Solved: The demand for subassembly S is 100 units in week 7. Each unit of S requires 1 unit of

a)Demand for subassembly S is during week 7

It requires a lead time of 2 weeks to get ready

Therefore S must start getting ready during week 7 – 2 = Week 5

Time phased product structures A, B and C represents   above. However time phased product structure D does not reflect above condition since it shows requirement of S in week 1 and getting ready in week 3 . Therefore Item D is eliminated from possible choices.

b)Since S must start getting ready in week 5, its input items i.e. T and U both must be delivered by week 5

T has a lead time of delivery of 1 week and therefore T must start getting ready by week 5 – 1 = week 4

Item T requires V, W and X as input items and therefore all three must be delivered by week 4

V has a lead time of delivery of 1 week, therefore V must start getting ready during week 4 -1 = 3^rd week

W has a lead time of delivery of 3 weeks , therefore W must start getting ready during 4 – 3 = 1st week

X has a lead time of delivery of 2 weeks , therefore X must start getting ready during 4 -2 = 2^nd week

Only time phased structures B and C represent above and therefore time phased structure A is rejected as possible option

c)Since U must be delivered by week 5 and it has a lead time of 2 weeks, U must start getting ready by week 3

The items Y and Z , both of which go into making of U , thus must be delivered by week 3

Since it takes 2 weeks to make Y , therefore Y must start getting ready by week 3 – 2 = Week 1

Since it takes 1 week to make Z , therefore Z must start getting ready by week 3 – 1 = week 2

Out of option B and C of time phased product structure, it is only option C which satisfies above condition.

Therefore FINAL CHOICE IS “OPTION C”