Question

Apparently something I should be able to inspect...

Answer 1

\[s=\frac{ (1+\alpha)(g+\delta) }{ (1+g)(1+\frac{ 1 }{ \beta }) }\]

Answer 2

Apparently s increases with n, g, delta and beta and decreases with alpha

Answer 3

Anyone help on how I should be able to inspect this?

Answer 4

@iambatman @eliassaab @satellite73 @wio

Answer 5

@jim_thompson5910 @ganeshie8 @dumbcow can you help?

Answer 6

where is n?

it increases with alpha , beta, and delta

g is indeterminate because it effects both numerator and denominator

Answer 7

@dumbcow apologies, it should look like this: \[\frac{ (1-\alpha)(\lambda+\delta) }{ (1+\lambda) (1+\frac{ 1 }{ \beta }) }\] where lamda is (1+n)(1+g)

Answer 8

@dumbcow I see how it (now) decreases with alpha, and increases with beta and increases with delta, but the lamda is confusing me

Answer 9

lambda could go either way depending on the other variables

Answer 10

If one assumes delta is between 0 and 1 can we be more sure?

Answer 11

for example, if delta < 1
then s increases with lambda

Answer 12

@dumbcow okay thanks I just can't see that my inspection myself

Answer 13

it may help to just assign values to the variables, then vary 1 and see how it changes

Answer 14

Okay if that's a trustworthy method I'll use that - thanks! How did you see if delta <1 it must increase?

Answer 15

well 2 ways

1) plug in value, let delta = .5 , lambda = .5
\[\frac{\lambda +\delta}{1 + \lambda} = \frac{1}{1.5} = \frac{2}{3}\]
increase lambda to 1
\[= \frac{1.5}{2} = .75\]
s has increased

2) look at the limit when delta < 1
the numerator will always be less than denominator but for large lambda that difference gets smaller and smaller
thus limit of ratio = 1
when lambda = 0, ratio = delta , thus s has increased as lambda increases

Answer 16

@dumbcow That is a fantastic response thank you so much for your help it really makes sense!!

Answer 17

yw