1. Throughput Key Problems
Effective Capacity
&
Utilization
Based on the book: Managing Business Process Flows.

2. Please read the material up to slide 16
Please read the material up to slide 16. The solution to the problem on slides 9-15 is on slide 8.

3. Problem 1
Example: The average arrival rate to a GAP store is 6 customers per hour. The average service time is 5 min per customer.
R = 6 customers per hour
Rp =1/5 customer per minute, or 60(1/5) = 12/hour
U= R/Rp = 6/12 = 0.5
a) How long does a customer stay in the processor (with the server)?
Tp = 5 minutes
b) On average how many customers are there with the server?
RTp = Ip = 6(5/60) = 0.5
Alternatively; Ip = cU =1(0.5) = 0.5

4. Problem 2
What if the arrival rate is 11 per hour? Processing rate is still Rp=12
U= R/Rp
U=11/12
a) How long does a customer stay in the processor (with the server)?
Tp = 5 minutes
b) On average how many customers are there with the server?
RTp = Ip = (11/60)(5) = 11/12
Alternatively
Ip = cU = 1(11/12) = 11/12

5. Problem 3
A local GAP store on average has 10 customers per hour for the checkout line. The store has two cashiers. The average service time for checkout is 5 minutes .
Arrival rate: R = 10 per hour
Average inter-arrival time: Ta = 1/R = 1/10 hr = 6 min
Average service time: Tp = 5 min
Number of servers: c =2
Rp = c/Tp = 2/5 per min or 24 per hour
U= R/Rp = 10/24 = 0.42

6. Problem 3
a) How long does a customer stay in the processors (with the servers)?
Average service time: Tp = 5 min
b) On average how many customers are there with the servers?
Ip =?
RTp = Ip = (1/12)(10) = 0.84
Alternatively
Ip = cU = 2(0.42) = 0.84

7. Problem 4
A call center has 11 operators. The average arrival rate of calls is 200 calls per hour. Each of the operators on average can serve 20 customers per hour. U = 200/220 =0.91.
a) How long does a customer stay in the processors (with the servers)?
Average service time: Tp = 60/20 = 3 min
b) On average how many customers are there with the servers?
Ip =?
RTp = Ip = (200/60)(3) = 10
Alternatively
Ip = cU = 11(200/220) = 10

8. Lecture on the Next Problem

9. Problem 4. Problem 5.1 in the book
A law firm processes shopping centers and medical complexes contracts. There are four Paralegals, three Tax lawyers, and two Senior partners. The unit loads of the resources to handle one standard contract is given below. Assume 8 hours per day, and 20 days per month.
It takes a Paralegal 20 hours to complete 3 contract.
That is 20/3 = hours to complete a contract.
It takes a Tax lawyer 2 hours to complete a contract.
It takes a Senior partner 2 hours to complete a contract.
a) What is the Theoretical Flow Time of a contract? =
Flow Time = Theoretical Flow Time + Waiting times
Flow Time = Waiting times
Compute the Capacity of each of the three Resource Pools

10. Paralegals Operation 1 Operation 2 Operation 3 Tp=20/3 hr., c= 4
Tax Lawyers
Senior Partner
A Paralegal can complete 1 contract in 20/3 = hour
How many contracts in one hour? 1/ = 0.15
How many contracts all the Paralegals can complete in one hour.
There are 4 Paralegals: c = 4
Four Paralegals 4(0.15) = 0.6 contracts per hours
We could have also said Tp =
Capacity of one resource unit is 1/Tp.
Capacity of one resource unit is 1/6.667 = 0.15.
Capacity of all resource units: Rp=c/Tp where c=4 and Tp = 6.667
Rp = 4/6.667 = 0.6 per hour
Capacity of the resource pool is 0.6 contracts per hour.
It is 8(0.6) = 4.8 contracts per day

11. Tax Lawyers A Tax Lawyer can complete 1 contract in 2 hour
How many contracts in one hour?
1/2 = 0.5
How many contracts all the Tax Lawyers can complete in one hour.
There are 3 Tax Lawyers: c = 3
There Tax Lawyers 3(0.5) = 1.5 contracts per hours
We could have also said Tp = 2.
Capacity of one resource unit is 1/Tp.
Capacity of one resource unit is 1/2 = 0.5.
Capacity of all resource units: Rp=c/Tp where c=3 and Tp = 2
Rp = 3/2 = 1.5 per hour
Capacity of the resource pool is 1.5 contracts per hour.
It is 8(1.5) = 12 contracts per day

