Chapter 6 Solution Ops Management Essay

Section 6 Solution Ops Management Essay Section 6 Solution Ops Management Essay Section 6 Procedure DESIGN AND Facility LAYOUT Answers to Problems 1. Longest errand = 2.4 minutes Absolute undertaking times = 18 minutes OT = 450 minutes out of each day a. Least process duration = length of longest assignment, which is 2.4 minutes. a. Most extreme process duration = ï  task times = 18 minutes.CT = 450/180 = 2.50 minutes for each unit N = 18/2.5 = 7.2, round to 8 b.b. CT = 450/125 = 3.6 minutes for every unit Scope of yield: e. c I. yield = 450/9 = 50 units for each day ii. yield = 450/15 = 30 units for each day 2. Wanted yield = 33.33 units every hour Working time = an hour out of each hour CT = Working time = an hour for each hr. = 1.80 82 minutes for each unit Wanted yield 33.33 units per hr. a. Station Time left Qualified Will fit Relegate (time) Inactive 1 1.82 0.42 a b an a (1.4) 0.42 2 1.82 1.32 0.52 b c, d, e c, d b c, d, e b (0.5) e* (0.8) 0.52 3 1.82 1.12 0.52 .02 c, d c, g, f g c, d c, g f d* (0.7) c** (0.6) f (0.5) .02 4 1.82 0.82 0.32 g h g h g (1.0) h (0.5) 0.32 1.28 * is tied in no. of devotees, yet is longer(longest) ** has more devotees b. Efficiency = 1 †[1.28/4(1.82)] = .82 or 82%. 3. Wanted yield = 4 units for every hour Working time = 56 minutes of great importance CT = Working time = 56 minutes for every hr. = 14 minutes for every unit Wanted yield 4 units for every hr. a. Station Time left Qualified Will fit Dole out (time) Inert 1 14 9 6 a, d, f a, d, g b, d, g a, d, f a, d, g b, g f* (5) a** (3) g* (6) 0 2 14 7 5 1 b, d b, e c, e c b, d b, e c, e d* (7) b** (2) e*** (4) 1 3 14 10 1 c h I c h c (4) h (9) 1 4 14 9 I I (5) 9 11 * is tied for no. of adherents, yet is longer (longest) ** has progressively (most) no. of adherents *** tied in no. of adherents and time; pick haphazardly b. Effectiveness = 1 †Complete inert time = 1 †11 = 80.4% CT x no. of stations 14(4) 1. CT = 1.3 minutes per unit Time [no. followers] .3 [3] a. I .2 [4] .4 [3] 1.3 [2] 1.2 [0] .1[3] .8[2] .3[1] a. ii Station Time left Eligible Will fit Allot (time) Inert time 1 1.3 a, c, e a, c, e a* (.2) 1.1 b, c, e b, c, e b** (.4) .7 c, e c, e c** (.3) .4 d, e (.1) .3 d, f 0.3 2 1.3 d, f d, f d** (1.3) 0.0 f 0.0 3 1.3 f (.8) .5 g (.3) .2 h 0.2 4 1.3 h (1.2) 0.1 0.1 0.6 * most supporters ** tied in no. of supporters, however more (longest) a. iii Rate inert time: ï (idle time) = .6 = 11.5% N Ãâ€"CT 4(1.3) a. iv Yield: OT = 420 min./day = 323.1 units/day CT 1.3 min./unit b. I. Absolute time = 4.6 min., CT = Absolute time = 4.6 = 2.3 minutes. N 2 Allot a, b, c, d, and e to workstation 1: 2.3 minutes Allot f, g, and h to workstation 2: 2.3 minutes ii. Rate inert time = 0 iii. Yield = OT = 420 = 182.6 units every day. CT 2.3 5. Output rate = 240 units for every eight-hour day a. b. CT = OT = 480 min/day = 2 minutes for each unit yield 240 units/day c. N = ï t = 4.6 = 2.3 (gather together to 3) workstations CT 2.0 d. Station Time Left Qualified Will Fit Dole out (time) Inert time 1 2 a, c, e a, c, e a* (.2) 1.8 b, c, e b, c, e e** (1.2) .6 b, c, f b, c b** (.4) .2 c, f c (.2) 0 .0 2 2 d, f d, f f** (1.2)