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

2. A Second Example on Throughput Part 2a
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 = 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
What is the average CWF
11.9(1-CWF) = 9  = 11.9CWF  CWF = 0.24

3. A Second Example on Throughput Part 2a
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) = Min(0.175, 0.3, 0.35) = /hr Min[1/(4/0.7),1/(3/0.9), 1/(2/0.7)]=Min(0.7/4,0.9/3,0.7/2) = /hr or 8(0.175)= 1.4 per day 5. Compute the cycle time. 1/1.4 = day or 8(0.7143) = 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

4. A Second Example on Throughput Part 2a
U = Throughput/Capacity 0.8 = R/1.4 Throughput = 1.12 per day 7. Compute the takt time. 1/1.12 = day OR 8(0.893) = 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 1.4, 2.4, 2.8 per day.

5. A Second Example on Throughput Part 2a
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

6. Throughput Improvement Mapping
Throughput ≤ Effective Capacity ≤ Theoretical Capacity
Throughput << Capacity – External Bottleneck
External blockage (demand) - ↑sales efforts, ↑ advertising budget, ……. Make them an offer they can’t refuse. ↓prices, ↑quality, ↓time ↑variety.
External starvation (supply) - identifying additional suppliers, more reliable suppliers.
Throughput = Capacity – Internal Bottleneck
Increase financial capacity - modifying the product mix.
Increase physical capacity - ↑ c ↓ Tp
if Capacity ≈ Theoretical Capacity
If Capacity << Theoretical Capacity.

7. 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.

8. 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.

9. 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.

10. 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

11. Setup Batch Size: Throughput or Flow Time
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

12. Throughput or Flow Time
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)

13. Throughput & Cost or Flow Time
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)/2/2 = 30 = 0.5 hour
Capacity = 8/0.5 = 16 aggregate units per day.
Each aggregate unit is 0.5 A and 0.5 B (total of 8A and 8B)
8A and 8B need 4*30*2 = 4 hrs Setup + 8*10+8*20 = 4 hrs Production,