Decision Tables

[andy4919]

I have a little problem with decision tables. I do under stand that for every 2 conditions there should be 4 possible outcomes or true/false condidtions.

I have this telephone problem where the telephone company charges .10 a minute for all calls outside the callers area code (2 outcomes) and charges .13 cents a minute for all other calls (1 outcome?) The phone company wants a report showing the callers area code and number, the area code called and number, number of minutes and the charge per minute.

I only have 3 outcomes and there are supposed to be 4. I am totally stuck, does anyone know a little about desicion tables.

[mittelgeek]

Sorry about the formatting. I tried to use LaTeX formatting codes, but to no avail... =8(

[TEX]
\begin{tabular}{| c | c | c | c |}
\hline
Area Code & \$0.00 & \$0.10 & \$0.13 \\ \hline
None & X & & \\ \hline
Outside & & X & \\ \hline
Others & & & X \\ \hline
\end{tabular}
[/TEX]

If you look at this table, which I tried to layout what issues the problem states, there are only two possible types of calls that can be made: outside the caller's area code and any other call. The only other possibility is that there is no call made, as indicated.

Not sure if laying it out this way gives you a better understanding of decision tables or helps at all with this problem. Try looking up logic or truth tables and/or Algebraic/Boolean Logic on the 'net. It might give you a better understanding.

[leg1]

The problem is that the conditions are not independent. You can use conditions that have more than 2 states: ex: area code: local, within US, England. In the condition cell you could use "L", "U", and "E", say, for the 3 states.

I hope this helps.

(the reason I changed your area code example was because I don't understand what "outside" and "other" refer to.)