Question

set-up time reduction
demand: 4,800 pints/week for each of 5 flavors
production rate: 6 pints/minute = 28,800 pints/week
setup time: 0.2 weeks, regardless of flavor

setup time | min cycle length | batch size (pints) | avg. inv. per flavor (pints) | total avg. inv. (pints)
16 hrs | 6 wks | 28,800 | 12,000 | 60,000
8 hrs | 3 wks | | |
4 hrs | | | |
2 hrs | | | |
1 hr | | | |
30 min | | | |
15 min | | | |

note: recall that the line operates 80 hours per week!
ops223 | operations and strategy
setups | 20
numeric
use: 0123456789 - , ^ e
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Explanation:

Step1: Convert setup time to weeks

First, convert setup time from hours to weeks using 80 hrs/week.
For 8 hrs: $\frac{8}{80} = 0.1$ weeks
For 4 hrs: $\frac{4}{80} = 0.05$ weeks
For 2 hrs: $\frac{2}{80} = 0.025$ weeks
For 1 hr: $\frac{1}{80} = 0.0125$ weeks
For 30 min: $\frac{0.5}{80} = 0.00625$ weeks
For 15 min: $\frac{0.25}{80} = 0.003125$ weeks

Step2: Calculate min cycle length

Cycle length = $\text{Setup Time (weeks)} + \frac{\text{Demand per flavor per week} \times \text{Cycle Length}}{\text{Production Rate per week}}$
Rearrange to solve for cycle length ($CL$):
$CL = \frac{\text{Setup Time (weeks)}}{1 - \frac{\text{Demand per flavor}}{\text{Production Rate}}}$
Given demand per flavor = 4800 pints/week, production rate = 28800 pints/week, so $\frac{4800}{28800} = \frac{1}{6}$
Thus $CL = \frac{\text{Setup Time}}{1 - \frac{1}{6}} = \frac{\text{Setup Time}}{\frac{5}{6}} = 1.2 \times \text{Setup Time}$
- 8 hrs: $1.2 \times 0.1 = 0.12$ weeks? *Correction: Use standard batch cycle formula $CL = \frac{S}{1 - \frac{D}{P}}$, where $S$=setup time in weeks, $D$=demand per flavor, $P$=production rate.
Wait, alternative: Batch size ($Q$) is determined by $Q = D \times CL$, and production time for batch is $\frac{Q}{P}$. So $CL = S + \frac{Q}{P} = S + \frac{D \times CL}{P}$, solving for $CL$:
$CL - \frac{D}{P}CL = S$
$CL(1 - \frac{D}{P}) = S$
$CL = \frac{S}{1 - \frac{D}{P}}$
Plug in values:
- 8 hrs ($S=0.1$): $CL = \frac{0.1}{1 - \frac{4800}{28800}} = \frac{0.1}{\frac{5}{6}} = 0.12$ weeks? No, the first row has 16 hrs setup = 0.2 weeks, $CL=6$ weeks, which matches $\frac{0.2}{1 - 4800/28800} = \frac{0.2}{5/6}=0.24$? No, correction: First row batch size=28800, which is full production rate. So batch size $Q$ is the amount produced in a cycle, so $Q = P \times (CL - S)$. And $Q = D \times CL$, so $P(CL - S) = D CL$
$P CL - P S = D CL$
$CL(P - D) = P S$
$CL = \frac{P S}{P - D}$
This matches first row: $P=28800$, $S=0.2$, $D=4800$:
$CL = \frac{28800 \times 0.2}{28800 - 4800} = \frac{5760}{24000} = 0.24$ weeks? No, first row says 6 weeks. Oh, note: setup time is per flavor, 5 flavors. So total setup time per cycle is $5 \times S$.
First row: 16 hrs per flavor, 5 flavors = 80 hrs = 1 week setup time. Then $CL = \frac{5S}{1 - \frac{5D}{P}}$, since total demand is $5 \times 4800=24000$.
$1 - \frac{24000}{28800} = 1 - \frac{5}{6} = \frac{1}{6}$
$CL = \frac{1}{\frac{1}{6}} = 6$ weeks, which matches. Correct formula: Total setup time $S_{total}=5 \times S_{per flavor}$, total demand $D_{total}=5 \times 4800=24000$
$CL = \frac{S_{total}}{1 - \frac{D_{total}}{P}} = \frac{5S_{per flavor}}{1 - \frac{24000}{28800}} = 6 \times 5S_{per flavor} = 30 S_{per flavor}$ (since $1/(1-5/6)=6$)
$S_{per flavor}$ in weeks:
- 16 hrs: $\frac{16}{80}=0.2$ weeks, $CL=30 \times 0.2=6$ weeks (matches)
- 8 hrs: $\frac{8}{80}=0.1$ weeks, $CL=30 \times 0.1=3$ weeks (matches red note)
- 4 hrs: $\frac{4}{80}=0.05$ weeks, $CL=30 \times 0.05=1.5$ weeks
- 2 hrs: $\frac{2}{80}=0.025$ weeks, $CL=30 \times 0.025=0.75$ weeks
- 1 hr: $\frac{1}{80}=0.0125$ weeks, $CL=30 \times 0.0125=0.375$ weeks
- 30 min: $\frac{0.5}{80}=0.00625$ weeks, $CL=30 \times 0.00625=0.1875$ weeks
- 15 min: $\frac{0.25}{80}=0.003125$ weeks, $CL=30 \times 0.003125=0.09375$ weeks

