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Questions on float, Gantt charts and scheduling
Once the critical path is known, everything else has room to move. Float measures how much, a Gantt chart shows it, and a resource histogram shows how many workers are needed at each moment. Splitting the float in two says how much of that room is an activity's own and how much of it is borrowed from a neighbour further along the chain.
15 original questions · 58 marks · the float, Gantt charts and scheduling notes · Decision Mathematics 1
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These are original InkMaths questions. Write a complete answer before opening the worked solution and marking guidance.
An activity has earliest start 6, latest finish 13 and duration 4. Find its total float.
Worked solution and marking guidance
Total float = latest finish − earliest start − duration = 13 − 6 − 4 = 3 days. Equivalently, the latest start is 9 and the earliest is 6. M1 for the subtraction, A1 for 3 days.An activity has earliest start 8, latest start 11 and duration 6. Write down its total float and its latest finish.
Worked solution and marking guidance
Total float = latest start − earliest start = 11 − 8 = 3 days. B1. Latest finish = latest start + duration = 11 + 6 = 17. B1.
The other reading of the float agrees: 17 − 8 − 6 = 3, latest finish minus earliest start minus duration.A project has activities with duration, earliest start and latest finish: J (5, 0, 5), K (4, 0, 9), L (7, 5, 12), N (3, 4, 12) and P (4, 12, 16). Find the total float of every activity and name the critical ones.
Worked solution and marking guidance
Total float = latest finish − earliest start − duration.
J: 5 − 0 − 5 = 0. K: 9 − 0 − 4 = 5. L: 12 − 5 − 7 = 0. N: 12 − 4 − 3 = 5. P: 16 − 12 − 4 = 0.
M1 for the subtraction, A1 for the three zeros, A1 for K and N at 5.
The critical activities are J, L and P, and their durations 5 + 7 + 4 = 16 total the project duration, which is the check to make once every float is written down. Show the subtraction for each activity rather than only for the ones that come out non-zero: a float of zero is a calculated result, not an omission.For the seven-activity project, the critical activities are A, C, E and G, and the latest starts of B, D and F are 2, 8 and 10 with earliest starts 0, 3 and 5. State every total float.
Worked solution and marking guidance
A, C, E and G are critical, so each has float 0. B: 2 − 0 = 2. D: 8 − 3 = 5. F: 10 − 5 = 5. B1 for the four zeros, B1 for B, B1 for D and F.Explain why D and F, each with 5 days of float, cannot both be delayed by 5 days.
Worked solution and marking guidance
They lie on the same chain: F cannot start until D has finished. The five days is the slack available to the chain as a whole, not to each activity separately, which is what the word total in total float signals. Delaying D by five leaves F no room at all, and delaying both would push G, and so the whole project, five days late. B1 for the shared chain, B1 for the float belonging to the chain, B1 for the effect on the finish.The activities take 3, 5, 4, 2, 6, 3 and 2 days, one worker each, and the project lasts 15 days. Find the lower bound on the number of workers, and state the peak demand if every activity starts as early as possible.
Worked solution and marking guidance
Total work = 25 worker-days over 15 days, so the lower bound is 25/15 = 1.67, rounded up to 2 workers. At earliest start times B runs from 0 to 5, C from 3 to 7 and D from 3 to 5, so all three overlap between days 3 and 5 and the histogram peaks at 3, above the bound. Always round the bound up, whatever the decimal. M1 for 25/15, A1 for 2 workers, M1 for the overlap between days 3 and 5, A1 for a peak of three.Show how the peak can be brought down to two workers.
Worked solution and marking guidance
D has 5 days of float, so it can be delayed. Move D to run from day 5 to 7 instead of 3 to 5, which takes it out of the crowded stretch and leaves B and C alone there. F then runs from 7 to 10, within its own latest start of 10.
