2021-11-17
Nov 17 In-Class Exercise Thread .
Please post your solution to the Nov 17 In-Class Exercise to this thread.
Best,
Chris
Please post your solution to the Nov 17 In-Class Exercise to this thread.
Best,
Chris
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2021-11-20
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2021-11-21
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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
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@
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