Leasable area



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    Stephen Starks



    The way this question seems to work is:

    1. We are given a target "Net" SF to meet / calculate for = 4500 SF.
    2. We are informed that the the target existing office building's "Efficiency" = 0.75
    3. We are informed that the building load factor = 1.20.

    If the building was 100% efficient (1.00) we could multiply Net SF by building load factor and be done - but the target building does not use available space as well as that, and this must be accounted for by an additional step.

    From "target Net" and "existing Efficiency" we then determine that the building's actual "Gross" SF = 6000 SF.

    We then multiply 6000 SF by 1.20 and the product = 7200 SF (Answer C).

    Alternately: (4500 SF / 0.75) x 1.2 = 7200 SF. (this method is a bit less cumbersome).


    As always, check up on this your self though.

    Thanks for the exercise :)

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    Shanglin Li

    Thanks Stephen,

    but I still couldn’t understand this, cuz first, I understand this loading formula is relative to Rentable and Usable space, there’s nothing about Gross area. Second, bases on the answer, is that means the Rentable Area is larger than Gross area? How is that possible?


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    Francis Pham

    Hi All,

    Was anyone able to follow up on this? I also dont understand this as R/U = rentable/useable , Gross area is not mentioned anywhere. So why solve for gross area ? 

    Thanks ahead of time!

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