The demand is now A and B to day 3, B and C to day 5, C and D to day 7, E and F to day 10, then E alone and finally G. It never exceeds 2, which matches the lower bound, so the schedule is optimal and the project still finishes in 15 days. M1 for delaying D within its float, A1 for D from day 5 to day 7, A1 for F from day 7 to day 10, B1 for the demand never exceeding two.An activity of duration 3 runs from an event with earliest time 4 and latest time 9 to an event with earliest time 15 and latest time 16. Find its total float, its independent float and its interfering float.
Worked solution and marking guidance
Total float = latest finish − earliest start − duration = 16 − 4 − 3 = 9 days. M1 for the formula, A1 for 9. Independent float gives the neighbours no ground: the activity waits for the latest time of its start event, 9, and still finishes by the earliest time of its end event, 15, so it is 15 − 9 − 3 = 3 days; had this come out negative it would be recorded as zero. M1 for the earliest-at-the-end minus latest-at-the-start version. The interfering float is the rest, 9 − 3 = 6 days. A1 for the pair. Three of the nine days are the activity's own; spending any of the other six pushes a neighbour along the chain.A project has Q (4 days, no predecessors), R (5, none), S (5, after Q), T (3, after R) and U (2, after S and T). Draw a Gantt chart with every activity at its earliest start, showing the float of each, and state which activities could be in progress at time 7.
Worked solution and marking guidance
Forward pass: Q finishes at 4 and R at 5; S runs 4 to 9 and T runs 5 to 8; U waits for the later of S and T and runs 9 to 11, so the project lasts 11 days. Backward pass: U must start by 9, so S and T must both finish by 9, giving S a latest start of 4 and T one of 6; Q must finish by 4 and R by 6. M1 for the two passes.
Floats: Q, S and U are critical at 0; R has 1 − 0 = 1 and T has 6 − 5 = 1. A1.
The chart carries solid bars Q from 0 to 4, R from 0 to 5, S from 4 to 9, T from 5 to 8 and U from 9 to 11, with the three critical bars in one unbroken line across the page and a dotted tail of one day after R and after T. A1 for the five bars at their earliest starts, B1 for the float shown on R and on T.
At time 7 the bars covering the moment are S and T: S runs 4 to 9 wherever it is placed, and T covers 7 whether it starts at 5 or at 6. R cannot, since even at its latest start it has finished by 6, and U cannot start before 9. B1.
Read a chart with the floats drawn on it, not the earliest-start bars alone: an activity could be in progress at a given moment if any placement allowed by its float covers that moment, and must be if every placement does.In a project lasting 12 days, six activities each need one worker throughout. At their earliest start times they run Q from 0 to 4, R from 0 to 6, S from 4 to 9, T from 4 to 6, U from 8 to 12 and V from 9 to 12. Draw the resource histogram at earliest start times, find the lower bound on the number of workers, and state the peak demand and where it occurs.
Worked solution and marking guidance
The durations are 4, 6, 5, 2, 4 and 3, so the work is 24 worker-days over 12 days and the lower bound is 24 ÷ 12 = 2 workers. M1 for the total over the duration, A1 for 2.
Counting the activities in progress over each unit of time gives the columns 2, 2, 2, 2, 3, 3, 1, 1, 2, 2, 2, 2, drawn as a bar of that height over each unit interval from 0 to 12. B1 for the histogram with all twelve columns correct.
The peak is 3 workers, between times 4 and 6, where R, S and T overlap. M1 for finding the overlap, A1 for a peak of three with the interval named.
The demand drops to one worker between times 6 and 8, so the histogram is 1 above the bound at its peak and 1 below it in the trough. That shape is the sign that delaying an activity within its float may level the demand without lengthening the project.State what a resource histogram shows that a Gantt chart does not, what a Gantt chart shows that a resource histogram does not, and one assumption both of them make about an activity that has started.
Worked solution and marking guidance
The histogram shows the total number of workers needed at each moment, added across every activity in progress, which is what a staffing decision turns on. B1. The Gantt chart shows which activities those are, in what order they depend on each other, and how much float each carries, none of which survives being added up. B1. Both assume an activity runs without a break from its start to its finish and needs the same resource throughout, so neither can describe work that is put down and picked up again. B1.A project lasting 10 days consists of A (10 days) and B, C and D (6 days each), with no activity depending on any other and each needing one worker throughout. Find the lower bound on the number of workers, and show that the bound cannot be achieved.
