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How do you determine 'cost' when selling returns?

  • avatar

    by: Muleears
    SaleHoo Junior Member
    10 posts

    How do you determine 'cost' when selling returns?

    Posted 26 Nov 08 10:13 pm.

    I'm not new to Ebay but I am new to selling Jimmy's stuff. How does one determine what their cost is for each item? I try to determine retail of the items then divide my total cost by the total retail value which give me a percentage. I use this percentage to determine my cost. Sound reasonable? Is there a simpler/better way? Please share it with me!
  • avatar

    by: jimmy_huber
    SaleHoo Master Member
    2868 posts

    Re: How do you determine 'cost' when selling returns?

    Posted 6 Feb 09 5:32 am.

    Hmm Im not sure how your doing it if you are trying to find out a cost per item you would just divide the cost of the pack or pallet by the number of units you have. Lets say you got a $45 sample and you had 8 items in the box, that would mean your paying $5.65 for each item. If you bought a pallet for lets say $600 and you had 85 units on the pallet your cost for the items on the pallet would be $7.05 for each unit.

    Example. Macys has a High end Houswares pallet that is worth $5,900 Retail
    There are 89 Units on this pallet

    Buyer cost is $1,273

    Buyer percentage is $14.30 per unit.

    Also meaning the average retail value of each item is $66.29

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  • avatar

    by: Tildorf
    SaleHoo Junior Member
    11 posts

    Re: How do you determine 'cost' when selling returns?

    Posted 6 Feb 09 12:03 pm.

    I have not bought or sold any pallets yet so take it for what its worth. :)

    I came up with the same equation Muleears did. Here's why. If you buy a $5,900 pallet some items may retail at $400 and some may retail at. $20. They aren't equal. With Muleear's equation you can calculate the profit or loss of each item. With Jimmy's calculation you can only determine the profit or loss per pallet. The end result is the same, but the more information you can get the better IMO. :)

    Jason