Repair (equipment)

You can repair damaged equipment by visiting a vendor, and paying a fee that depends on the amount of damage the equipment has suffered (this fee may vary depending on the vendor).

Repair Vendors
Most vendors repair equipment at a cost of Repair Cost (pp)= 0.1 * (points of damage to durability) / (current maximum durability) * (item's original undamaged base price in pp) Which can be expanded to Repair Cost (pp)= 0.1 * (current maximum durability - current durability) / (current maximum durability) * (item's current base price in pp) * (item's original undamaged durability) / (current maximum durability)

That is, it costs 10% of the item's original base price to repair it from being fully damaged (i.e., Broken).

However, it is more expensive to repair at a few locations:
 * Expert repair from:
 * Marraenoloth, the Yugoloth Ferryman in The Shroud raid (10x the normal cost, Expert-level repair for all item types)
 * Haddad, the Efreeti Forgemaster summoned by the repair anvil item available in the DDO Store (10x the normal cost, Expert-level repair for all item types)
 * Jaidene Forgemaiden in Zawabi's Refuge in the Demon Sands (5x the normal cost, Expert-level repair for weapons and armor only, requires 400 Free Agent Favor)


 * Apprentice repair from:
 * Jaidene Forgemaiden's apprentice in Zawabi's Refuge in the Demon Sands (2x the normal cost, Apprentice-level repair for all item types, no Favor requirement)
 * Rechard Sorl's apprentice in The Erstwhile Emporium in House Phiarlan (2x the normal cost, Apprentice-level repair for clothing and jewelry, no Favor requirement)


 * Normal repair at inflated price from:
 * Eldin Thestral, Guild General Vendor in House Kundarak costs 500% more for no benefit (scheduled to be reduced in Update 7, Sep 2010)
 * Ostler Caulstone, in The Foothold in the Restless Isles costs 25% more for no benefit.
 * Hargrun, General Vendor in Gianthold costs 25% more for no benefit.
 * Belline Toulia, General Vendor in Meridia costs 25% more for no benefit.
 * Manar'ak, General Vendor in Reaver's Refuge costs 25% more for no benefit.
 * Steppy, Barkeep and Repairs in Delver's Canteen in Threnal costs 20% more for no benefit.

In most of those cases, there are other vendors (Potion, Barkeep, etc) nearby that repair at the normal rate. Use them instead.


 * Normal-quality repair services are 10% cheaper from:
 * Sorvile Smythe, Free Agent Vendor in the Harbor (requires 75 Free Agent Favor)
 * Tarvin d'Deneith, House Deneith Vendor in House Deneith (requires 75 House Deneith Favor)

When repairing any non-bound piece of equipment, there is always some chance of permanently damaging it.

Permanent damage can never be removed from an item, but there are a few ways to avoid or mitigate it:
 * 1) Bound items (Bound to Character, Bound to Account, or Bound and Attuned via the Stone of Change ritual) never suffer permanent damage when being repaired.
 * 2) Items that have sustained temporary damage can be Bound and Attuned before repairing to avoid the chance of permanent damage.
 * 3) Items that are replaced by a different version of themselves do not keep their permanent damage (i.e., upgrading an item to its Epic version).
 * 4) The Adamantine Rituals at the Stone of Change add durability to an item, which can bring it back from unusability (i.e., a +3 CHA Tome that has been reduced to 0 maximum durability can have 10 durability added to it, to make it possible to use the Tome).

The normal chance of permanent damage is given by P(permanent damage) = Percent of durability being repaired = 1 - (current durability / current maximum durability) = current temporary damage / current maximum durability That is, if you repair something that is 1% damaged, you have a 1% chance of getting permanent damage; if you repair something that is 50% damaged, you have a 50% chance of getting permanent damage; if you repair something that is 100% damaged (i.e., Broken), you have a 100% chance of getting permanent damage.

This is for Normal-quality repair services. For Expert-quality repair services, this chance is halved. For Apprentice-quality repair services, this chance is reduced by ~10% (88.20% of the 551 broken items tested were permanently damaged, [link]).

Additionally, the amount of permanent damage taken when repairing is given by max( 1, trunc( sqrt( x ) ) ) where x is the amount of permanent damage already taken by the item. So, the amount of permanent damage taken by any item will always follow this sequence:

Proof that it does not matter when you repair
By looking at the above formula, an interesting question is: when is it optimal to repair so as to maximize an item's duration? It turns out that it does not matter when you repair, as is shown below (using some latex-like notations for maths), and under some necessary assumptions also presented below.

Before going into the maths, a simple English formulation of the result would be the following:

''Repairing often does not change the average lifespan of your items. It tends to make their effective lifespan vary more than if you repair less frequently: if you're lucky, your item will last longer, otherwise it will last less, both having equal chance of happening.''

The required assumptions for this to be true are: If you have made any experiment supporting (or denying) any of these two assumptions, please tell me about them!
 * The amount of temporary damage you get when hitting a mob must not depend on the current amount of temporary damage. This was confirmed to be true by developer Eladrin on January 31st, 2008.
 * The formula above is computed exactly (or at least, to a sufficient precision). If rounding occurs, then the result is not obvious (and the "best" strategy now depends on the current maximum durability of the item). In particular, a specific way of rounding could result in a zero probability of getting permanent damage when repairing an item with a single point of temporary damage and a maximum durability higher than 100 (see the discussion page for details).

If you have made experiments to verify this, please tell me about it!

Denote by d the current maximum durability of the item, and by x the amount of damage taken before you decide to repair it. We are interested in how long it will take before the item gets permanent damage.

Each time you repair, the probability of suffering permanent damage is x/d, and the probability of not suffering permanent damage is thus 1 - x/d. Since each repair event is independent, the probability of getting permanent damage exactly on the n-th repair is equal to P(permanent damage on n-th repair) = (1 - x/d)^(n-1) * x/d The expected number of repairs before taking your first permanent damage is thus: E[number of repairs before taking first permanent damage] = sum_{n = 1}^infinity n * P(permanent damage on n-th repair) = sum_{n = 1}^infinity n * (1 - x/d)^(n-1) * x/d To compute this sum, consider any y in ]0,1[ and the desired sum sum_{n = 1}^infinity n * y^(n-1). It can also be written (skipping some maths technical details to move the derivative from inside the sum to outside the sum): sum_{n = 1}^infinity n * y^(n-1) = sum_{n = 1}^infinity d(y^n)/dy = d/dy ( sum_{n = 1}^infinity y^n ) = d/dy ( 1 / (1 -y) - 1 ) = 1 / (1 - y)^2 Plugging this result into the precent equation, we obtain E[number of repairs before taking first permanent damage] = x/d * 1 / (1 - (1 - x/d))^2 = x/d * d^2 / x^2 = d/x But what we are interested in is the expectation of the amount of total temporary damage taken before getting the first permanent damage (i.e. also the amount of time before getting the first permanent damage). This is also equal to the number of repairs multiplied by x, since x is the amount of temporary damage taken before we repair. Thus E[total temporary damage (or time elapsed) before getting first permanent damage] = d/x * x = d which does not depend on x!

About the second part of the result, namely that repairing often introduces more variance in an item's lifespan, it can be shown by computing the variance just like we computed the expectation. I won't go into the details unless someone asks me to, but I obtained a simple formula for the variance: V[number of repairs before taking first permanent damage] = d * (d - x)

I have verified the above mathematical results with a computer simulation. See the talk page for details.