Question

(Operations Management) Identify a process in a bank and one in a retail store that control charts can be used to monitor. If possible, please also draw a chart to help explain.

Answer 1

1. In a retail store, a number of different categories are analysed for their turnover rate. Perishable items such as vegetables, fruit and meats, need to be discarded f not sold. One such observation is the fraction of pears in kilograms spolit ( and thus discarded) per kilogram of receipt. If the store orders 1000 kgs each week, the p chart can be useful in identifying the fraction spoilt. A table shows observations taken at the end of consecutive weeks.

Week	Kgs spoilt	p
1	12	0.012
2	19	0.019
3	8	0.008
4	7	0.007
5	11	0.011
6	14	0.014
7	9	0.009

p s the fraction of spoil to the total receipt

mean p value pbar =0.01142

The 3 sigma upper and lower control limits can be calculated as per formula given below

UCLp = pbar + 3*[ pbar*(1-pbar)/n]^0.5

= 0.01142+3*[ 0.01142*(1-0.01142)/1000]^0.5 =0.02149

LCLp = pbar - 3*[ pbar*(1-pbar)/n]^0.5

= 0.01142-3*[ 0.01142*(1-0.01142)/1000]^0.5 =0.00134

Since all values are in control, the process is in statistical control.

(ii) In another application in a bank, the incidents of detection of an account mismatch in a remote branch of a bank is obserevd for straight 10 days and following observations are found.

Day	Incidents c
1	0
2	1
3	2
4	0
5	2
6	5
7	0
8	0
9	1
10	3

c chart is the right tool to identify the pattern of such incidents over a time and to analyse if the process has some assignable cause

cbar is the mean value of c , which are the number of incidents of detection of account mismatch.

cbar =1.4

3 sigma control limits for c

UCLc = cbar+ 3*( cbar)^0.5 = 1.4 + 3*(1.4)^0.5 = 4.9496

LCLc = cbar- 3*( cbar)^0.5 = 1.4 - 3*(1.4)^0.5 =0

Since one value is higher than cbar, the process can be said to be out of control.