Table 3 Results of applying optimal departure procedure on two scenarios.

Scenario 1

o Delivery times:

18, 24, 27, 41, 47

o Late items L = {8}

o Early items E = {1,5}

o Timely items G = {4,7}

\(t=e=1\)

\(\rho =\left|E\left|-\frac{t}{e}\right|L\right|=1\)

o Step 5 is initiated

Because \(\rho >0\)

\(A=\left\{\text{1,3}\right\}\)

\(B=\left\{\text{1,2}\right\}\)

\(\Delta {e}_{1}=\text{min}A=1\)

\(\Delta {e}_{2}=\text{max}A=3\)

\(\Delta {s}_{1}=\text{min}B=1\)

\(\Delta {x}_{1}=\text{min}\left(\Delta {e}_{1},\Delta {s}_{1}\right)=1\)

o H = {1}

\(\Delta {d}_{\nu }^{\mu in}=\text{min}H=1\)

o Updated \({d}_{v}={d}_{v}+\Delta {d}_{\nu }^{\mu in}\) = 11

o New delivery times:

19, 25, 28, 42, 48

o Late items L = {8}

o Early items E = {5}

o Timely items G = {1,4,7}

\(\rho =0\)

o Procedure terminates with

\({d}_{v}^{*}=11\)

Scenario 2

o Delivery times:

18, 24, 34, 43

o Late items L = {3,8}

o Early items E = {1}

o Timely items G = {4,7}

\(t=e=1\)

\(\rho =\left|E\left|-\frac{t}{e}\right|L\right|=-1\)

o Since \(\rho <0\),\({d}_{v}^{*}=10\)