Extended Warranty, a good deal?

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Scenario

We are purchasing a $500 device (a printer for example). Should we take the $80 5-year extended warranty?

Let’s take the following assumptions:

• the device has 8% chance of failing over 5 years.
• the cost of repairing the device is $0 if the device is under warranty or $250 if it is not the case.

We can draw the following decision tree:

• We have 2 alternatives: we take the warranty or not.
• For each alternative, the device can fail or not.

Cost of each alternative

“Not taking the warranty” is the cheapest alternative.

If we take the warranty, the cost will be $80 whether or not the device fails. \[ \begin{align*} Cost_{Warranty} &= Prob_{DeviceFails} \times Cost_{DeviceFails} + Prob_{DeviceDoesNotFail} \times Cost_{DeviceDoesNotFail} \\ Cost_{Warranty} &= 8\% \times \$80 + 92\% \times \$80 \\ Cost_{Warranty} &= \$80 \end{align*} \]

If we do not take the warranty the cost is $20. \[ \begin{align*} Cost_{NoWarranty} &= Prob_{DeviceFails} \times Cost_{DeviceFails} + Prob_{DeviceDoesNotFail} \times Cost_{DeviceDoesNotFail} \\ Cost_{NoWarranty} &= 8\% \times \$250 + 92\% \times \$0 \\ Cost_{NoWarranty} &= \$20 \end{align*} \]

What if the repair cost was more expensive?

The “no warranty” alternative is becoming cheaper if the repair cost exceeds $1,000 which does not make sense for a $500 device.

We will calculate what should be the cost of repair (outside of warranty) so that the cost of the “no warranty” alternative equals $80 (i.e. cost of the “warranty” alternative).

\[ \begin{align*} Cost_{NoWarranty} &= Prob_{DeviceFails} \times Cost_{DeviceFails} + Prob_{DeviceDoesNotFail} \times Cost_{DeviceDoesNotFail} \\ Cost_{DeviceFails} &= \frac{Cost_{NoWarranty} - Prob_{DeviceDoesNotFail} \times Cost_{DeviceDoesNotFail}}{Prob_{DeviceFails}} \\ Cost_{DeviceFails} &= \frac{\$80 - 92\% \times \$0}{8\%} \\ Cost_{DeviceFails} &= \$1000 \end{align*} \]

What if the probability of failure was higher?

The "no warranty alternative is becoming cheaper if the probability of failure over 5 years exceeds 32%. We should run away from such an unreliable device.

We will calculate what should be the probability of failure so that the cost of the “no warranty” alternative equals $80 (i.e. cost of the “warranty” alternative).

\[ \begin{align*} Cost_{NoWarranty} &= Prob_{DeviceFails} \times Cost_{DeviceFails} + Prob_{DeviceDoesNotFail} \times Cost_{DeviceDoesNotFail} \\ Cost_{NoWarranty} &= Prob_{DeviceFails} \times Cost_{DeviceFails} + (1 - Cost_{DeviceFails} ) \times Cost_{DeviceDoesNotFail} \\ Prob_{DeviceFails} &= \frac{Cost_{NoWarranty} - Cost_{DeviceDoesNotFail}}{Cost_{DeviceFails} - Cost_{DeviceDoesNotFail}} \\ Prob_{DeviceFails} &= \frac{\$80 - \$0}{\$250 - \$0} \\ Prob_{DeviceFails} &= 32 \% \end{align*} \]

What if we do not take the warranty and the device fails?

We need to make a distinction between a “decision” and the corresponding “outcome”. We can make a good decision and get a bad outcome and vice versa.