## 4 The Parametric PERT/CPM Method for Finding the Optimum Schedule
### The Prohibited Intervals Method B
These two inequalities give the following prohibited interval with respect to \(T\): \(T\
otin(Z_{5}-Z_{0}-p_{5},\,Z_{5}-Z_{0}+p_{0})\). Because \(Z_{0}=p_{0}\), we obtain that  
\[T\
otin(Z_{5}-p_{0}-p_{5},\,Z_{5}) \tag{7}\]Workstation \(M_{2}\) performs operations \(O_{2}\) and \(O_{4}\). Using the same arguments as described above, we obtain the following prohibited interval:  
\[T\
otin(Z_{4}-Z_{2}-p_{4},Z_{4}-Z_{2}+p_{2}) \tag{8}\]  
The Prohibited Intervals Rule for the 1-cycle case. The set of the prohibited intervals (5)-(8) is denoted by \(F\). The main idea of the Prohibited Intervals Method is as follows: According to the definition of the set \(F\), and due to the periodicity of the screening process under study, a schedule is feasible (that is, a real number \(T\) satisfying inequalities (5)-(8) exists) if the time interval between any two microplates introduced into the screening process does not belong to the set \(F\). Formally, this rule is as follows:  
Since all microplates enter the screening system periodically at equal time period \(T\), it follows that the interval between them must be one of the numbers \(T\), \(2T\), \(3T\), \(\ldots\), \(mT\). The Prohibited Intervals Rule can now be re-formulated as follows: A cyclic schedule with cycle time \(T\) exists if the values \(T\), \(2T\), \(\ldots\), \(mT\) do not belong to any of the prohibited intervals of the set \(F\). The minimum \(T\), such that \(T\), \(2T\), \(3T\), \(\ldots\), \(mT\) do not belong to any of the prohibited intervals in the set \(F\), is the optimal 1-cyclic time [15]. The set \(U\) is defined as a complementary set to the set \(F\): \(U=(-\infty,\ \infty)\backslash F\), and it is known as the set of allowed intervals. Let \(T\) be the cycle time of a one-cyclic schedule, and \(m\) be the number of workstations. Then a one-cyclic schedule is feasible if  
\[T\in U,2T\in U,3T\in U,\ldots,mT\in U. \tag{9}\]  
_The Modified Prohibited Intervals Rule for the 2-cycle case._ Since we are searching for an optimal two-cyclic schedule, instead of the conditions described in (9), we will have the following:  
\[T_{1}\in U,T_{2}\in U,T\in U,T+T_{1}\in U,T_{2}+T\in U,2T\in U,2T+T_{1}\in U,\ldots,\lceil m/2\rceil T\in U \tag{10}\]  
where \(T=T_{1}+T_{2}\) is the cycle time and \(T_{avr}=(T_{1}+T_{2})/2\). The proof directly follows from the periodicity of the screening process under study. Numerical example. Consider the set of numerical data used in subalgorithm \(A\), corresponding to the real-world HTS system in [14]. The processing times of the operations are the following numbers, obtained as the optimal one-cyclic solution of the corresponding critical path problem \(\mathbf{Q}\), solved above by the PERT/CPM subalgorithm \(A\): \(p_{0}=32\), \(p_{1}=20\), \(p_{2}=54\), \(p_{3}=210\), \(p_{4}=54\), and \(p_{5}=34\). The completion times \(Z_{k}\) are the following: \(Z_{0}=32\), \(Z_{1}=71\), \(Z_{2}=148\), \(Z_{3}=378\), \(Z_{4}=452\), and \(Z_{5}=506\). According to Expression (6), we obtain the following prohibited intervals:  
Set S1 = {(20, 62), (54, 97), (210, 250), (54, 94), (97, 136), (284, 327), (284, 324), (327, 366), (358, 401), (401, 440)}. According to Expressions (5), (7) and (8), we obtain, in addition, the following prohibited intervals:  
Set S2 = {(-\(\infty\), 54), (440, 506), (250, 358)}. At the next step, we merge the prohibited intervals for the sets S1 and S2 and finally obtain:  
\(F=\{(-\infty,\ 97), (97,136), (210, 250), (250, 401), (401, 440), (440, 506)\}.\)  
Thus, the set of allowed intervals \(U=(-\infty,\ \infty)\backslash F\) has the following form:  
\(U=\{[97,\ 97],\ [136,\ 210],\ [250,\ 250],\ [401,\ 401],\ [440,\ 440],\ [506,\ \infty).\)  
Remark. Using the Prohibited Intervals Rule formulated above and the polynomial algorithm of the authors in [15], we can find the optimal 1-cyclic solution. However, in this study, there is no need to use this algorithm for finding such a 1-cyclic solution, since this value has already been found by Algorithm A, namely, \(T=200.5\).