An amazing formula was provided to me by a super member on this site, it functions very well for my needs until a recent discovery, the formula; =LET(s,SEQUENCE(3),IF(J$7="Yes",IF(D18>4500,"over",XLOOKUP(D18,1730*(s-1),s,,-1)),0)). I’m looking to add further functionality to this formula by adding yet another constant.
This formula results in the number 3 when D18 is set to 3660. The constant in the formula 1730 has a dependent cell, F42 with the formula; =IFERROR(($D$18-4)/H42,0) resulting in the number 1219. The result in F42 in this case has 3 parts being reduced to 1219 from 1730, each having a waste factor of 511, at 3 parts a total of 1533 wasted material. I work with one more constant that I’d like to add to the formula to reduce the large amount of waste, 2500 is the new constant. In the case that D18 is 3660, if the formula could consider 1-2500/2=1250-1219=31 of waste x 2, and 1-1730-1219=511 of waste to total 573 of waste it would be a more efficient waste factor. If D18 were set at 4450 it would consider 2-2500 parts that would result in less waste than if it selected 3-1730 parts. I can’t understand if a formula can effectively choose from the two constants to minimize 1) waste per part and 2) the total number of parts needed without the need of a complicated algorithm and macro.
Anyone’s input would be appreciated.
Thank you.
This formula results in the number 3 when D18 is set to 3660. The constant in the formula 1730 has a dependent cell, F42 with the formula; =IFERROR(($D$18-4)/H42,0) resulting in the number 1219. The result in F42 in this case has 3 parts being reduced to 1219 from 1730, each having a waste factor of 511, at 3 parts a total of 1533 wasted material. I work with one more constant that I’d like to add to the formula to reduce the large amount of waste, 2500 is the new constant. In the case that D18 is 3660, if the formula could consider 1-2500/2=1250-1219=31 of waste x 2, and 1-1730-1219=511 of waste to total 573 of waste it would be a more efficient waste factor. If D18 were set at 4450 it would consider 2-2500 parts that would result in less waste than if it selected 3-1730 parts. I can’t understand if a formula can effectively choose from the two constants to minimize 1) waste per part and 2) the total number of parts needed without the need of a complicated algorithm and macro.
Anyone’s input would be appreciated.
Thank you.