XB>n+l = 0.3333AC 01 + 0.30OQAT^ + 0.2308AT 0 , (b)
Xc>n+l = 0.2500A^n + 0.200OY^n + 0.6154AfCn (c)
- Calculate the expected values for 1 year hence
Apply Eqs. a, b, and c with n = O and XAO = 460, X 80 = 400, Xco = 340. Then XA ^ =
(0.4167)460 + (0.5000)400 + (0.1538)340 = 444; XB\ = (0.3333)460 + (0.3000)400 +
(0.2308)340 = 352; XCil = (0.2500)460 + (0.2000)400 + (0.6154)340 = 404. Record the
results in Table 36. - Calculate the expected values for 2 years hence
Apply Eqs. a, b, and c with n = 1 and the values ofXA \,XBl, and Xcc l9 shown in Table - Then XA2 = (0.4167)444 + (0.5000)352 + (0.1538)404 = 423; XB2 = (0.3333)444 +
(0.3000)352'+ (0.2308)404 = 347; Xca = (0.2500)444 + (0.2000)352 + (0.6154)404 - - Record the results in Table 36.
- Calculate the expected values for 3 years hence
Apply Eqs. a, b, and c for the third cycle. Then XA3 = (0.4167)423 + (0.5000)347 +
(0.1538)430 = 416; XS3 = (0.3333)423 + (0.3000)347 + (0.2308)430 = 344; XC3 =
(0.2500)423 + (0.2000)347 + (0.6154)430 = 440. Record the results in Table 36. - Determine the expected values with the aid of a diagram
Refer to Fig. 37, which shows the expected manner in which units of a given model will
be replaced. Each value of X is recorded in the appropriate box. Multiply the values of
XAf0, XBt0, and AfCj0 by the corresponding probabilities to find the expected replacements
during the first year. Thus, with reference to the 460 units of model A, the expected re-
placements are as follows: model A, 460(0.4167) = 192; model B, 460(0.3333) = 153;
model C, 460(0.2500) = 115. Record all values in Fig. 37. Then XA l = 192 + 200 + 52 =
444; xB%l = 153 + 120 + 79 = 352; XCtl = 115 + 80 + 209 =,404. Repeat the cycle of cal-
culations for the second and third years. The values in Fig. 37 agree with those in Table
Related Calculations: Matrix multiplication provides a compact procedure for
solving problems pertaining to a Markov process. Let P denote the matrix in Table 35 and
Rn denote a row vector consisting ofXAn, X 8 ^n, andXCn. The values of these variables ap-
pear in Table 36. Then
[
0.4167 0.3333 0.25001
0.5000 0.3000 0.2000 = [444 352 404]
0.1538 0.2308 0.6154J
Similarly, R 2 = R 1 P = R 0 P^2 , and R 3 = R 2 P = R 0 P^3. In general, Rn = R 0 P".
TABLE 36. Expected Number of Units in Use
Elapsed time
years XA XB Xc
0 46 0 40 0 34 0
1 44 4 35 2 40 4
2 42 3 34 7 43 0
3 41 6 34 4 44 0