Friday, February 24, 2012

The logic of too many orders is better than no order

"Businesses should look at the positive side of the new regulations on foreign workers. They are complaining that they cannot fulfil orders because they don't have enough workers. This is a happy problem. Which would you prefer - too many orders, or no orders at all?" - MP Gan Thiam Poh

Assume number of orders received = x, number of workers = y, number of workers to fulfill one order = z.

Thus number of orders that can be fulfilled = y/z
If x = y/z, number of orders that can be cashed in = x
If x > y/z, number of orders that can be cashed in = y/z
If x = 0 then number of orders that can be cashed in = 0

So Mr MP is right that bosses should be happy if they have more orders than they can handle because they still can cash in y/z orders.

However... Mr MP forgot this equation too...

Assume number of customers lost = m,
Number of potential orders lost b = m x n, where n = number of potential order per customer
Number of customers lost to competitors c = m
Number of peers from lost customers that moved over to competitors d = m x p, where p = number of peers per customer

So definitely Mr MP feels that businesses should still be happy and contented even if their number of customers and potential customers who moved over to their competitors is more than their remaining number of customers, because they could not fulfill their orders.

Roger that. A business should be contented to keep y/z constant.

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