Bottleneck Model Example

(Bottleneck Model Example )

19.1A flexible manufacturing cell consists of two machining workstations plus a load/unload station. The load/unload station is station 1. Station 2 performs milling operations and consists of one server (one CNC milling machine). Station 3 has one server that performs drilling (one CNC drill press). The three stations are connected by a part handling system that has one work carrier. The mean transport time is 2.5 min. The FMC produces three parts, A, B, and C. The part mix fractions and process routings for the three parts are presented in the table below. The operation frequency fijk = 1.0 for all operations. Determine

(a) maximum production rate of the FMC,

(b) corresponding production rates of each product,

(d) number of busy servers at each station.

Part j / Part mix pj / Operation k / Description / Station i / Process time tijk
A / 0.2 / 1 / Load / 1 / 3 min
2 / Mill / 2 / 20 min
3 / Drill / 3 / 12 min
4 / Unload / 1 / 2 min
B / 0.3 / 1 / Load / 1 / 3 min
2 / Mill / 2 / 15 min
3 / Drill / 3 / 30 min
4 / Unload / 1 / 2 min
C / 0.5 / 1 / Load / 1 / 3 min
2 / Drill / 3 / 14 min
3 / Mill / 2 / 22 min
4 / Unload / 1 / 2 min

Solution:

(a) WL1 = (3+2)(0.2)(1.0) + (3+2)(0.3)(1.0) + (3+2)(0.5)(1.0) = 5.0 min

WL2 = 20(0.2)(1.0) + 15(0.3)(1.0) + 22(0.5)(1.0) = 19.5 min

WL3 = 12(0.2)(1.0) + 30(0.3)(1.0) + 14(0.5)(1.0) = 18.4 min

nt = 3(0.2)(1.0) + 3(0.3)(1.0) + 3(0.5)(1.0) = 3, WL4 = 3(2.5) = 7.5 min

Bottleneck station has largest WLi/si ratio:

Station / WLi/si ratio
1 (load/unload) / 5.0/1 = 5.0 min
2 (mill) / 19.5/1 = 19.5 min /  Bottleneck
3 (drill) / 18.4/1 = 18.4 min
4 (material handling) / 7.5/1 = 7.5 min

Bottleneck is station 2: Rp* = 1/19.5 = 0.05128 pc/min = 3.077 pc/hr

(b) RpA = 0.05128(0.2) = 0.01026 pc/min = 0.6154 pc/hr

RpB = 0.05128(0.3) = 0.01538 pc/min = 0.9231 pc/hr

RpC = 0.05128(0.5) = 0.02564 pc/min = 1.5385 pc/hr

U2 = (19.5/1)(0.05128) = 1.0 = 100%

U3 = (18.4/1)(0.05128) = 0.944 = 94.4%

U4 = (7.5/1)(0.05128) = 0.385 = 38.5%

(d) BS1 = (5.0)(0.05128) = 0.256 servers

BS2 = (19.5)(0.05128) = 1.0 servers

BS3 = (18.4)(0.05128) = 0.944 servers

BS4 = (7.5)(0.05128) = 0.385 servers

(Extended Bottleneck Model Example)

19.2Use the extended bottleneck model to solve problem 19.1 with the following number of parts in the system:

(a) N = 2 parts and

(b) N = 4 parts. Also determine the manufacturing lead time for the two cases of N in (a) and (b).

Solution: See solution to Problem 19.1.

MLT1 = 5.0 + 19.5 + 18.4 + 7.5 = 50.4 min

Rp* = 0.05128 pc/min from solution to Problem 19.1.

N* = 0.05128(50.4) = 2.58

(a) For N = 2 < N* = 2.58, apply case 1 (Table 19.5)

MLT1 = 50.4 min, and Tw = 0

Rp = 2/50.4 = 0.0397 pc/min = 2.38 pc/hr

RpA = 0.0397(0.2) = 0.00794 pc/min = 0.476 pc/hr

RpB = 0.0397(0.3) = 0.01191 pc/min = 0.714 pc/hr

RpC = 0.0397(0.5) = 0.01985 pc/min = 1.191 pc/hr

U1 = (5.0/1)(0.0397) = 0.198 = 19.8%

U2 = (19.5/1)(0.0397) = 0.774 = 77.4%

U3 = (18.4/1)(0.0397) = 0.730 = 73.0%

U4 = (7.5/1)(0.0397) = 0.298 = 29.8%

BS1 = (5.0)( 0.0397) = 0.198 servers

BS2 = (19.5)( 0.0397) = 0.774 servers

BS3 = (18.4)( 0.0397) = 0.730 servers

BS4 = (7.5)( 0.0397) = 0.298 servers

(b) N = 4 > N* = 2.58, apply case 2 (Table 19.5)

Rp* = 0.05128 pc/min = 3.077 pc/hr

MLT2 = 4/0.05128 = 78.0 min, and Tw = 78.0 - 50.4 = 27.6 min

RpA = 0.05128(0.2) = 0.01026 pc/min = 0.6154 pc/hr*

RpB = 0.05128(0.3) = 0.01538 pc/min = 0.9231 pc/hr*

RpC = 0.05128(0.5) = 0.02564 pc/min = 1.5385 pc/hr*

U1 = (5.0/1)(0.05128) = 0.256 = 25.6%*

U2 = (19.5/1)(0.05128) = 1.0 = 100%*

U3 = (18.4/1)(0.05128) = 0.944 = 94.4%*

U4 = (7.5/1)(0.05128) = 0.385 = 38.5%*

BS1 = (5.0)(0.05128) = 0.256 servers*

BS2 = (19.5)(0.05128) = 1.0 servers*

BS3 = (18.4)(0.05128) = 0.944 servers*

BS4 = (7.5)(0.05128) = 0.385 servers*

* These answers are the same as in Problem 19.1