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MARKOV CHAIN EXAMPLE Personnel Modeling. DYNAMICS Grades N1..N4 Personnel exhibit one of the following behaviors: –get promoted –quit, causing a vacancy.

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Presentation on theme: "MARKOV CHAIN EXAMPLE Personnel Modeling. DYNAMICS Grades N1..N4 Personnel exhibit one of the following behaviors: –get promoted –quit, causing a vacancy."— Presentation transcript:

1 MARKOV CHAIN EXAMPLE Personnel Modeling

2 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

3 STATE SPACE S = {N1, N2, N3, N4, V} V for Vacancy Every time period, the employee moves according to a probability

4 MODELED AS A MARKOV CHAIN Discrete time periods Stationarity –transitions stay constant over time –transitions do not depend on time in grade

5 TRANSITION DIAGRAM 1 2 3 4V 0.1 0.2 0.1 0.6 0.5 0.3 1.0 0.3 0.6 0.8

6 PROBABILITY TRANSITION MATRIX 1234V 10.10.6 0.3 20.10.50.3 0.1 3 0.60.20.1 4 0.80.1 V1 = P

7 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

8 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

9 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

10 COMPUTATION STRATEGY  P  N1  N2   N3   N4   V Substitute stochastic equation for first component of  P Solve Linear System via Gaussian Elimination

11 ...more COMPUTATION STRATEGY Start with arbitrary a0 calculate a1, a2, a3,... will converge to   

12 CONVERGENCE TO 

13 CONVERGENCE IS QUICK

14 FOR GRINS Changed P N4,V to 0.0  = [0.09, 0.16, 0.23, 0.46, 0.06]

15 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

16 LABOR COSTS Salaries –C N1 = $12,000 –C N2 = $21,000 –C N3 = $25,000 –C N4 = $31,000 Total Workforce = 180,000 Cost = 180K * (C *  ) = $3.7B

17 EXCURSION Promotion probabilities unchanged Allow attrition to reduce workforce –P V,N1 = 0.6 results in workforce of 108,000 How much $ saved? How fast does it happen?

18 LABOR COSTS

19 CONVERGENCE TO 75% WORKFORCE (135K)

20 CONVERGENCE TO 60% WORKFORCE (108K)

21 BUILDING AN N4 FROM AN N1 CUMULATIVE

22 BUILDING AN N4 FROM AN N1 MARGINAL


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