Answer :

$($c$)$

Precedence Network Diagram with forword and backword pass computation:$-$

Es Acc Ef

Ls Dur Lf

$0$ a $0\mathrm{.}2$

$00\mathrm{.}20\mathrm{.}2$

X

$0\mathrm{.}2$ b $0\mathrm{.}6$

$0\mathrm{.}40\mathrm{.}40\mathrm{.}8$

$0\mathrm{.}2$ e $1\mathrm{.}2$

$01\mathrm{.}01\mathrm{.}2$

$0\mathrm{.}6$ c $1\mathrm{.}1$

$0\mathrm{.}80\mathrm{.}51\mathrm{.}3$

$1\mathrm{.}2$ f $1\mathrm{.}7$

$1\mathrm{.}20\mathrm{.}51\mathrm{.}7$

Critical Path : a$--$e$--$f$--$g$--$i$--$k$--$l$--$m$--$n

Porject duration$($Cycle Time$)=6\mathrm{.}7$ Minutse

$($a$)$

Total Available Time per shift $=8$ Hrs $-(10*2+40)$ Min $=8$ Hrs $-60$ Min $=7$ Hrs $=7*60$ Min $=420$ Min

The number of widgets that can be produced per shift with one workstation$=$ Total Available Time $/$ Cycle time $=420/6\mathrm{.}7=62\mathrm{.}68=63$ Widgets

Hence, it is not possible to accomplish the objective of $300$ widgets per shift with one workstation.

$($b$)$

Maximum possible output with $14$ workstation $=63*14=883$ Widgets

$($d$)$

Theoretical minimum number of workstations to achieve $300$ pieces per shift $=300/63=4\mathrm{.}76=5$