calculation of a table in maximax approach

I assume that you have downloaded the free lecture notes and therefore have the lecture example in front of you.

John is selling to the contract customer and also to ‘normal’ customers. All sales to the contract customer will make a contribution of 9 – 6 = $3 per unit. All sales to normal customers will make a contribution of 11 – 6 = $5 per unit.

If the sign the contract to sell them 300 units a month, then they have to see them 300 and on these they will make a contribution of 300 x $3 = $900. In addition they will sell to normal customers, but how many they sell depends on the demand by normal customers.
If normal customer demand 400 units, then they will make extra contribution of 400 x $5 = $2,000. So a total of $2,900 (900 + 2,00).

If normal customers demand 500 units, then they will make extra contribution of 500 x $5 = $2,500. So a total of $3,400 (900 + 2500).

It is the same logic for all the combinations of the amount they sign the contract for, and the amount demanded by normal customers.

The one thing to be careful of is the fact that the question says that the factory capacity is 1,200 units. They have to sell to contract customers the amount they have signed for, and so the most that can be sold to normal customers is whatever is left out of the 1,200.

So, for example, if they sign a contract to sell 800 units, then the most they can ever sell to normal customers is 400 (1,200 – 800), regardless of how many units normal customers want to buy.

You have headed up this thread ‘maximax’, but it is the same profit table that is required for all of the different decision making criteria – maximax, maximin, expected values, and minimax regret.

You are welcome 🙂