-- Nov 17 In-Class Exercise Thread
R = 4
d = 3
`\lambda = 21`
Range = (0, 100)
For `d = 1\ Split => {0, 33, 66, 100} => best value = 33`
For `d = 2\ Split => {13.2, 26.4, 39.6, 52.8 } => best value = 26.4`
For `d = 3\ Split => {13.2 + 1*c, 13.2 + 2*c, 13.2 + 3*c, 13.2 + 4*c} => {18.48, 23.76, 29.04, 34.32}` where `c = (39.6-13.2)/(k+1) = 5.28`
Therefore for d = 3, Split => `{18,48, 23.76, 29.04, 34.32}`
`\lambda = 21, \lambda - 18.48 < 23.76 - \lambda.` Therefore, `\lambda` is closer to `18.48 -> ` `Best value = 18.48`
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Edited: 2021-11-22)
R = 4
d = 3
@BT@\lambda = 21@BT@
Range = (0, 100)
For @BT@d = 1\ Split => {0, 33, 66, 100} => best value = 33@BT@
For @BT@d = 2\ Split => {13.2, 26.4, 39.6, 52.8 } => best value = 26.4@BT@
For @BT@d = 3\ Split => {13.2 + 1*c, 13.2 + 2*c, 13.2 + 3*c, 13.2 + 4*c} => {18.48, 23.76, 29.04, 34.32}@BT@ where @BT@c = (39.6-13.2)/(k+1) = 5.28@BT@
Therefore for d = 3, Split => @BT@{18,48, 23.76, 29.04, 34.32}@BT@
@BT@\lambda = 21, \lambda - 18.48 < 23.76 - \lambda.@BT@ Therefore, @BT@\lambda@BT@ is closer to @BT@18.48 -> @BT@ @BT@Best value = 18.48@BT@