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# Production & Operation Management

Submitted By ckgini
Words 1195
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Production & Operations
Management
Session 3-2
More on Processes

1

Outline
 Multi-product, multi-flow process analysis
– So far: 1 product, 1 flow
– Differing process times, yield issues, machine breakdown

 Big Takeaway:
– Product-mix becomes critical in multiple flows
– Implications in capital investment, scaling business, and risk management  Calculating capacity when you have
– Multiple flows
• With the same processing time at each resource
• With different processing times at a single resource

– Yield issues
– Machine Breakdown
2

Measure: Implied Utilization
 Implied Utilization captures the mismatch between the capacity requested from a resource by demand and the capacity currently available at the resource

Capacity requsted by demand
Implied utilization =
Available capacity

3

Ex. 1 : Furniture Factory (I)
Chair
Assembly Line

PC
SC

Paint Shop

Inspection

Table
Assembly Line

PT
ST

Stain Shop

Four Products
– Painted Chairs (PC), Stained Chairs (SC), Painted Tables (PT), Stained Tables (ST)

Suppose

Chair Assembly Line : 2 min / chair
Table Assembly Line : 6 min / table
Paint Shop : 4 min / unit
Stain Shop : 3 min / unit
Inspection : 1 min / unit

Product

Demand
(units/hr)

% of Demand

PC

8

26.67%

SC

11

36.67%

PT

6

20.00%

ST

5

16.67%

What is capacity/implied utilization of each resource?
4

Capacity/Utilization (Furniture I)
 Chair Assembly
• Capacity =(1/2) unit/min*60 mins/hr=30 units/hr
• Implied Utilization =Demand/Capacity=(8+11)/30=0.63 < 1

 Table Assembly
• Capacity=(1/6) unit/min*60 mins/hr=10 units/hr
• Implied Utilization=Demand/Capacity=(6+5)/10=1.1 which is >1

 Paint Shop
• Capacity=(1/4) unit/min*60 mins/hr=15 units/hr
• Implied Utilization=Demand/Capacity=(8+6)/15=0.93 < 1

 Stain Shop
• Capacity=(1/3) unit/min*60 mins/hr=20 units/hr
• Implied Utilization=Demand/Capacity=(11+5)/20=0.80 < 1

 Inspection
• Capacity=1 unit/min*60 mins/hr=60 units/hr
• Implied Utilization=Demand/Capacity=(8+11+6+5)/60=30/60=0.5 < 1
5

Furniture Factory (I)
 Utilization cannot be larger than 100%!
• Factory cannot meet all the demand (capacity constrained) • How many can it serve (supposing product mix is constant)?  Factory capacity
• Let F be the factory capacity in units/hr
• The Table Assembly Line accounts for (5+6)=11 units out of a total of 30 units of demand, i.e., (11/30) of demand
• Its capacity is 10 units/hr and is the bottleneck
• So, (11/30)*F=10 units/hr
• F=27.27 units/hr
6

Calculating capacity
 Step 1. Find the bottleneck.

Need to know demand!!!

Single Flow

Multiple Flows

Bottleneck is resource with lowest capacity

Bottleneck is resource with highest utilization

For a single flow, the lowest capacity resource is the same as highest utilization resource

 Step 2. Bottleneck governs the process capacity.
Single Flow

Multiple Flows

Process capacity equals the bottleneck resource capacity

Depends on the bottleneck resource AND the demand mix 7

Your Turn: Change in Demand Mix
 Suppose overall demand remains the same, but the demand-mix changes
Product

Demand (units/hr)

% of Demand

PC

11

36.67%

SC

10

33.33%

PT

7

23.33%

ST

2

6.67%

 Which resource is the bottleneck?
 What is the capacity of the factory in units/hr?

8

A fortunate coincidence
PC
SC

Chair
Assembly Line

Paint Shop

Inspection

PT
ST

Table
Assembly Line

Stain Shop

 Furniture example was special
– Even though two different products at Paint Shop (chairs/tables), both take same time to paint
– Same with Stain shop

 Generally, this won’t be the case
– Different processing times
9

Recall: capacity with one processing time
 Workstation with a single product. Processing time: 10 min / unit

10 mins

10 mins

10 mins

10 mins

10 mins

 Flow Time (FT) = processing time= 10 mins for all units
 Workstation capacity = Flow Rate
= (1/10) units/min*60 mins/hr = 6 units/hr
10

Ex. 2: Capacity with two processing times
 Workstation makes two products
– Product 1 processing time 10 min / unit
– Product 2 processing time 12 min / unit

10 mins

10 mins

10 mins

12 mins

 Product mix is 3:1, i.e. (3/4)th of the output is Product 1 and
(1/4)th of the output is Product 2
 What is the capacity of the workstation?
11

Ex. 2: Capacity with two processing times
 PT1=10 mins/unit
 PT2=12 mins/unit
 Think of a “typical” unit:
– Weighted average of FT1 and FT2
= (3/4)*10+(1/4)12=10.5 mins/unit
– NOTE THIS RELIES ON DEMAND MIX

 Capacity of Resource
= (1/10.5)*60min/hr = 5.71 units/hr
Same as capacity of process because only one resource
12

Ex. 2 (Alternate way)
 An alternate method is to consider time to make a cycle, i.e. 3 units of product 1 and one unit of product
2
– Total Processing Time for a cycle = 3*10 + 1*12 = 42 mins
(see picture on previous slide)
– average processing time per unit = 42/4 = 10.5 mins/unit

 Workstation capacity
= (1/10.5)units/min*60 min/hr
= 5.71 units/hr
13

Your Turn: Capacity single product with breakdowns  Single product processing time:10 min/unit

10 mins

10 mins

10 mins

10 mins

10 mins

 Workstation breaks down every 5 hrs (on average). Repair time is 1 hr.
 What is the process capacity?
– Hint: Think about this as if it were two different products: the toy, and repair 14

Ex. 4 Capacity when there are yield issues
 Littlefield Labs is a specialized laboratory that performs low volume chemical tests for pharmaceutical companies.
For one of their products, there is a two-stage process.
– In the first stage (Centrifuge) chemicals are mixed into 125 vials and placed into a centrifuge. For each vial, it takes a technician about one minute to do the mixing and loading. Once all of the vials are in the centrifuge, it is operated for 10 minutes.
– The second stage (Testing) tests the vials for purity. This testing equipment requires 75 minutes to calibrate, and then a technician can test two vials per minute, for the next 125 vials.
Currently, 40% of the vials on average fail this test; i.e, the yield is
– What is the capacity of the process in ”good vials”/hour?

15

Finding capacity when there are yield issues-2  Centrifuge
• Total Processing Time
= 125*1 + 10 =135 mins for 125 vials

 Testing
• Total Processing Time
= 75 + 125*0.5 = 137.5 mins for 125 vials
• 40% fail  yield=60%  only 0.6*125=75 “good vials” are produced in 137.5 mins (50 vials are discarded)
 Capacity of Testing = (75/137.5)*60
= 32.73 “good vials”/hr
16

Finding capacity when there are yield issues-3  Note that this implies that the Cetrifuge produces only 75 “good vials” in 135 mins
 FR of Centrifuge in “good vials”/hr
= (75/135)*60 = 33.33 “good vials”/hr
 Testing is the bottleneck (MIN(33.33, 32.73))
 System capacity = 32.73 “good vials”/hr

17

Wrap-up
 We saw how to calculate process capacity in multiflow systems with different processing times
– Change in demand, risks?
– Capital investment?
– Revenue stream?

18

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