Rule 2 = {Lj i.e. the latest time for activities is the minimum of the Li Dij}, – latest time of all succeeding activities reducing their activity time. Rule 3 Repeat staring activity is reached. rule 2until Latest times for activities of the network are calculated below: By set = 19. Applying it is to determine and rule 1, L10 rule 2, L9, L8 L7, = {Lj = 19 4 = 15 for = 10. L9 D9, j} j – – = {Lj = = 15 3 = 12. (j contains only one node 9) L8 D8, j} L9 D8, 9 – – – = {Lj = = 15 5 = 10. (j contains node 9) L7 D7, j} L9 D7, 9 – – – Now consider node 6. For this node, there are two succeeding activities, namely 6 8 – and 6 7. – Hence, = [Lj = min = min = 8. L6 D6, j] – Similarly, = = 12 2 = 10, = = 10 4 = 6 L5 L8 D5, 8 L4 L7 D4, 7 – – – – = min = min = 2, = = 10 4 = 6 L3 L2 L5 D2, 5 – – = [Lj = min = 0. L1 – D1, j] Figure - 15 The minimum value of = 0 is no surprising result. Since, started with = it is L1 Li Ei, always possible to lave = 0. If this is not so, it means that some error is made in L1 calculations of forward pass or backward pass values. shows earliest and Figure 15 – latest times of each event. Recall that path 1 3 7 9 10 was defined as the critical path of this network. – – – – this path, it is observed that the latest and earliest times are the same implying that any activity along this path cannot be delayed without affecting the duration of the project. Step 3 Computation of Floats
Rule 2 Li = {Lj – Dij}, i.e. the latest time for activities is the minimum of the latest time of all succeeding activities reducing their activity time. Rule 3 Repeat rule 2until staring activity is reached. Latest times for activities of the network are calculated below: By rule 1, set L10 = 19. Applying rule 2, it is to determine L9, L8 and L7, L9 = {Lj – D9, j} = 19 – 4 = 15 for j = 10. L8 = {Lj – D8, j} = L9 – D8, 9 = 15 – 3 = 12. (j contains only one node 9) L7 = {Lj – D7, j} = L9 – D7, 9 = 15 – 5 = 10. (j contains node 9) Now consider node 6. For this node, there are two succeeding activities, namely 6 – 8 and 6 – 7. Hence, L6 = [Lj – D6, j] = min = min = 8. Similarly, L5 = L8 – D5, 8 = 12 – 2 = 10, L4 = L7 – D4, 7 = 10 – 4 = 6 L3 = min = min = 2, L2 = L5 – D2, 5 = 10 – 4 = 6 L1 = [Lj – D1, j] = min = 0. Figure - 15 The minimum value of L1 = 0 is no surprising result. Since, started with Li = Ei, it is always possible to lave L1 = 0. If this is not so, it means that some error is made in calculations of forward pass or backward pass values. Figure – 15 shows earliest and latest times of each event. Recall that path 1 – 3 – 7 – 9 – 10 was defined as the critical path of this network. this path, it is observed that the latest and earliest times are the same implying that any activity along this path cannot be delayed without affecting the duration of the project. Step 3 Computation of Floats