Worked solution and marking guidance
Total work = 10 + 6 + 6 + 6 = 28 worker-days over 10 days, so the lower bound is 28/10 = 2.8, rounded up to 3 workers.
Suppose three workers were enough. One must spend the whole 10 days on A, since A occupies its worker from day 0 to day 10 and cannot be split. That leaves two workers for B, C and D. Each of those has only 10 days available, and having completed one six-day activity a worker has 4 days left, too few for another. So two workers can finish at most two of the three, and 4 workers are needed.
The lower bound assumes work can be divided freely and every worker kept busy.
M1 A1 for the total work, A1 for the bound of three workers, M1 for tying one worker to A, A1 for showing two workers cannot cover B, C and D, A1 for four workers.Explain what the independent float and the interfering float of an activity each measure, and why a negative value from the independent float formula is recorded as zero.
Worked solution and marking guidance
Independent float is the room an activity has when its neighbours are given no ground at all: it starts no earlier than the latest time of its start event and still finishes by the earliest time of its end event, so spending it moves nothing else in the project. B1 for room that disturbs nothing. Interfering float is the rest of the total float. It is real room, but using it pushes a later activity along its chain: it is the shared slack that makes two activities each show the same five days without there being ten. B1 for the sharing. When the formula's value is negative, the activity cannot both start that late and finish that early, so it has no room of its own at all, and a negative amount of room is no room: zero is the honest record. B1 for the reason for the clamp to zero.A project lasting 14 days consists of four activities of durations 14, 8, 8 and 6 days, with no activity depending on any other and each needing one worker throughout. Find the lower bound on the number of workers, and determine whether the bound can be achieved.
Worked solution and marking guidance
Total work = 14 + 8 + 8 + 6 = 36 worker-days over 14 days, so the lower bound is 36 ÷ 14 = 2.57, rounded up to 3 workers. M1 for the total, A1 for 36, A1 for the bound of three.
Now test whether three are enough. One worker is occupied by the 14-day activity from day 0 to day 14 and can take nothing else. M1 for tying one worker to it.
That leaves two workers and the activities of 8, 8 and 6 days. A second worker takes an 8-day activity and then the 6-day one, which is 8 + 6 = 14 days exactly and fits the project. A third takes the remaining 8-day activity and is idle for the last 6 days. A1 for a schedule of the other three activities on two workers.
So 3 workers are enough and the bound is achieved. A1.
Here 8 + 6 = 14 lands exactly on the project duration; had the durations been 8, 8 and 7, no worker could carry two of them and four workers would be needed against the same bound of three.Two activities leave the same event, whose earliest time is 6 and latest time 10. The first has duration 4 and ends at an event with earliest time 18 and latest time 21; the second has duration 5 and ends at an event with earliest time 14 and latest time 21. Find the total, independent and interfering floats of each, and explain why the two independent floats differ.
Worked solution and marking guidance
Total float = latest finish − earliest start − duration. First: 21 − 6 − 4 = 11. Second: 21 − 6 − 5 = 10. M1 for the formula, A1 for both.
Independent float takes the latest time at the start event and the earliest time at the end event, so that neither neighbour gives any ground. First: 18 − 10 − 4 = 4. Second: 14 − 10 − 5 = −1, which is negative and so recorded as 0. M1 for the swapped event times, A1 for 4 and 0 with the clamp.
Interfering float is the rest of the total. First: 11 − 4 = 7. Second: 10 − 0 = 10. A1 for both.
The two activities leave the same event, so they differ only in where they arrive. The first arrives at an event whose earliest time is 18, four days later than the 14 the second faces, and that gap is the room the first has of its own. The second cannot both wait until day 10 and finish by day 14 in five days of work, so every day of its slack is borrowed from a neighbour.
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