Cost of Quality: A technical treatment

1. Cost of Quality: A technical treatment

2. Let’s think for a moment the Cost of Quality • Suppose there are S stages in a manufacturing or service process, indexed by s=1,…S • Denote defective rate of each as pi. • The variable cost at each stage is vi. • Let selling price be r, which is greater than vi. • Start with materials and labor to make n units, we ended up with n(1-p1).(1-p2)…(1-pS) units • At each stage, the cost is nv1, n(1-p1).v2, …, and n(1-p1).(1-p2)…(1-pS-1) vS, respectively.

3. Cost of Quality • So the total cost is nv1+ n(1-p1).v2+.. +n(1-p1).(1-p2)…(1-pS-1) vS • The total revenue is rn(1-p1).(1-p2)…(1-pS) • So it follows that the profit is [rn(1-p1).(1-p2)…(1-pS)] -[nv1+ n(1-p1).v2+.. +n(1-p1).(1-p2)…(1-pS-1) vS ]

4. Cost of Quality (S=2) • What if you have perfect quality? P1=p2=0  Profit = r.n – (v1+v2).n • Compare with sub-quality: profit = [rn(1-p1).(1-p2)]-[nv1+ n(1-p1).v2] • The difference is cost of quality (COQ) COQ = rn[1-(1-p1).(1-p2)] –v2np1 • Is COQ > 0? The answer is yes. Can you see why? • So what is the insight from above?

5. Comparing Cost per Good Unit

7. So which step should you correct first? • Think about the simplest case of s=2 first. • Recall that COQ = rn[1-(1-p1).(1-p2)] –v2np1 • The marginal impact can be obtained by taking the derivative of CQQ with respect to the ps • d(COQ)/dp1=(1-p2)rn-v2n • d(COQ)/dp2=(1-p1)rn • So you should correct step 1 first, if d(COQ)/dp1> d(COQ)/dp2, which is • (1-p2)rn-v2n>(1-p1)rn • Rn(p1-p2)>v2n  p1-p2>v2/r