2) Consider the Part Type Selection problem we covered in class. This problem helps us determine which parts we will produce in a new FMS. The problem assumes that we have candidate parts i = 1, 2, ...N, that the total savings achieved by producing part i in the FMS is s, and that we know that there is a 'key bottleneck' in the FMS, which has P available time units. Each part i requires a total processing time p, in the 'key bottleneck'. we saw that a way to solve this problem is the following Greedy Heuristic: Step 1: Order parts [1] to [N] such that P [1] P[2] P[N] Define remaining available time T = P Step 2: For i = 1...N select part [i] and add the part to the FMS if p, < T and s, > 0. Each time you add a new part i, update the remaining available time as T = T - P₁ Now suppose you had 3 'key bottlenecks' (that is 3 resources that each might be the bottleneck depending on the parts you choose to produce in the FMS). These resources have P, Q and R time units available respectively. And each part i requires a total processing time p, q, and r, in each of the three, respectively. Modify the algorithm above to solve the 3-bottleneck version of the problem. TIP: In Step 1 you need to consider the processing times at all 3 stations to do the ordering. In step 2, you must keep track of the remaining capacity at each station separately. 3) An FMS consists of five stations. Station 1 is a load/unload station with one server. Station 2 performs milling operations with one server. Station 3 performs drilling operations with two servers (two identical CNC drill presses). Station 4 is for performing grinding operations with three identical servers (three identical CNC grinding machines). Station 5 is an inspection station with one server that performs inspections on a sample of the parts. The stations are connected by a part handling system that has two work carriers and a mean transport time = 4.0 min. The FMS produces six parts A, B, C, D, E, and F. The part mix fractions and process routings for the six parts are presented in the table below. Note that the operation frequency at the inspection station (f_5jk) is less than 1.0 to account for the fact that only a fraction of the parts are inspected. For this system determine: (a) Maximum production rate of the FMS; (b) Corresponding production rate of each part; (c) Utilization of each station in the system; (d) Overall FMS utilization. Part Part mix Station Frequency J Pj Op k Description Process time tijk (min) fijk A 0.1 1 Load 1 3 1.0 2 Drill 3 12 1.0 3 Grind 4 28 1.0 4 Inspect 5 8 0.4 5 Unload 1 2 1.0 B 0.2 1 Load 1 3 1.0 2 Mill 2 12 1.0 3 Drill 3 12 1.0 4 Grind 4 33 1.0 5 Drill 3 11 1.0 6 Inspect 5 9 0.3 7 Unload 1 2 1.0 0.15 1 Load 1 3 1.0 2 Mill 2 14 1.0 3 Drill 3 11 1.0 4 Inspect 5 12 0.2 5 Unload 1 2 1.0 0.1 1 Load 1 3 1.0 2 Drill 3 17 1.0 3 Inspect 5 10 0.2 4 Unload 1 2 1.0 E 0.1 1 Load 1 3 1.0 2 Grind 4 32 1.0 3 Drill 3 13 1.0 4 Grind 4 37 1.0 5 Inspect 5 12 0.1 6 Unload 1 2 1.0 0.35 1 Load 1 3 1.0 2 Mill 2 26 1.0 3 Drill 3 13 1.0 4 Grind 4 39 1.0 5 Inspect 5 8 0.3 6 Unload 1 2 1.0

2) Consider the Part Type Selection problem we covered in class. This problem helps us determine which parts we will produce in a new FMS. The problem assumes that we have candidate parts i = 1, 2, ...N, that the total savings achieved by producing part i in the FMS is s, and that we know that there is a 'key bottleneck' in the FMS, which has P available time units. Each part i requires a total processing time p, in the 'key bottleneck'. we saw that a way to solve this problem is the following Greedy Heuristic: Step 1: Order parts [1] to [N] such that P [1] P[2] P[N] Define remaining available time T = P Step 2: For i = 1...N select part [i] and add the part to the FMS if p, < T and s, > 0. Each time you add a new part i, update the remaining available time as T = T - P₁ Now suppose you had 3 'key bottlenecks' (that is 3 resources that each might be the bottleneck depending on the parts you choose to produce in the FMS). These resources have P, Q and R time units available respectively. And each part i requires a total processing time p, q, and r, in each of the three, respectively. Modify the algorithm above to solve the 3-bottleneck version of the problem. TIP: In Step 1 you need to consider the processing times at all 3 stations to do the ordering. In step 2, you must keep track of the remaining capacity at each station separately. 3) An FMS consists of five stations. Station 1 is a load/unload station with one server. Station 2 performs milling operations with one server. Station 3 performs drilling operations with two servers (two identical CNC drill presses). Station 4 is for performing grinding operations with three identical servers (three identical CNC grinding machines). Station 5 is an inspection station with one server that performs inspections on a sample of the parts. The stations are connected by a part handling system that has two work carriers and a mean transport time = 4.0 min. The FMS produces six parts A, B, C, D, E, and F. The part mix fractions and process routings for the six parts are presented in the table below. Note that the operation frequency at the inspection station (f_5jk) is less than 1.0 to account for the fact that only a fraction of the parts are inspected. For this system determine: (a) Maximum production rate of the FMS; (b) Corresponding production rate of each part; (c) Utilization of each station in the system; (d) Overall FMS utilization. Part Part mix Station Frequency J Pj Op k Description Process time tijk (min) fijk A 0.1 1 Load 1 3 1.0 2 Drill 3 12 1.0 3 Grind 4 28 1.0 4 Inspect 5 8 0.4 5 Unload 1 2 1.0 B 0.2 1 Load 1 3 1.0 2 Mill 2 12 1.0 3 Drill 3 12 1.0 4 Grind 4 33 1.0 5 Drill 3 11 1.0 6 Inspect 5 9 0.3 7 Unload 1 2 1.0 0.15 1 Load 1 3 1.0 2 Mill 2 14 1.0 3 Drill 3 11 1.0 4 Inspect 5 12 0.2 5 Unload 1 2 1.0 0.1 1 Load 1 3 1.0 2 Drill 3 17 1.0 3 Inspect 5 10 0.2 4 Unload 1 2 1.0 E 0.1 1 Load 1 3 1.0 2 Grind 4 32 1.0 3 Drill 3 13 1.0 4 Grind 4 37 1.0 5 Inspect 5 12 0.1 6 Unload 1 2 1.0 0.35 1 Load 1 3 1.0 2 Mill 2 26 1.0 3 Drill 3 13 1.0 4 Grind 4 39 1.0 5 Inspect 5 8 0.3 6 Unload 1 2 1.0

Linear Algebra: A Modern Introduction

4th Edition

ISBN:9781285463247

Author:David Poole

Publisher:David Poole

Chapter2: Systems Of Linear Equations

Section2.4: Applications

Problem 28EQ

See similar textbooks