1. Based on the book: Managing Business Process Flows.
Throughput Part 2
Effective Capacity
&
Throughput Leverages
Based on the book: Managing Business Process Flows.

2. 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.

3. ThCapacity, Capacity, ThFlowTime, FlowTime
Theoretical Unit Load, Theoretical Activity Time
Theoretical Capacity is computed based on the Theoretical Unit Load (ThUL)
Theoretical Flow Time is NOT computed based on Theoretical Activity Time
Very Theoretical Flow Time is computed based on theoretical Activity Time
30 mins.
ThUL(1+CWF) = Unit Load (Tp), Activity time
Capacity is computed based on the Unit Load
Theoretical Flow Time is computed based on Activity Time
Then What is Flow Time?
10 mins.
30 mins.
Flow Time Ti + Tp
Flow time includes time in buffers
Capacity does not care about buffer times
3 days
10 mins.
30 mins.

4. 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)

5. 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. CWF at the three resource-s are 25%, 0%, and 50%, respectively.

6. Problem 4. Problem 5.1 in the book
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) =
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

7. 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
b) Compute the cycle time?
4.8 units in 8 hours.
Cycle time = 8/4.8 = 1.67 hours

8. Problem 4. Problem 5.1 in the book
c) Compute the flow time.
Vey Theoretical Flow time = = 8 !!!!!!
Theoretical Flow Time = 10.67
Flow Time = Waiting times
d) Compute the average inventory.
Lets look at the utilization of the 3 stations
Station Capacity Throughput Utilization
Station /4.8 = 1
Station /12 = 0.4
Station /8 = 0.6
On average 1 person with a resource in Station 1, 0.4 person with a resource in Station 2, and 0.6 person with a resource in Station 3

9. Problem 4. Problem 5.1 in the book
Inventory with the processors is = 2
On average there are 2 flow units with the processors; Inventoty in the processors (Ii)
Now let’s look from another angle; from the Little’s Law point of view
RT=I  R= 4.8 per 8 hours or 0.6 per hour
T =10.67 hours  I = 0.6(10.67) = 6.4
6.4 vs 2? Where is my mistake??
1(4)+ 0.4(3)+0.6(2) = 6.4
e) Can the company process all 150 cases in November?
4.8 (20) = 96 < 150.
150/20 = 7.5. The effective capacity of 4.8 /day is not sufficient.

10. Problem 4. Problem 5.1 in the book
f) If the firm wishes to process all the 150 cases available in November, how many professionals of each type are needed?
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
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 = minutes over time.

11. Do you Really Need Additional Problem?
A law firm handles several types of insurance policies. The theoretical unit loads for an aggregate policy (representing the combination of all policies) are as follows: 4 hours of Resource 1, 3 hours of Resource 2, and 2 hours of Resource 3. Capacity waste factor for resources 1 to 3 is 0.3, 0.1, and 0.3, respectively. Utilization of the bottleneck resource is 0.8. There are 8 working hours per day.
Compute the VERY theoretical flow time for an aggregate product.
4+3+2 = 9 hours
2. Compute the theoretical flow time for an aggregate product.
4/(1-0.3) + 3/(1-0.1) +2/(1-0.3) = 4/0.7+3/0.9+2/0.7
= = 11.9

12. Do you Really Need Additional Problem?
3. Compute the daily theoretical capacity of the process. Min(1/4,1/3,1/2) = 1/4 per hour 8(1/4) = 2 per day 4. Compute the daily capacity of the process. Min(1/5.71,1/3.33,1/2.86) = 1/5.71 /hr or 8(1/5.71)= 1.4 per day Min(0.175, 0.3, 0.35) = /hr or 8(1/5.71)= 1.4 per day 5. Compute the cycle time. 8/1.4 = 5.71 hour OR 1/0.175 = 5.71 hour OR 1/(1/5.71) = 5.71 hour 6. Compute the throughput per day. Capacity = 1.4 per day U=0.8

13. Do you Really Need Additional Problem?
U = Throughput/Capacity 0.8 = R/1.4 Throughput = 1.12 per day 7. Compute the takt time. 1/1.12 day OR 8/1.12 = 7.14 hours 8. Compute the utilization of the most utilized resource. The problem has already said that the utilization of the process is 0.8. That is the utilization of the most utilized resource. That is utilization of the bottleneck. 9. Compute the utilization of the least utilized resource. The three resources have capacity of 0.175, 0.3, 0.35 per hour OR OR 1.4, 2.4, 2.8 per day.

14. Do you Really Need Additional Problem?
Throughput is 1.12 per day. U1 = 1.12/ 1.4 =0.8 U2 = 1.12/ 2.4 =0.47 U3 = 1.12/2.8 = On average how many flow units are with the resources (in the processors) There are three resources with U1 = 0.8, U2=0.47 , U3 = = Suppose the number of flow units in all the waiting lines are 10 units. Compute the flow time. (A day is 8 hours.) R = 1.12 per day I = = RT = I  1.12T =  T = 11.67/1.12 = 10.42

15. 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
Setup batch (also lot size): number of units processed consecutively after a setup
EXAMPLE: painting cars->how many cars before you change paint color

16. 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.
Setup batch (also lot size): number of units processed consecutively after a setup
EXAMPLE: painting cars->how many cars before you change paint color

17. 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
Operation B
20 min/unit
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)

18. Capacity Analysis; Multiple Products - Single Stage
Compute the capacity under min flow time strategy.
In a minimum inventory strategy, we produce two product at a time then switch to the other.
10 min/unit A
Operation B
20 min/unit
1 machine
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

19. 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.

20. 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.

21. 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.

22. 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 non-bottleneck 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.

23. Total Unit Load for Product mix
STOP
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.

24. Effective Capacity of a Resource Unit
STOP
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