Question 10.69: Analyzing congestion at a shrinkwrap machine Unit loads arri...

Analyzing congestion at a shrinkwrap machine

Unit loads arrive randomly at a shrinkwrap machine. An average of 15 unit loads arrive per hour at a Poisson rate. The time required for a unit load to be processed through the shrinkwrapping operation is a constant of 2.5 minutes.

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The operating characteristics are obtained using Equations 10.17210.172 through 10.17510.175 by letting λ=15,μ=24,σ2=0\lambda=15, \mu=24, \sigma^{2}=0, and ρ=0.625\rho=0.625.

L=ρ+λ2σ2+ρ22(1ρ)L=\rho+\frac{\lambda^{2} \sigma^{2}+\rho^{2}}{2(1-\rho)}     (10.172)

Lq=λ2σ2+ρ22(1ρ)L_{q}=\frac{\lambda^{2} \sigma^{2}+\rho^{2}}{2(1-\rho)}     (10.173)

W=1μ+λ(σ2+1μ2)2(1ρ)W=\frac{1}{\mu}+\frac{\lambda\left(\sigma^{2}+\frac{1}{\mu^{2}}\right)}{2(1-\rho)}     (10.174)

Wq=λ(σ2+1μ2)2(1ρ)W_{q}=\frac{\lambda\left(\sigma^{2}+\frac{1}{\mu^{2}}\right)}{2(1-\rho)}     (10.175)

L=0.625+(15)2(0)+(0.625)22(10.625)=1.1458L=0.625+\frac{(15)^{2}(0)+(0.625)^{2}}{2(1-0.625)}=1.1458 unit load

Lq=(15)2(0)+(0.625)22(10.625)=0.5208L_{q}=\frac{(15)^{2}(0)+(0.625)^{2}}{2(1-0.625)}=0.5208 unit load

W=Lλ=0.07639W=\frac{L}{\lambda}=0.07639 hour per unit load

=4.583=4.583 minutes per unit load

Wq=W1μ=2.083W_{q}=W-\frac{1}{\mu}=2.083 minutes per unit load

10-69a
10-69b

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