IE-348: Assignment #7

Name: Date:

Instructions: Complete each of the questions below. Answers may be typed in the space provided. If needed, you can attach a separate Excel or Minitab file to your assignment submission (just make sure to refer me to that attachment in your answer).

Data Set

A loan agency has promised that decisions about loan approval will be made in 24 hours. The table below presents the data on the cycle time (in hours) to process loan applications.

SampleCycle Time

SampleCycle Time

SampleCycle Time

SampleCycle Time

1 16.3 11 22.0 21 19.3 31 16.4

2 16.3 12 14.7 22 14.6 32 18.2

3 19.3 13 18.0 23 17.8 33 19.4

4 15.1 14 18.9 24 15.6 34 14.1

5 22.2 15 19.1 25 22.5 35 16.4

6 19.1 16 10.6 26 17.6 36 19.6

7 18.5 17 18.1 27 17.2 37 17.5

8 18.3 18 19.6 28 20.9 38 17.1

9 18.7 19 20.8 29 14.8 39 21.7

10 20.2 20 16.5 30 18.2 40 20.8

1) For the data set above, do the following:

Create a tabular CUSUM in Excel (use a shift of 1σ)

In the following table, I have used the original process standard deviation.

cycle time

Upper Cusum C+ C-

Lower Cusum N+ N- Target Stdev K

16.3 12.5 0.0 -0.5 -12.5 0 1 182.49563

71.24781

916.3 12.5 0.0 -0.9 -12.5 0 2 1819.3 12.5 0.1 0.0 -12.5 1 0 1815.1 12.5 0.0 -1.7 -12.5 0 1 1822.2 12.5 3.0 0.0 -12.5 1 0 1819.1 12.5 2.8 0.0 -12.5 2 0 1818.5 12.5 2.1 0.0 -12.5 3 0 1818.3 12.5 1.1 0.0 -12.5 4 0 1818.7 12.5 0.6 0.0 -12.5 5 0 1820.2 12.5 1.5 0.0 -12.5 6 0 18

22 12.5 4.3 0.0 -12.5 7 0 1814.7 12.5 0.0 -2.1 -12.5 0 1 18

18 12.5 0.0 -0.8 -12.5 0 2 1818.9 12.5 0.0 0.0 -12.5 0 0 1819.1 12.5 0.0 0.0 -12.5 0 0 1810.6 12.5 0.0 -6.2 -12.5 0 1 1818.1 12.5 0.0 -4.8 -12.5 0 2 1819.6 12.5 0.4 -2.0 -12.5 1 3 1820.8 12.5 1.9 0.0 -12.5 2 0 1816.5 12.5 0.0 -0.3 -12.5 0 1 1819.3 12.5 0.1 0.0 -12.5 1 0 1814.6 12.5 0.0 -2.2 -12.5 0 1 1817.8 12.5 0.0 -1.1 -12.5 0 2 1815.6 12.5 0.0 -2.3 -12.5 0 3 1822.5 12.5 3.3 0.0 -12.5 1 0 1817.6 12.5 1.6 0.0 -12.5 2 0 1817.2 12.5 0.0 0.0 -12.5 0 0 1820.9 12.5 1.7 0.0 -12.5 1 0 1814.8 12.5 0.0 -2.0 -12.5 0 1 1818.2 12.5 0.0 -0.5 -12.5 0 2 1816.4 12.5 0.0 -0.9 -12.5 0 3 1818.2 12.5 0.0 0.0 -12.5 0 0 1819.4 12.5 0.2 0.0 -12.5 1 0 1814.1 12.5 0.0 -2.7 -12.5 0 1 1816.4 12.5 0.0 -3.0 -12.5 0 2 1819.6 12.5 0.4 -0.2 -12.5 1 3 18

17.5 12.5 0.0 0.0 -12.5 0 0 1817.1 12.5 0.0 0.0 -12.5 0 0 1821.7 12.5 2.5 0.0 -12.5 1 0 1820.8 12.5 4.0 0.0 -12.5 2 0 18

And I got the following chart, which suggest that process is well with in control limit.

