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OPIM 201 BUSINESS PROCESSES Homework 2 Instructor: Zhou Yangfang (Helen) GROUP MEMBERS: DIAN ATIQA DAYANA BINTE MD PAWAZI NGUYEN HUY PHUC NOELLE CHUA PEI YI SESSION: G1- MONDAY, 815 AM GROUP: 05
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Page 1: Hw 2

OPIM 201 BUSINESS PROCESSES

Homework 2

Instructor: Zhou Yangfang (Helen)

GROUP MEMBERS: DIAN ATIQA DAYANA BINTE MD PAWAZI NGUYEN HUY PHUC

NOELLE CHUA PEI YI

SESSION: G1- MONDAY, 815 AM

GROUP: 05

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Question 1 Scenario 1 Average CT: (3+3)/2 = 3mins/unit Average Flow Rate: ⅓ unit/min Scenario 2 Average CT: (3+4)/2=3.5 mins/unit Average Flow Rate: 1/ 3.5 = 0.2857 unit/min Scenario 3 Average CT: 3 mins/unit Average Flow Rate: ⅓ unit/min Average Flow Rate in descending order → 1 & 3, 2 Although there is a degree of uncertainty in scenario one, the discreteness of the uncertainty causes the cycle time to be 3 minutes/unit regardless of the time taken in Task B. Thus, the average flow rate in Scenario 1 and Scenario 3 is the same. Question 2

a) Desired Cycle Time: (8X60X60)/370 = 77.83 ≈ 70s (Rounded down to nearest multiple of 10s) b) RPW:

A B C D E F G H

220 55 180 75 80 30 60 5

In descending order: A, C, E, D, G, B, F, H Station Tasks Task Time (s) Eligible Tasks

1 A,C 40+30=70 A, B, C, D, E

2 E,D 20+45= 65 B, D, E, G, F

3 G 55 B, G, F

4 B 50 B, F

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5 F,H 25+5=30 F, H Line Efficiency = 220 / (70X5) = 0.7714 = 77.14% c) Using Trial & Error, we maximise the task time for each station (70s), Station Tasks Task Time (s) Eligible Tasks

1 A,C 40+30=70 A, B, C, D, E

2 B,E 50+20= 70 B, D, E, G

3 D,F 45+25=70 D, G, F

4 G,H 55+5=60 G, H

Question 3

a) Ṝ = 0.36 UCL = D4Ṝ = 1.864 x 0.358 = 0.67 LCL = D3Ṝ = 0.136 X 0.358 = 0.05

Sta$on  1:  A  +  C  

Sta$on  2:  E  +  B  

Sta$on  3:  D  +  F  

Sta$on  4:  G  +  H  

Sample Observations within Sample Sample Average

Sample Range

1 2 3 4 5 6 7 8 1 8.08 7.92 7.93 8.08 7.82 8.03 7.91 7.9 7.95875 0.26 2 8.33 8.23 7.87 8.06 8.13 8.31 7.98 8.03 8.1175 0.46 3 8.13 7.89 8.13 8.14 8.14 8.14 8.11 8.12 8.1 0.25 4 7.99 7.87 8.12 7.88 8.13 7.88 7.9 7.83 7.95 0.3 5 7.94 7.98 7.81 8.11 8.14 7.88 8.02 7.68 7.945 0.46 6 7.87 7.76 8.1 8.14 8.18 7.89 7.82 8.12 7.985 0.42

Grand Average 8.009375 0.358333333

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As seen from the graph above, all the points are either below or above the UCL and LCL respectively. Therefore, the range of the process is in statistical control.

b) =𝑋   = 8.01 3 𝜎 = A2Ṝ = 0.373 X 0.358 = 0.13 UCL =

=𝑋   + 3 𝜎 = 8.009375 + 0.133534 = 8.14

LCL = =𝑋   - 3 𝜎 = 8.009375 - 0.133534 = 7.88

As seen from the graph above, all the points are either below or above the UCL and LCL respectively. Therefore, the range of the process is in statistical control.

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Question 4

a) CP  ≥ 1.3

!"#!!"!!!

≥ 1.3

8.5− 7.56 1.3 ≥ 𝜎

𝜎   ≤ 0.128 (3sf)

Therefore, the largest value of standard deviation to achieve process capability ratio of 1.3 is 0.128.

b) 𝜎 = Ṝ!!

= !.!"#!!.!"#

= 0.126

c) CP  =𝑈𝑆𝐿−𝐿𝑆𝐿

6𝜎 =   8.5−7.56 0.126 = 1.32

d) CPk  = min{𝑈𝑆𝐿−=𝑋  3𝜎 ,

=𝑋−𝐿𝑆𝐿3𝜎 }

           =  min{ 8.5−8.0093753(0.12585) ,8.009375  −  7.53(0.12585) } = min{ 1.29, 1.34} = 1.29

e) In order to achieve 6-sigma quality • The current Cpk is less than the Cp. Therefore, the nominal value of product specification must be

made closer to the process mean. Hence 8.01 must be reduced to 8.00. • As A =

=𝑋  , Cp = Cpk hence we only need to make sure that Cp = Cpk = 2 to achieve 6𝜎

2 = !6𝜎 ⇒ 𝜎 = !

!∗!=   !

!"= 0.0833

Process deviation must be reduced to 0.0833

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Question 5

The specification limit for late flights is 2%.

𝑝 = 2.4222%

𝜎 = 𝑝(1 − 𝑝)/𝑛 =   0.024222 ∗ (1 − 0.024222)/300 = 0.008876

𝑈𝐶𝐿 = 𝑝 + 3𝜎 = 0.05085

𝐿𝐶𝐿 = 𝑝 − 3𝜎 = −0.002406

We can tell the management of SIA that:

• The process is not in statistical control as there are some outliers outside the process control limits. The process may have assignable causes of variation and management needs to find out that cause to improve conformance quality of the process.

• There are a lot of occasions where the number of late flights exceeds the specification limit. All the data points above the 2% horizontal line are below the standard of 98% SIA sets. Therefore SIA needs to improve its performance quality. There are common causes of variations which are unavoidable with the current process. SIA needs to change the current process in order to improve on-time performance.


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