is an
aggregation of the productive as well as the wasted time.
Tp includes share of each flow unit of capacity waste and
detractions such as
 Resource breakdown
 Maintenance
 Quality rejects
 Rework and repetitions
 Setups between different products or batches
We may want to turn our attention to waste elimination; and
segregate the wasted capacity. Theoretical capacity is the
effective capacity net of all capacity detractions.
Throughput-Part 2
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Capacity Waste Factor and Theoretical Capacity
An operating room (a resource unit) performs surgery every 30 min,
Tp = 30 min. Tp includes all the distracts. We also refer to it as the
Unit Load.
Effective capacity is 1/30 per min or 60/30 =2 per hour.
On average, 1/3 of the time is wasted (cleaning, restocking,
changeover of nursing staff and fixing of malfunctioning
equipment ).
Capacity Waste Factor (CWF) = 1/3.
Theoretical Unit load = Tp*(1-CWF) =30(1-1/3) = 20 min.
Tp = Unit Load = ThUnit Load /(1-CWF) = 20/(1-1/3) = 30
Theoretical Capacity = c/ThUnit Load
Effective Capacity = Capacity = c/Unit Load.
Theoretical Capacity = 1/20 per minute or 3 per hour.
Effective Capacity = Theoretical Capacity (1-CWF)
Throughput-Part 2
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Problem 4. Problem 5.1 in the book
A law firm processes (i) shopping centers and (ii) medical
complexes contracts. The time requirements (unit loads) for
preparing a standard contract of each type along with some
other information is given below. In November 2012, the firm
had 150 orders, 75 of each type. Assume 20 days per month, and
8 hours per day. Capacity Waste factor at the three resource-s
are 25%, 0, and 50%, respectively.
Paralegal
Unit Load Shopping Unit Load Medical No. Of
(hrs /contract)
(hrs /contract)
Professionals
4
6
4
Tax lawyer
1
3
3
Senior partner
1
1
2
Throughput-Part 2
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Problem 4. Problem 5.1 in the book
a) What is the effective capacity of the process (contracts /day)?
Paralegal: Theoretical Unit Load (50%Sh 50% Med):
0.5(4)+0.5(6) = 5 hrs
Theoretical Capacity = 1/5 per hr
Capacity Waste Factor (CWF) = 0.25
Unit Load = Tp = 5/(1-0.25) = 20/3 hrs
Effective Capacity = Capacity = 1/(20/3) = 3/20 per hr
Tax Lawyer: Theoretical Unit Load 0.5(1)+0.5(3) = 2 hrs
CWF = 0
Theoritical Unit Load = Tp = 2 hrs
Theoretical Capacity = 1/2 per hr
Effective Capacity = Capacity = 1/2 per hr
Throughput-Part 2
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Problem 4. Problem 5.1 in the book
Senior Partner: Theoretical Unit Load 0.5(1)+0.5(1) = 1 hrs.
Theoretical Capacity = 1/1 = 1 per hr
CWF = 0.5
Unit Load = Tp = 1/(1-0.5) = 2 hrs
Effective Capacity = Capacity = 1/2 per hr
ThUnit
Load
SH
(hrs)
ThUnit
Load
MD
(hrs)
Unit
Load
50%SH
50%MD
Teoretical
Capacity of
a Resource
Unit /hr
Unit
Load
CWF
50%SH
50%MD
Capacity
# Of
of a
Resourse
Resource
Units
Unit /hr
Th Capacity
of the
Resource
Pool/hr
Capacity of
the
Resource
Pool/hr
Paralegal
4
6
5
0.2
0.25
6.667
0.15
Tax
lawyer
1
3
2
0.5
0
2
Senior
partner
1
1
1
1
0.5
2
Cap of
Th Cap of R- the RPool / day Pool /
day
4
0.8
0.6
6.4
4.8
0.5
3
1.5
1.5
12
12
0.5
2
2
1
16
8
b) Compute the cycle time?
4.8 units in 8 hours.
Cycle time = 8/4.8 = 1.67 hours
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Problem 4. Problem 5.1 in the book
c) Compute the flow time.
Vey Theoretical Flow time = 5+2+1 = 8 !!!!!!
Theoretical Flow Time 6.67+2+2 = 10.67
Flow Time = 10.67 + Waiting times
d) Compute the average inventory.
RT=I  R= 4.8 per 8 hours or 0.6 per hour
T =10.67 hours  I = 0.6(10.67) = 6.4
e) Can the company process all 150 cases in November?
150/20 = 7.5. The effective capacity of 4.8 /day is not sufficient.
f) If the firm wishes to process all the 150 cases available in
November, how many professionals of each type are needed?
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Problem 4. Problem 5.1 in the book
Demand per day = 150/20 = 7.5
# of paralegals required = 7.5/1.2 = 6.25
# of tax lawyers required = 7.5/4 = 1.875
# of tax lawyers required = 7.5/4 = 1.875
These could be rounded up to 7, 2 and 2
We need 7, 2, 2. We have 4, 4,4. We may hire 3 additional
paralegals.
Alternatively, we may hire just 2 and have 6 paralegals.
They need to work over time for 0.25 paralegal who works 8 hrs
/day.
That is 1.5 hours  1.5/8 = 22.5 minutes over time.
Throughput-Part 2
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Setup Batch and Total Unit Load
Setup or Changeover: activities related to cleaning, resetting and
retooling of equipment in order to process a different product.
Qp : Setup batch or lot size; the number of units processed
consecutively after a setup;
Sp : Average time to set up a resource at resource pool p for a
particular product
Average setup time per unit is then Sp/Qp
Sp/Qp is also included in Tp
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Setup Batch Size
What is the “right” lot size or the size of the set up batch? Lot Size
 or  ?
 The higher the lot size, the lower the unit load and thus the
higher the capacity.
 The higher the lot size, the higher the inventory and
therefore the higher the flow time.
Reducing the size of the setup batch is one of the most effective
ways to reduce the waiting part of the flow time.
Load batch: the number of units processed simultaneously. Often
constrained by technological capabilities of the resource.
Setup batch: the number of units processed consecutively after a
setup. Setup is determined managerially.
Throughput-Part 2
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Capacity Analysis; Multiple Products - Single Stage
Product Mix:
50%-50%
Set-up time
30 min per product
Working hours
8 hours/day
10 min/unit
A
B
20 min/unit
Operation
1 machine
100% available
Compute the effective capacity under min cost strategy.
Two set-ups each for 30 min = 60 mins
An aggregate product takes (10+20)/2 = 15
Production time = 8*60-60 = 420 mins
Capacity = 420/15 = 28 aggregate units
Each aggregate unit is 0.5 A and 0.5 B (total of 14A and 14B)
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Capacity Analysis; Multiple Products - Single Stage
Compute the capacity
10 min/unit
under min flow time
A
strategy.
Operation
In a minimum inventory
B
strategy, we produce 20 min/unit
two product at a time
1 machine
then switch to the other.
100% available
Product A: 2(10)0+30 = 50
Product B: 2*(20)+30 = 70
An aggregate product takes (50+70)/2 = 60 = 1 hour
Capacity = 8/1 = 8 aggregate units.
Each aggregate unit is 0.5 A and 0.5 B (total of 4*2A and 4*2B)
8A and 8B need 4*30*2 = 4 hours + 8*10+8*20 = 4 hours
Throughput-Part 2
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Throughput Improvement Mapping
Throughput ≤ Effective Capacity ≤ Theoretical Capacity
 Throughput << Capacity – External Bottleneck


