Calculation Of Float

By far the quickest way to calculate the float of a particular activity is to do it manually. In practice, one does not require to know the float of all activities at the same time. A list of floats is, therefore, unnecessary. The important point is that the float of a particular activity which is of immediate interest is obtainable quickly and accurately.

Consider the string of activities in a simple construction process. This is shown in Figure 24.1 in Activity on Arrow (AoA) format and in Figure 24.2 in the simplified Activity on Node (AoN) format.

Figure 24.1

Pour foundations

Figure 24.3

It can be seen that the total duration of the sequence is 34 days. By drafting the network in the method shown, and by using the day numbers at the end of each activity, including dummies, an accurate prediction is obtained immediately and the float of any particular activity can be seen almost by inspection. It will be noted that each activity has two dates or day numbers - one at the beginning and one at the end (Figure 24.3). Therefore, where two (or more) activities meet at a node, all the end day numbers are inserted (Figure 24.4). The highest number is now used to calculate the overall project duration, i.e. 30 + 3 = 33, and the difference between the highest and the other number immediately gives the float of the other activity and all the activities in that string up to the previous node at which more than one activity meet. In other words, 'set pumps' (Figure 24.1) has a float of 30 - 26 = 4 days, as have all the activities preceding it except 'deliver pump', which has an additional 24 - 20 = 4 days float.

If, for example, the electrical engineer requires to know for how long he can delay the cabling because of an emergency situation on another part of the site, without delaying the project, he can find the answer right away. The float is 33 - 28 = 5 days. If the labour he needs for the emergency can be drawn from the gang erecting the starters, he can gain another 28 - 23 = 5 days. This gives him a total of 10 days' grace to start the starter installation without affecting the total project time.

A few practice runs with small networks will soon emphasize the simplicity and speed of this method. We have in fact only dealt in this exposition with small - indeed, tiny - networks. How about large ones? It would appear that this is where the computer is essential, but in fact, a well-drawn network can be analysed manually just as easily whether it is large or small. Provided the very simple base rules are adhered to, a very fast forward pass can be inserted. The float of any string can then be seen by inspection, i.e. by simply subtracting the lower node number from the higher number of the node which forms the termination point of the string in question. This point can best be illustrated by the example given in Figure 24.5. For simplicity, the activities have been given letters instead of names, since the importance lies in understanding the principle, and the use of letters helps to identify the string of activities. In this example there are 50 activities. Normally, a practical network should have between 200 and 300 activities maximum (i.e. four to six times the number of activities shown) but this

24 Set 26 Connect 33

24 Set 26 Connect 33

20 Lay 30

20 Lay 30

0 3 9 16 20 30
Figure 24.5

does not pose any greater problem. All the times (day numbers) were inserted, and the floats of activities in strings A, B, C, E, F, G and H were calculated in 5 minutes. A 300-activity network would, therefore, take 30 minutes.

It can in fact be stated that any practical network can be 'timed', i.e. the forward pass can be inserted and the important float reported in 45 minutes. It is, furthermore, very easy to find the critical path. Clearly, it runs along the strings of activities with the highest node times. This is most easily calculated by working back from the end. Therefore the path runs through Aj, Ah, dummy, Dh, Dg, Df, De, Dd, Dc, Db, Da.

An interesting little problem arises when calculating the float of activity Ce, since there are two strings emanating from the end node of that activity. By conventional backward pass methods - and indeed this is how a computer carries out the calculation - one would insert the backward pass in the nodes starting from the end (see Figure 24.6). When arriving at Ce, one finds that the latest possible time is 40 when calculating back along string Cg and Cf, while

it is 38 when calculating back along string Ag, Af. Clearly, the actual float is the difference between the earliest date and the earliest of the two latest dates, i.e. day 38 instead of day 40. The float of Ce is therefore 38 - 21 = 17 days.

As described above, the calculation is tedious and time consuming. A far quicker method is available by using the technique shown in Figure 14.5, i.e. one simply inserts the various forward passes on each string and then looks at the end node of the activity in question - in our case, activity Ce. It can be seen that by following the two strings emanating from Ce that string Af, Ag joins Ah at day 36. String Cf, Cg, on the other hand, joins Ah at day 34. The float is, therefore, the smallest difference between the highest day number and one of the two day numbers just mentioned. Clearly, therefore, the float of activity Ce is 53 - 36 = 17 days. Cf and Cg, of course, have a float of 53 - 34 = 19 days.

The time to inspect and calculate the float by the second method is literally only a few minutes. All one has to do is to run through the paths emanating from the end node of the selected activity and note the highest day number where the strings meet the critical path. The difference between the day number of the critical string and the highest number on the tributary strings (emanating from the activity in question) is the float. Supposing we now wish to find the float of activity Gb:

Follow string Fd, Fe, Follow string Gc, Gd, Ge, Follow string Gf, Gg, Gh, Follow string Ef, Eg, Ah.

Fe and Gd meet at Ge, therefore they can be ignored.

String Gf-Gh meets Aj at day 45 String Ef-Eg meets Ah at day 36 Therefore float is either 56 - 45 = 11 or 53 - 36 = 17

Clearly, the correct float is 11 since it is the smaller. The time taken to inspect and calculate the float was exactly 21 seconds!

All the floats calculated above have been total floats. Free float can only occur on activities entering a node when more than one enters that node. It can be calculated very easily by subtracting the total float of the incoming activity from the total float of the outgoing activity, as shown in Figure 24.7. It should be noted that one of the activities entering the node must have zero free float.

When more than one activity leaves a node, the value of the free float to be subtracted is the lowest of the outgoing activity floats, as shown in Figure 24.8.

TF = Total float FF = Free float a = 28 - 20 = 8 b = 25-20 = 5

TF = Total float FF = Free float

TF = Total float FF = Free float

Figure 24.8

TF = Total float FF = Free float

Figure 24.8

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Project Management Made Easy

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