Step3: Calculate batch size per flavor

Batch size ($Q$) = $D_{per flavor} \times CL$
- 8 hrs: $4800 \times 3 = 14400$ pints
- 4 hrs: $4800 \times 1.5 = 7200$ pints
- 2 hrs: $4800 \times 0.75 = 3600$ pints
- 1 hr: $4800 \times 0.375 = 1800$ pints
- 30 min: $4800 \times 0.1875 = 900$ pints
- 15 min: $4800 \times 0.09375 = 450$ pints

Step4: Calculate avg inv per flavor

Average inventory per flavor = $\frac{Q}{2} \times (1 - \frac{D_{per flavor}}{P})$
- 8 hrs: $\frac{14400}{2} \times (1 - \frac{4800}{28800}) = 7200 \times \frac{5}{6} = 6000$ pints
- 4 hrs: $\frac{7200}{2} \times \frac{5}{6} = 3600 \times \frac{5}{6} = 3000$ pints
- 2 hrs: $\frac{3600}{2} \times \frac{5}{6} = 1800 \times \frac{5}{6} = 1500$ pints
- 1 hr: $\frac{1800}{2} \times \frac{5}{6} = 900 \times \frac{5}{6} = 750$ pints
- 30 min: $\frac{900}{2} \times \frac{5}{6} = 450 \times \frac{5}{6} = 375$ pints
- 15 min: $\frac{450}{2} \times \frac{5}{6} = 225 \times \frac{5}{6} = 187.5$ pints

Step5: Calculate total avg inventory

Total avg inv = $\text{Avg inv per flavor} \times 5$
- 8 hrs: $6000 \times 5 = 30000$ pints
- 4 hrs: $3000 \times 5 = 15000$ pints
- 2 hrs: $1500 \times 5 = 7500$ pints
- 1 hr: $750 \times 5 = 3750$ pints
- 30 min: $375 \times 5 = 1875$ pints
- 15 min: $187.5 \times 5 = 937.5$ pints

Answer:

Setup time	Min Cycle Length (wks)	Batch Size (pints)	Avg. Inv. per Flavor (pints)	Total Avg. Inv. (pints)
8 hrs	3	14,400	6,000	30,000
4 hrs	1.5	7,200	3,000	15,000
2 hrs	0.75	3,600	1,500	7,500
1 hr	0.375	1,800	750	3,750
30 min	0.1875	900	375	1,875
15 min	0.09375	450	187.5	937.5