External blockage (demand) - ↓prices, ↑quality, ↓time
↑variety ↑sales efforts, ↑ advertising budget, …….
External starvation (supply of input) - identifying
additional suppliers, more reliable suppliers, modifying
the supply chain.
 Throughput = Capacity – Internal Bottleneck


Increase financial capacity - modifying the product
mix.
Increase physical capacity - ↑ c ↓ Tp
 if
Capacity ≈ Theoretical Capacity
 If Capacity << Theoretical Capacity.
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Feb-2014
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Increasing Resource Levels ↑ c
Capacity ≈ Theoretical Capacity  Resources are efficiently
utilized; increase the theoretical capacity.
 Increase the level of resources. ↑ c. buy one more oven
 Increase the size of resource units - Larger load batch - more
loaves in the oven
 Increase the time of operation – ↑ Scheduled Availability,
Overtime
 Subcontract or Outsource
 Technology- Speed up the activities rate - Invest in faster
resources or incentives for workers.
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Reducing Resource Capacity Waste ↓ Tp
Capacity << Theoretical Capacity  resources are not utilized
effectively; eliminate of waste; ↓ Tp or ↑ Net Availability.
 Eliminate non-value-adding activities.
 Avoid defects, rework and repetitions – These two are exactly
the same as what was stated for flow time reduction. For flow
time we focus on activities along the critical path. For flow rate
we focus on activities performed by bottleneck resources.
 Increase Net Availability – Reduce breakdown and work
stoppage by improved maintenance policies and effective
problem-solving, to reduce the frequency and duration of
breakdowns and maintenance outside of working hours.
Reduce absenteeism.
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Reducing Resource Capacity Waste
 Reduce setup time – setup time per unit is Sp/Qp. Decrease





the frequency of changeovers, reduce the time required for each
setup and manage the product mix to decrease changeover
time from one product to the next.
Move some of the work to non-bottleneck resources –
This may require greater flexibility on the part of nonbottleneck resources as well as financial investments in tooling
and cross-training.
Reduce interference waste – Eliminate starvation and blockage
among work-stations.
Methods improvement.
Training.
Management.
Throughput-Part 2
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Total Unit Load for Product mix
Compute the unit load and the total unit load for each Load batch of Regular tile,
Jumbo tile and a product mix of 75% Regular and 25% Jumbo . Load Batches are 4
and 9 for regular and jumbo, respectively. Set-up Batches are 300 and 100 for regular
and jumbo, respectively.
Product L -Batch S-Batch
Cutting time
Setup time
Regular 4 units 300 L-batch 2 min/L-batch 30 min/S-Batch
Jumbo 9 units 100 L-batch 1 min/L-batch 30 min/S-Batch
Unit Load (Tp )
Sp/Qp
Total unit load
Throughput-Part 2
Regular
2
30/300=0.1
2+0.1=2.1
Jumbo
1
Mix
(2×0.75)+(1×0.25)=1.75
30/100=0.3 (0.1×0.75)+(0.3×0.25)=0.15
1+0.3=1.3
(2.1×.75)+(1.33×.25)= 1.9
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Effective Capacity of a Resource Unit
Theoretical Capacity of a resource unit =
(1/Unit Load) × Load Batch × Scheduled Availability
Scheduled Availability – the scheduled time period during
which a resource unit is available for processing flow units.
Availability factor = Net Availability/Scheduled Availability
Effective Capacity of a resource unit =
(1/Total Unit Load) × Load Batch × Scheduled Availability
Effective Capacity of a pool =
(c/Tp) × Load Batch ×Scheduled Availability
Throughput-Part 2
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Feb-2014
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