# Operations Management Question: Cycle Counting Leemin Enterprises carries 9,000 items of outdoor garden décor annually. It’s owner, Peter Ng, uses an ABC classification system. 7% of the items...

Operations Management Question: Cycle Counting

Leemin Enterprises carries 9,000 items of outdoor garden décor annually. It’s owner, Peter Ng, uses an ABC classification system. 7% of the items are classified as A. 33% of the items are classified as B. The remaining items are classified as C. Peter wants his operations staff to count the A items monthly, the B items quarterly, and the C items semi-annually.

**How many items of each classification and in total need to be counted each day? (Assume the handling of items is approximately constant throughout the year)**

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There are 9,000 decor items in total annually. Assume that the items are handled at a constant rate throughout the year and that counts for items are made on individual days at regular intervals throughout the year, where the interval is determined by the type of item being counted. The proportion that is type A, B and C equals, respectively, 7%, 33% and (100-(7+33) = 100-40 = ) 60%.

This implies that the total number of decor items that are type A is 0.07 x 9000 = 630,

the total that are type B is 0.33 x 9000 = 2970,

and the total that are type C is 0.6 x 9000 = 5400.

Now, the A items are counted monthly. Assuming a constant handling rate of decor items, there would be 630/12 (dividing by the 12 months in the year) = 52.5 of type A items per month. At the same point in each month, so that the counting is always done a month apart, 53 (rounding up) items of type A need to be counted.

The B items are counted quarterly, that is every 3 months, for example in March, June, September and December. As there are 2970 type B items handled in the year, 2970/4 = 742.5 would be handled each quarter. At the same point in every third month then, 742.5 (approx 743) B items would need to be counted. If the day of counting coincides with the day of counting of type A items then 53+743 = 796 items would need to be counted at that point in the month, every third month.

Finally, the C items are calculated semi-annually, that is, twice a year. As there are 5400 type C items, 5400/2 = 2700 would need to be counted at the same point in two months in the year that are 6 months apart. Again, the counting of type C items could coincide with the time in the month when type A items are counted, and perhaps also when type B items are counted.

If all 3 types of item A, B and C are calculated at the same point in the month, when it is in a month in which they are calculated, then in a particular date in two months of the year that are six months apart (say on 10th June and 10th December) 53+743+2700 = 3496 items would need to be counted. In two months of the year equidistant between these points (10th March and 10th September, continuing with the example given) 53+743 = 796 items would need to be counted, and for the remaining 8 other months, on the 10th of the month 53 items would need to be counted. If the dates did not coincide, 12 equidistant dates a year 53 items would need to be counted; 4 equidistant dates a year 743 items would need to be counted; and 2 equidistant date a year 2700 items would need to be counted. On the majority of days (365-12 = 353 in the case of coincident dates in the first example, and 365-(12+4+2) = 347 in the case of non-coincident dates, on non-leap years) no items need to be counted.

The maximum number of items potentially needing to be counted on any particular day is 3496, and this would be semi-annually. The minimum number of items needing to be counted on any particular day, and this would be the case for the majority of days in the year, would be 0. Contrastingly, the number that would need to be counted each day if the counting of the items were spread entirely evenly over each and every day of the year would be 9000/365 = 24.7, that is approximately 25 items per day. Efficiency considerations regarding employee capabilities and pay and workspace costs would determine how the timetable for counting items would sit between these two most extreme scenarios, that is the most uneven spread in the first case, and the most even spread in the second.