MARKOV CHAIN EXAMPLE Personnel Modeling. DYNAMICS Grades N1..N4 Personnel exhibit one of the...

22
MARKOV CHAIN EXAMPLE Personnel Modeling

Transcript of MARKOV CHAIN EXAMPLE Personnel Modeling. DYNAMICS Grades N1..N4 Personnel exhibit one of the...

MARKOV CHAIN EXAMPLEPersonnel Modeling

DYNAMICS

• Grades N1..N4

• Personnel exhibit one of the following behaviors:– get promoted– quit, causing a vacancy

that is filled during the next promotion period

– remain in grade– get demoted

STATE SPACE

•S = {N1, N2, N3, N4, V}

• V for Vacancy

• Every time period, the employee moves according to a probability

MODELED AS A MARKOV CHAIN

• Discrete time periods

• Stationarity– transitions stay constant

over time– transitions do not depend

on time in grade

TRANSITION DIAGRAM

1

2

3

4V

0.1

0.1

0.1

0.10.1

0.20.1

0.1

0.60.5

0.31.0

0.30.6

0.8

PROBABILITY TRANSITION MATRIX

  1 2 3 4 V

1 0.1 0.6     0.3

2 0.1 0.5 0.3   0.1

3   0.1 0.6 0.2 0.1

4     0.1 0.8 0.1

V 1        

= P

MEASURES OF INTEREST

• Proportion of the workforce at each level

• Expected labor costs per year

• Expected annual cost of Entry-level training

• PDF of passage from N1 to N4

TRANSITION PROBABILITY CALCULATION

• Start with employee in N1• a0 = [1, 0, 0, 0, 0]• a1 = a0 * P• a1 = [0.1, 0.6, 0, 0, 0.3]• a2 = a1 * P

37.0

1

1

1

2

1,

1,22

1,11

1

VV

N

N

N

Pa

Pa

Pa

a

STEADY STATE PROBABILITIES

• a0 * P * P * P * P * ....

• P is singular (rank 4)

• P is stochastic– rows sum to 1

is the stationary probability distribution

N1 is the proportion of the time spent in state N1

COMPUTATION STRATEGY

PN1N2 N3 N4

V

• Substitute stochastic equation for first component of P

• Solve Linear System via Gaussian Elimination

...more COMPUTATION STRATEGY

• Start with arbitrary a0

• calculate a1, a2, a3, ...

• will converge to

CONVERGENCE TO

0

0.2

0.4

0.6

0.8

1

1.2

0 5 10 15

ITERATION

N1

N2

N3

N4

V

CONVERGENCE IS QUICK

0

0.5

1

1.5

iteration

sqrt

(sum

sqr

err

or)

FOR GRINS

• Changed PN4,V to 0.0

= [0.09, 0.16, 0.23, 0.46, 0.06]

ENTRY-LEVEL TRAINING

• 12% of the time we are in state V

• Cost of ELT = – 12%– times the Workforce size– times the cost of training

LABOR COSTS

• Salaries– CN1 = $12,000

– CN2 = $21,000

– CN3 = $25,000

– CN4 = $31,000

• Total Workforce = 180,000

• Cost = 180K * (C * ) = $3.7B

EXCURSION

• Promotion probabilities unchanged

• Allow attrition to reduce workforce– PV,N1 = 0.6 results in

workforce of 108,000

• How much $ saved?

• How fast does it happen?

LABOR COSTS

3.35

3.4

3.45

3.5

3.55

3.6

3.65

3.7

100000 120000 140000 160000 180000 200000

WORKFORCE

$B

CONVERGENCE TO 75% WORKFORCE (135K)

0

0.05

0.1

0.15

0.2

0.25

0.3

0 2 4 6 8 10

ITERATION

N1

N2

N3

N4

V

CONVERGENCE TO 60% WORKFORCE (108K)

0

0.05

0.1

0.15

0.2

0.25

0.3

0 2 4 6 8 10

ITERATION

N1

N2

N3

N4

V

BUILDING AN N4 FROM AN N1CUMULATIVE

0

0.05

0.1

0.15

0.2

0.25

0 5 10 15 20

BUILDING AN N4 FROM AN N1MARGINAL

0

0.005

0.01

0.015

0.02

0.025

0.03

0.035

0.04

0 5 10 15 20