1 3 5 7 9 11 13 15 17 19 21 23 25 27 29 31 33 35 37 39-15.0

-10.0

-5.0

0.0

5.0

10.0

15.0

Cycle Time

Upper Cusum C+ C- Lower Cusum

Period

cusu

m

And also using the Minitab software we get the following cusum chart. Which also suggest the process to be within control

37332925211713951

15

10

5

0

-5

-10

-15

Sample

Cum

ulat

ive Su

m

0

UCL=13.47

LCL=-13.47

Cusum

But if we decrease the sigma shift to one, we get the following table and control chart.

cycle time

Upper Cusum C+ C-

Lower Cusum N+ N- Target Stdev K

16.3 5.0 0.0 -1.2 -5.0 0 1 18 1 0.516.3 5.0 0.0 -2.4 -5.0 0 2 1819.3 5.0 0.8 -0.6 -5.0 1 3 1815.1 5.0 0.0 -3.0 -5.0 0 4 1822.2 5.0 3.7 0.0 -5.0 1 0 1819.1 5.0 4.3 0.0 -5.0 2 0 1818.5 5.0 4.3 0.0 -5.0 3 0 1818.3 5.0 4.1 0.0 -5.0 4 0 1818.7 5.0 4.3 0.0 -5.0 5 0 1820.2 5.0 6.0 0.0 -5.0 6 0 18

22 5.0 9.5 0.0 -5.0 7 0 1814.7 5.0 5.7 -2.8 -5.0 8 1 18

18 5.0 5.2 -2.3 -5.0 9 2 1818.9 5.0 5.6 -0.9 -5.0 10 3 1819.1 5.0 6.2 0.0 -5.0 11 0 1810.6 5.0 0.0 -6.9 -5.0 0 1 1818.1 5.0 0.0 -6.3 -5.0 0 2 1819.6 5.0 1.1 -4.2 -5.0 1 3 1820.8 5.0 3.4 -0.9 -5.0 2 4 1816.5 5.0 1.4 -1.9 -5.0 3 5 1819.3 5.0 2.2 -0.1 -5.0 4 6 1814.6 5.0 0.0 -3.0 -5.0 0 7 1817.8 5.0 0.0 -2.7 -5.0 0 8 1815.6 5.0 0.0 -4.6 -5.0 0 9 1822.5 5.0 4.0 0.0 -5.0 1 0 1817.6 5.0 3.1 0.0 -5.0 2 0 1817.2 5.0 1.8 -0.3 -5.0 3 1 1820.9 5.0 4.2 0.0 -5.0 4 0 1814.8 5.0 0.5 -2.7 -5.0 5 1 1818.2 5.0 0.2 -2.0 -5.0 6 2 1816.4 5.0 0.0 -3.1 -5.0 0 3 1818.2 5.0 0.0 -2.4 -5.0 0 4 1819.4 5.0 0.9 -0.5 -5.0 1 5 1814.1 5.0 0.0 -3.9 -5.0 0 6 1816.4 5.0 0.0 -5.0 -5.0 0 7 1819.6 5.0 1.1 -2.9 -5.0 1 8 1817.5 5.0 0.1 -2.9 -5.0 2 9 1817.1 5.0 0.0 -3.3 -5.0 0 10 1821.7 5.0 3.2 0.0 -5.0 1 0 1820.8 5.0 5.5 0.0 -5.0 2 0 18

1 3 5 7 9 1 1 1 3 1 5 1 7 1 9 2 1 2 3 2 5 2 7 2 9 3 1 3 3 3 5 3 7 3 9-8.0

-6.0

-4.0

-2.0

0.0

2.0

4.0

6.0

8.0

10.0

12.0

Cycle TimeUpper Cusum C+ C- Lower Cusum

Sample

cusu

m

Also the result from the Minitab.

37332925211713951

10

5

0

-5

Sample

Cum

ulat

ive

Sum

0

UCL=5

LCL=-5

Cusum

The above results suggest the process to be out of control. But with calculation from the actual process the process is with control because all the values are within the range of H.

Provide an interpretation for the loan agency manager (what does the CUSUM show about the loan application cycle time and process).

The process suggest that, there process is within control, there is no mean shift occurring in the process, all the loan are processed within time.

2) For the data set above, do the following:

Create an EWMA control chart in Excel (use λ=0.1 and L=2.7)

1 2 3 4 5 6 7 8 9 1011121314151617181920212223242526272829303132333435363738394015.8

16.3

16.8

17.3

17.8

18.3

18.8

19.3

19.8

20.3

UCL

19.90

CL 18.05

LCL

16.20

EWMA

EWMA

EWM

A

2) Create an EWMA control chart in Minitab (use λ=0.1 and L=2.7)

37332925211713951

20

19

18

17

16

Sample

EWM

A __X=18.05

UCL=19.904

LCL=16.196

EWMA Chart of Cycle Time

Does the process seem in statistical control? Why or why not?

The process is well within the control, all the points are inside the limit.

Provide an interpretation for the loan agency manager (what does the EWMA show about the loan application cycle time and process).

Its shows that the loans are processed within 24 hours of time, and from the taken sample there is no sudden shift in the process.

Is the interpretation of the EWMA control chart consistent with the CUSUM?

Yes the results we got from the both the charts are consistent and similar as both of them suggest the process to be within the control.