Answers to the May 22 Weekend Puzzler on Drag Cost

My recent Weekend Puzzler blog, to calculate drag and drag cost (in human lives) attracted quite a bit of attention. A total of 241 different readers viewed it over the weekend, generating a total of 288 views and pushing the total for the blog thus far this year to just over 6,900. This is sufficient encouragement to make me believe that there is a readership for relatively complex project management information, and therefore to keep doing this. This is despite the fact that only 5 people entered a total of 16 answers in the “poll” multiple choice selection format. Oh, well, perhaps many other people figured out their answers but just didn’t enter them into the poll. Of the 16 answers entered into the poll, 10 were correct (and I have a sneaking feeling that one person got them ALL correct, in which case congratulations to that person!). First, from the last article, the final network diagram schedule with all the schedule computations on which the drag cost questions were based: Fig 4 All FS network lives for blog Now here are the answers and explanations. *****

  1. If Activity B takes 10D longer than planned, how many more lives will it cost? A= 7 B= 10 C=26 D=40 E=50 (4 correct answers out of 5)

If Activity B takes 10D longer than planned, it will take 15D, use up all 8D of its float, and migrate to the critical path with 2D of drag. Those 2D will push the project duration to 69D and cost 2 * 5 lives, or 10 more lives lost.

  1. If we can shorten Activity D to 3D instead of 7D, how many more lives would that save? A= 0 B= 7 C=10 D=14 E=20 (1 correct answer out of 3)

If Activity D is shortened to 3D instead of 7D, the first 2D will be its drag and the next 2D will increase its float to 2D. Those first 2D will pull the project duration in to 65D and save 2 * 5 lives, or 10 more lives saved.

  1. If we take resources away from Activity P so that it takes 30D instead of 15D, how many more lives will it cost? A= 0 B= 5 C=10 D=12 E=75 (2 correct answers out of 3)

If Activity P’s duration is increased to 30D instead of 15D, that will be an increase of 15D. But P had float of 17D. Therefore the increase in P’s duration will only reduce its float to 2D and will not impact the critical path and the project duration. There will be no change in the number of lives saved/lost.

  1. If Activity P is now scheduled to take those 30D, AND we also shorten Activity K to 12D, how many more lives would it save OR cost? A= 0 B= 7 more lost C=10 more lost D=10 more saved E=50 more saved (2 correct answers out of 3)

With Activity P now having just 2D of float, it means that K’s drag has been reduced to 2D. Now if K is shortened by 10D, the first 2D will be drag and the next 8D will be float. Those first 2D will pull the project duration in to 65D and save 2 * 5 lives, or 10 more lives saved.

  1. With Activity K now planned to take 12D, if we now decide to give some resources back to Activity P so that it only takes 20D, how many more lives would be saved than in Question 4? A= 0 B=10 C=16 D=22 E=40 (1 correct answer out of 2)

With K now having 8D of float, Activity P is on the critical path with 8D of drag. If P is now reduced by 10D, the first 8D will be its drag and the next 2D will increase its float to 2D. The critical path will have changed and now will again go through Activities K and Q. This will also pull the project duration in by 8D, from 65D to 57D, and save 2 lives for each day or a total of 8 * 2 lives, or 16 more lives saved. ****** If you understand this whole process but find drag and drag cost difficult to compute, then please push the powers that be (i.e., PMI, IPMA, APM and all the software companies) to start incorporating these data in their critical path analysis computations. Because the calculations are dead easy for a computer, and this information is critical! Fraternally in project management, Steve the Bajan

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