The numbers in Problems 5 and 6 have been changed. Thus, please do not copy the
ID: 432467 • Letter: T
Question
The numbers in Problems 5 and 6 have been changed. Thus, please do not copy the authors' solution books. Otherwise, you are deemed to get wrong answers answers from the 5. (20 points) 6.13 In an assembly plant, material handling berween departments is per performed using a uni- directional closed-loop conveyor. The figure below showrs the layout for the modular Conveyor fiow PO#5 facility, which consists of three equal-sized assembly modules (A, B, and C), one admin- istrative module (D), and one warehouse module f. P/D points for each module are also shown in the figure. The administrative and warehouse activities are not to be moved; however, assembly areas A, B, C can be relocated. The distance between P/D points and the number of pallet loads moved berween departments are given below. Pallet How per Day C DE 0 50 From To PD2 PDS 25 o 1535 PD PDS P. D 5 P.Di Using the pairwise exchange method, determine new locations for assembly modules A, B, and C that minimize the sum of the products of pallet flows and conveyor travel The combination will produce 6 cases. Please calculate only the following two cases: Case 1: RD#3 190 PID#2 | 60 Conveyor flow PIDE! PID#4 90Explanation / Answer
Case 1:
Module
No. of Pallets between
Location
Distance to (in feet)
Pallets x Distance (in feet)
A
C: 5
P/D 3
P/D 5: 30 + 90 = 120
5 x 120 = 600
E: 30
P/D 1: 30 + 90 + 60 = 180
30 x 180 = 5,400
B
A: 10
P/D 2
P/D 3: 90
10 x 90 = 900
C: 25
P/D 5: 90 + 30 + 90 = 210
25 x 210 = 5,250
C
A: 25
P/D 5
P/D 3: 60 + 60 + 90 = 210
25 x 210 = 5,250
B: 5
P/D 2: 60 + 60 = 120
5 x 120 = 600
D
C: 15
P/D 4
P/D 5: 90
15 x 90 = 1,350
E: 35
P/D 1: 90 + 60 = 150
35 x 150 = 5,250
E
A: 5
P/D 1
P/D 3: 60 + 90 = 150
5 x 150 = 750
B: 20
P/D 2: 60
20 x 60 = 1,200
C: 5
P/D 5: 60 + 90 + 30 + 90 = 270
5 x 270 = 1,350
Total
27,900
Case 2:
Module
No. of Pallets between
Location
Distance to (in feet)
Pallets x Distance (in feet)
A
C: 5
P/D 3
P/D 2: 30 + 90 + 60 + 60 = 240
5 x 240 = 1,200
E: 30
P/D 1: 30 + 90 + 60 = 180
30 x 180 = 5,400
B
A: 10
P/D 5
P/D 3: 60 + 60 + 90 = 210
10 x 210 = 2,100
C: 25
P/D 2: 60 + 60 = 120
25 x 120 = 3,000
C
A: 25
P/D 2
P/D 3: 90
25 x 90 = 2,250
B: 5
P/D 5: 90 + 30 + 90 = 210
5 x 210 = 1,050
D
C: 15
P/D 4
P/D 2: 90 + 60 + 60 = 210
15 x 210 = 3,150
E: 35
P/D 1: 90 + 60 = 150
35 x 150 = 5,250
E
A: 5
P/D 1
P/D 3: 60 + 90 = 150
5 x 150 = 750
B: 20
P/D 5: 60 + 90 + 30 + 90 = 270
20 x 270 = 5,400
C: 5
P/D 2: 60
5 x 60 =300
Total
29,850
Module
No. of Pallets between
Location
Distance to (in feet)
Pallets x Distance (in feet)
A
C: 5
P/D 3
P/D 5: 30 + 90 = 120
5 x 120 = 600
E: 30
P/D 1: 30 + 90 + 60 = 180
30 x 180 = 5,400
B
A: 10
P/D 2
P/D 3: 90
10 x 90 = 900
C: 25
P/D 5: 90 + 30 + 90 = 210
25 x 210 = 5,250
C
A: 25
P/D 5
P/D 3: 60 + 60 + 90 = 210
25 x 210 = 5,250
B: 5
P/D 2: 60 + 60 = 120
5 x 120 = 600
D
C: 15
P/D 4
P/D 5: 90
15 x 90 = 1,350
E: 35
P/D 1: 90 + 60 = 150
35 x 150 = 5,250
E
A: 5
P/D 1
P/D 3: 60 + 90 = 150
5 x 150 = 750
B: 20
P/D 2: 60
20 x 60 = 1,200
C: 5
P/D 5: 60 + 90 + 30 + 90 = 270
5 x 270 = 1,350
Total
27,900
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.