Southwestern University (SWU), a large state college in Stephenville, Texas, 30 miles southwest of the Dallas/Fort Worth metroplex, enrolls close to 20,000 students. In a typical town-gown relationship, the school is a dominant force in the small city, with more students during fall and spring than permanentresidents.
A longtime football powerhouse, SWU is a member of the Big Eleven conference and is usually in the top 20 in college football rankings. To bolster its chances of reaching the elusive and long-desired number-one ranking, in 2012, SWU hired the legendaryPhil Flamm as its head coach.
One of Flamm’s demands on joining SWU had been a new stadium. With attendance increasing, SWU administrators began to face the issue head-on. After 6 months of study, much political arm wrestling, and some serious financial analysis, Dr. Joel Wisner, president of Southwestern University, had reached a decision to expand the capacity at its on-campus stadium.
Adding thousandsof seats, includingdozens of luxury skyboxes, would not please everyone. The influential Flamm had argued the need for a first-class stadium, one with built-in dormitory rooms for his players and a palatial office appropriate for the coach of a future NCAA champion team. But the decision was made, and everyone, including the coach, would learn to live with it.
The job now was to get construction going immediately after the 2019 season ended. This would allow exactly270 days until the 2020 season opening game. The contractor, Jefferson Construction, signed his contract. Malick Jefferson looked at the tasks his engineers had outlined and looked President Wisner in the eye. “I guarantee the team will be able to take the field on schedule next year,” he said with a sense of confidence. “I sure hope so,” replied Wisner. “The contract penalty of $10,000 per day for running late is nothing compared to what Coach Flamm will do to you if our opening game with Penn State is delayedor canceled.” Jefferson, sweating slightly, did not need to respond.In football-crazy Texas, JeffersonConstruction would be mud if
the 270-day target was missed.
Back in his office, Jefferson again reviewed the data (see the following table) and noted that optimistic time estimates can be used as crash times. He then gatheredhis supervisors. “Folks, if we’re not 75% sure we’ll finish this stadium in less than 270 days, I want this project crashed! Give me the cost figures for a target date of 250 days—also for 240 days. I want to be early, not just on time!”
Time Estimates(Days)
| Activity | Description | Predecessor(s) | Optimistic | Most Likely | Pessimistic | Crash Cost/Day |
| A | Bonding, insurance, tax structuring | — | 20 | 30 | 40 | $1,500 |
| B | Foundation, concrete footingsfor boxes | A | 20 | 65 | 80 | 3,500 |
| C | Upgrading skybox stadiumseating | A | 50 | 60 | 100 | 4,000 |
| D | Upgradingwalkways, stairwells, elevators | C | 30 | 50 | 100 | 1,900 |
| E | Interior wiring, lathes | B | 25 | 30 | 35 | 9,500 |
| F | Inspectionapprovals | E | 0.1 | 0.1 | 0.1 | 0 |
| G | Plumbing | D, F | 25 | 30 | 35 | 2,500 |
| H | Painting | G | 10 | 20 | 30 | 2,000 |
| I | Hardware/AC/metal workings | H | 20 | 25 | 60 | 2,000 |
| J | Tile/carpet/windows | H | 8 | 10 | 12 | 6,000 |
| K | Inspection | J | 0.1 | 0.1 | 0.1 | 0 |
| L | Final detail work/cleanup | I, K | 20 | 25 | 60 | 4,500 |
Part 1: What is the length of the critical path as computed from the table?
Part 2: What is the standard deviation of the critical path?
Part 3: What is the probability of completing the project in 270 days?
Part 4: What is the cost of crashing to 250 days?
To determine the length of the critical path, we need to identify the sequence of activities that, when combined, take the longest time to complete. Any delay in these critical activities will directly impact the overall project timeline.
Here are the activities and their respective durations:
Activity A: 30 days (Most Likely) Activity B: 65 days (Most Likely) Activity C: 60 days (Most Likely) Activity D: 50 days (Most Likely) Activity E: 30 days (Most Likely) Activity G: 30 days (Most Likely) Activity H: 20 days (Most Likely) Activity I: 25 days (Most Likely) Activity J: 10 days (Most Likely) Activity K: 0.1 days (Most Likely) Activity L: 25 days (Most Likely)
Now, let’s identify the critical path by adding up the durations of activities with no slack (total float) and selecting the longest path:
Critical Path: A -> B -> D -> G -> H -> I -> J -> K -> L
The length of the critical path is the sum of the most likely durations of the activities on this path:
Critical Path Length = 30 + 65 + 50 + 30 + 20 + 25 + 10 + 0.1 + 25 = 265.1 days
So, the length of the critical path is 265.1 days.
