Question

Formula for Problem

W1 in year 1; W2 in year 2 only to those retained W1 < W2

• Value of E-types applying = W1 + q•W2 + (1–q)•WE

• Alternative for E-types = 2•WE

• Value of D-types applying = W1 + (1-q)•W2 + q•WD

• Alternative for D-types = 2•WD

1. Lorne Roberts Corp. has invested a lot of money in their employee screening process over the last few years for computer technicians. This company can assess whether to promote employees on probation accurately 90% of the time. But 10% of the time the company still inaccurately assesses employees on probation by terminating an employee who fits successfully in the job or promoting an employee who does not fit well in the job.

The WD for computer technicians is $30,000 per year. The WE for this position is $40,000.

Lorne Roberts Corp. is considering which W1 and W2 will encourage E’s to apply while discouraging D’s from applying. Out of the three options below, which is the best salary structure for the first two years:

A) W1 = $30,000; W2 = $42,000

B) W1 = $40,000; W2 = $52,000

C) W1 = $24,000; W2 = $59,000

Answer 1

using the above formulas, ie

Value of E-types applying = W1 + q•W2 + (1–q)•WE

Alternative for E-types = 2•WE

Value of D-types applying = W1 + (1-q)•W2 + q•WD

Alternative for D-types = 2•WD

where, q is 90% ie the no. of time a correct person is promoted / whose job is terminated, while 1-q is 10% ie the no. of time an incorrect person is promoted / whose job is terminated. and WD = $30,000 and WE = $40,000

To encourage E's to apply, the value of e types applying should be greater than alternative for e types, while to discourage D's from applying, value of d types applying should be less than the alternative for d types.

as per the given information,

Value of E-types applying = W1 + 90% * W2 + 10% * 40,000 = W1 + .9 * W2 + 4,000 [ equation 1 ]

Alternative for E-types = 2 * 40,000 = 80,000 [ equation 2 ]

Value of D-types applying = W1 + 10% * W2 + 90% * 30,000 = W1 + .1 * W2 + 27,000   [ equation 3 ]

Alternative for D-types = 2 * 30,000 = 60,000 [ equation 4 ]

on substituting the values of W1 and W2 in equation 1 and 3 , the answer that is derived should be greater than equation 2 and less than equation 4, respectively.

option	value of W1 as per option (in $)	value of W2 as per option (in $)	solving equation 1 ie W1 + .9 * W2 + 4,000 (in $)	equation 2 (in $)	whether answer of equation 1 greater than equation 2?	solving equation 3 ie W1 + .1 * W2 +27,000 (in $)	equation 4 (in $)	whether answer of equation 3 less than equation 4?
A)	30,000	42,000	= 30,000 + .9*42,000 + 4,000 = 34,000 + 37,800 = 71,800	80,000	NO	= 30,000 + .1*42,000 + 27,000 = 57,000 + 4,200 = 61,200	60,000	YES
B)	40,000	52,000	= 40,000 + .9*52,000 + 4,000 = 44,000 + 46,800 = 90,800	80,000	YES	= 40,000 + .1*52,000 + 27,000 = 67,000 + 5,200 = 72,200	60,000	NO
C)	24,000	59,000	= 24,000 + .9*59,000 + 4,000 = 28,000 + 53,100 = 81,100	80,000	YES	= 24,000 + .1*59,000 + 27,000 = 51,000 + 5,900 = 56,900	60,000	YES

The option under which both the columns have the answer as yes, is the best salary structure for the first two years, i.e W1 = $ 24,000 and W2 = $ 59,000 is the best salary structure.

Hope this helps.