Answer:
[tex]N_c \approx 5.0[/tex]
Step-by-step explanation:
From the question we are told that:
Waiting time for a container of the wheels during production [tex]T_w=0.25day[/tex]
Average processing time is [tex]T_a=0.15 day per container[/tex]
Wheel per container [tex]N_w=200[/tex]
Total wheel per day [tex]D=2000[/tex]
Policy variable of [tex]X=5%[/tex]
Generally the equation for Total number of container [tex]N_c[/tex] is mathematically given by
[tex]N_c=DT(\frac{1+X}{N_w} )[/tex]
Where
Total time [tex]T=T_w+T_a[/tex]
[tex]T=0.15+0.25[/tex]
Therefore
[tex]N_c=DT(\frac{1+X}{N_w} )[/tex]
[tex]N_c=2000*0.40(\frac{1+0.5}{200} )[/tex]
[tex]N_c=(\frac{840}{200} )[/tex]
[tex]N_c=4.2[/tex]
[tex]N_c \approx 5.0[/tex]
Therefore the number of Kanban containers needed for the wheels is
[tex]N_c \approx 5.0[/tex]