To calculate the standard deviation of the critical path, we need to use the standard deviation formula for a sequence of dependent activities. The standard deviation of a critical path can be determined by considering the variances of individual activities along that path. However, it’s essential to note that not all activities have variances; some are deterministic (with no variability).
Here are the activities with their variances (variance = [(Pessimistic – Optimistic) / 6]^2):
Activity A: Variance = [(40 – 20) / 6]^2 = (20 / 6)^2 ≈ 11.11 Activity B: Variance = [(80 – 20) / 6]^2 = (60 / 6)^2 = 100 Activity D: Variance = [(100 – 30) / 6]^2 = (70 / 6)^2 ≈ 81.94 Activity G: Variance = [(35 – 25) / 6]^2 ≈ (10 / 6)^2 ≈ 2.78 Activity H: Variance = [(30 – 20) / 6]^2 = (10 / 6)^2 ≈ 2.78 Activity I: Variance = [(60 – 25) / 6]^2 = (35 / 6)^2 ≈ 68.06 Activity J: Variance = [(12 – 8) / 6]^2 = (4 / 6)^2 ≈ 0.44 Activity K: No Variance (Deterministic) Activity L: Variance = [(60 – 25) / 6]^2 = (35 / 6)^2 ≈ 68.06
Now, we sum the variances along the critical path and take the square root to find the standard deviation:
Standard Deviation of Critical Path = √(11.11 + 100 + 81.94 + 2.78 + 2.78 + 68.06 + 0.44 + 0 + 68.06) ≈ √(335.17) ≈ 18.30 days
So, the standard deviation of the critical path is approximately 18.30 days.
To calculate the probability of completing the project in 270 days, we can use the critical path length and standard deviation. Assuming a normal distribution of project completion times, we can find the Z-score for 270 days and then use a standard normal distribution table (z-table) to find the corresponding probability.
Z-score formula: Z = (X – μ) / σ
Where:
Z = (270 – 265.1) / 18.30 ≈ 0.268
Now, we consult the z-table to find the probability associated with a Z-score of 0.268. Looking up this value in the table, we find that the probability is approximately 0.6071.
So, the probability of completing the project in 270 days is approximately 60.71%.
To calculate the cost of crashing the project to 250 days, we need to determine the additional cost incurred for each activity on the critical path when its duration is reduced. Crashing involves reducing the duration of critical activities to meet the target deadline.
First, let’s identify the critical activities that will be crashed:
Critical Path: A -> B -> D -> G -> H -> I -> J -> K -> L
Now, let’s calculate the cost of crashing each of these activities from their most likely duration to 250 days:
Activity A (Optimistic Duration: 20 days, Target Duration: 250 days) Additional Cost = (250 – 20) * Crash Cost/Day = 230 * $1,500 = $345,000
Activity B (Optimistic Duration: 65 days, Target Duration: 250 days) Additional Cost = (250 – 65) * Crash Cost/Day = 185 * $3,500 = $647,500
Activity D (Optimistic Duration: 50 days, Target Duration: 250 days) Additional Cost = (250 – 50) * Crash Cost/Day = 200 * $1,900 = $380,000
Activity G (Optimistic Duration: 25 days, Target Duration: 250 days) Additional Cost = (250 – 25) * Crash Cost/Day = 225 * $2,500 = $562,500
Activity H (Optimistic Duration: 20 days, Target Duration: 250 days) Additional Cost = (250 – 20) * Crash Cost/Day = 230 * $2,000 = $460,000
Activity I (Optimistic Duration: 25 days, Target Duration: 250 days) Additional Cost = (250 – 25) * Crash Cost/Day = 225 * $2,000 = $450,000
Activity J (Optimistic Duration: 10 days, Target Duration: 250 days) Additional Cost = (250 – 10) * Crash Cost/Day = 240 * $6,000 = $1,440,000
Activity K (Optimistic Duration: 0.1 days, Target Duration: 250 days) No crashing cost for K as it’s already at its minimum duration.
Activity L (Optimistic Duration: 25 days, Target Duration: 250 days) Additional Cost = (250 – 25) * Crash Cost/Day = 225 * $4,500 = $1,012,500
Now, sum up the additional costs for all these activities to find the total cost of crashing the project to 250 days:
Total Cost of Crashing to 250 Days = $345,000 + $647,500 + $380,000 + $562,500 + $460,000 + $450,000 + $1,440,000 + $1,012,500 = $5,297,500
The cost of crashing the project to 250 days is $5,297,500.
In conclusion, by identifying the critical path, calculating its standard deviation, determining the probability of completing the project in 270 days, and evaluating the cost of crashing to 250 days, project managers can make informed decisions to optimize project timelines and costs while considering the associated risks.
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