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Can someone explain to me how do we define what week to place a number for the required dates?
I understand part A but the other parts seem confusing.
For example, why do part B and C both have a required date in week 7?
Why do part E and D both have a required date in week 6?
Gross Requirements Plan Gross Material Requirements Plan for 50 Awesome Speaker Kits (As) with Order Release Dates Also Shown TABLE 14.3 WEEK TIME A. Required date 50 Order release date 50 1 week B. Required date 100 Order release date 100 2 weeks C. Required date 150 Order release date 150 1 week E. Required date 200 Order release date 200 300 2 weeks F. Required date 300 Order release date 300 3 weeks D. Required date 600 200 Order release date 600 200 1 week G. Required date 300 Order release date 300 2 weeks © 2014 Pearson EducationInc 14-34 ,
Expert Answer
Note that as per the product structure,
two(2) units of B are fitted in one(1) unit of A
& three(3) units of C are fitted in one(1) unit of A.
Since B. and C are just at the lower hierarchy of ‘A’ in the product structure, the week when the order release of A happens, the required date of B and C happens.
So, since the order release of A happens in week 7, the required date of B, and C is in the same week (i.e. week 7). The numbers are as per the units required i.e. No. of B = 2 x No. of A = 2 x 50 = 100 and No. of C=3 x 50 = 150.
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E, and D are more complicated. But the logic is the same as above.
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Consider ‘E’ first.
Two(2) units of E are fitted in one(1) unit of B
Since E is just at the lower hierarchy of B in the product structure, the week when the order release of B happens, the required date of E happens.
So, since the order release of B happens in week 5, the required date of E is in the same week (i.e. week 5). The numbers are as per the units required i.e. No. of E = 2 x No. of B = 2 x 100 = 200
So, for making ‘B’ only, we need 200 units of E in week 5 ————————–1
Two(2) units of E are fitted in one(1) unit of C
Since E is just at the lower hierarchy of C in the product structure, the week when the order release of C happens, the required date of E happens.
So, since the order release of C happens in week 6, the required date of E is in the same week (i.e. week 6). The numbers are as per the units required i.e. No. of E = 2 x No. of C = 2 x 150 = 300
So, for making ‘C’ only, we need 300 units of E in week 6 ————————– 2
Combine 1 and 2 to get the overall requirement of E i.e. 200 in week 5 and 300 in week 6
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Consider ‘D’ now
Two(2) units of D are fitted in one(1) unit of B
Since D is just at the lower hierarchy of B in the product structure, the week when the order release of B happens, the required date of D happens.
So, since the order release of B happens in week 5, the required date of D is in the same week (i.e. week 5). The numbers are as per the units required i.e. No. of D = 2 x No. of B = 2 x 100 = 200
So, for making ‘B’ only, we need 200 units of D in week 5 ————————–3
Two(2) units of D are fitted in one(1) unit of F
Since D is just at the lower hierarchy of F in the product structure, the week when the order release of F happens, the required date of D happens.
So, since the order release of F happens in week 3, the required date of D is in the same week (i.e. week 3). The numbers are as per the units required i.e. No. of D = 2 x No. of F = 2 x 300 = 600
So, for making ‘F’ only, we need 600 units of D in week 3 ————————–4
Combine 3 and 4 to get the overall requirement of D i.e. 200 in week 5 and 600 in week 3
