Handbook of Civil Engineering Calculations

TABLE 35. Probabilities for Two Successive

Purchases

Next model

Present

model ABC

A .4167 .3333 .2500

B .5000 .3000 .2000

C .1538 .2308 .6154

2. Determine the monetary value represented
   by the standard deviation
   Since the given monetary values are expressed in thousands of dollars, the value of the
   standard deviation = 6.9 ($ 1000) - $6900.

SHORT-TERM FORECASTING

WITH A MARKOV PROCESS

The XYZ Company manufactures a machine that is available in three models, A, B, and
C. There are currently 1200 such machines in use, divided as follows: model A, 460;
model B, 400; model C, 340. On the basis of a survey, the XYZ Company has established
probabilities corresponding to two successive purchases, and they are recorded in Table

35. Thus, if a firm currently owns model B, there is a probability of 0.5000 that its next
    model will be A; if a firm currently owns model C, there is a probability of 0.6154 that its
    next model will also be C. Assume that each machine will remain in service for precisely
    1 year, after which it will be replaced with another machine manufactured by the XYZ
    Company. Also assume that the XYZ Company will not acquire any new customers in the
    foreseeable future. Estimate the number of units of each model that will be in use 1 year,
    2 years, and 3 years hence.

Calculation Procedure:

1. Set up the basic equations that link two successive years
   Assume that a trial will be performed repeatedly and that the outcome of one trial directly
   influences the outcome of the succeeding trial. A trial of this type is called a Markov
   process. In this situation, the purchase of a machine is a Markov process because the
   model that a firm selects on one occasion has a direct bearing on the model it selects on
   the following occasion. The probabilities in Table 35 are termed transition probabilities,
   and the table itself is called a transition matrix.
   Let XA>n = expected number of units of model A that will be in use n years hence. Mul-
   tiply the expected values for n years hence by their respective probabilities to obtain the
   expected values for n + 1 years hence, giving

XA** = 0.4167JC 01 + 0.5000XBilt + 0.1538XCn (a)