\[D_i\]
is low and intermittent. The total monthly demand is a random variable with the following probability distribution. \[D_i\] |
0 | 1 | 2 | 3 |
\[P(D_i)\] |
1/9 | 6/9 | 1/9 | 1/9 |
We need to complete the table.
Month \[n+1\] \Month \[n\] |
0 | 1 | 2 | 3 |
0 | ||||
1 | ||||
2 | ||||
3 |
\[n +1\]
is zero, and if demand is zero, (with probability 1/9) the state in month \[n+1\]
is 1. The state in month \[n+1\]
cannot be 2 or 3, so the entries must be zero. We can complete the first column.Month \[n+1\] \Month \[n\] |
0 | 1 | 2 | 3 |
0 | 8/9 | |||
1 | 1/9 | |||
2 | 0 | |||
3 | 0 |
\[n +1\]
is zero, and if demand is 1, (with probability 6/9) the state in month \[n+1\]
is 1. If demand is zero (with probability 1/9) the state in month \[n+1\]
is 2. The state in month \[n+1\]
cannot be 3, so the entry must be zero. We can complete the second column.Month \[n+1\] \Month \[n\] |
0 | 1 | 2 | 3 |
0 | 8/9 | 2/9 | ||
1 | 1/9 | 6/9 | ||
2 | 0 | 1/9 | ||
3 | 0 | 0 |
\[n +1\]
is zero, and if demand is 2, (with probability 1/9) the state in month \[n+1\]
is 1. If demand is 1 (with probability 6/9) the state in month \[n+1\]
is 2. If demand is 0 (with probability 1/9) the state in month \[n+1\]
is 3. We can complete the third column.Month \[n+1\] \Month \[n\] |
0 | 1 | 2 | 3 |
0 | 8/9 | 2/9 | 1/9 | |
1 | 1/9 | 6/9 | 1/9 | |
2 | 0 | 1/9 | 6/9 | |
3 | 0 | 0 | 1/9 |
\[n +1\]
is zero, and if demand is 2, (with probability 1/9) the state in month \[n+1\]
is 1. If demand is 1 (with probability 6/9) the state in month \[n+1\]
is 2. If demand is 0 (with probability 1/9) the state in month \[n+1\]
is 3. We can complete the last column.Month \[n+1\] \Month \[n\] |
0 | 1 | 2 | 3 |
0 | 8/9 | 2/9 | 1/9 | 1/9 |
1 | 1/9 | 6/9 | 1/9 | 1/9 |
2 | 0 | 1/9 | 6/9 | 6/9 |
3 | 0 | 0 | 1/9 | 1/9 |