12. Senior Partners A Senior Partners can complete 1 contract in 2 hour
How many contracts in one hour?
1/2 = 0.5
How many contracts all the Senior Partnerss can complete in one hour.
There are 2 Senior Partnerss: c = 2
There Senior Partnerss 2(0.5) = 1 contracts per hours
We could have also said Tp = 2.
Capacity of one resource unit is 1/Tp.
Capacity of one resource unit is 1/2 = 0.5.
Capacity of all resource units: Rp=c/Tp where c=2 and Tp = 2
Rp = 2/2 = 1 per hour
Capacity of the resource pool is 1 contracts per hour.
It is 8(1) = 8 contracts per day

13. Problem 4. Problem 5.1 in the book
c) Compute the capacity of the process.
d) Compute the cycle time?
4.8 units in 8 hours.
Cycle time = 8/4.8 = 1.67 hours
Alternatively 1/0.6 =1.67 hours

14. Problem 4. Problem 5.1 in the book
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

15. Problem 4. Problem 5.1 in the book
Inventory with the processors is = 2
On average there are 2 flow units with the processors; Inventory 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) There are 150 cases in November can the company process all 150 cases?
150/20 = 7.5 per day  4.8 (Capacity) < 7.5 (Demand).

16. 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?
# 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, and 2. We have 4, 3, and 2. 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 (8) = 2 hours total over time.
There will be 6 paralegals; over time pf each = 2/6 = 1/3 hour
Or 20 minute per paralegal. PLUS some safety Capacity.

17. A Little More Practice
Now suppose throughput is 3.6 contracts per day.
Compute the average inventory.
Station Capacity Throughput Utilization
Station /4.8 = 0.75
Station /12 = 0.3
Station /8 = 0.45
Managerial Observation: Note that the utilization of bottleneck resource is not necessarily 100%.
On average
0.75 person with a resource in Station 1
0.3 person with a resource in Station 2
0.45 person with a resource in Station 3

18. A Little More Practice Station Rp R U c Ip
Lets check it through the Little’s Law
Theoretical Flow Time = =
RT=I  3.6( /8) = I  I = 4.8
(3.6/8)( ) = 4.8
Now suppose there are 16.8 contracts are waiting in different waiting lines? What is the Flow Time ?
RT=I 3.6*T= = 21.6  3.6T = 21.6 T=6 days

19. 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.

20. ThCapacity, Capacity, VThFlowTime, ThFlowTime, FlowTime
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.
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 ThUL(1+CWF) = Unit Load (Tp),
30 mins.

21. ThCapacity, Capacity, VThFlowTime, ThFlowTime, FlowTime
Unit Load Activity time
Capacity Theoretical Flow Time
10 mins.
30 mins.
Flow Time Capacity does not care about buffer times
3 days
10 mins.
30 mins.
Theoretical Unit Load Theoretical Activity Time
Theoretical Capacity Very Theoretical Flow Time
30 mins.

22. 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)

23. Problem 4. Problem 5.1 in the book=CWF
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.

24. Problem 4. Problem 5.1 in the book-CWF
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

25. Problem 4. Problem 5.1 in the book-CWF
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

26. A Second Example on Throughput Part 2a
The theoretical unit loads for an aggregate policy (combination of all policies) in a law firm are: 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. 1. 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 UL(1-CWF) = ThUL 11.9(1-CWF) = 9  = 11.9CWF  CWF = 0.24

27. 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 if process utilization is 80%. Capacity = 1.4 per day U=0.8

28. 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.

29. 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

30. Multiple Choice Which of the following 2 statements is true?
I. A process can have more than 1 bottleneck resource.
II. Having flexible equipment can increase utilization.
Only I
Only II
Both I and II
Neither I nor II
Cannot be determined
Which of the following statement is false?
Throughput rate is always smaller than or equal to the capacity
Customers may wait even if the utilization rate of the service process is smaller than 100%
Bottleneck resource(s) always has 100% utilization rate
Increasing WIP may increase utilization rate
None of the above

31. Multiple Choice To improve the utilization rate, we can
I. Cross-train the workers
II. Adopt flexibility equipment
III. Shift from MTS systems to MTO system
Choose the most appropriate. I II
III
I and II
I, II, and III

32. 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.

33. 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.

34. 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.

35. 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.
Stop

36. 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

37. 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

38. 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)

39. 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,