The mean time between fallures and mean time to repair in a certain department of the factory are $400$ hours and $8$ hours, respectively. The department operates $25$ machines during one $8-$hour shift per day, $5$ days per week, $50$ weeks per year. Each time a machine breaks down, it costs the company $200$TL per hour $($per machine$)$ in Lost Revenue.

A proposal has been submitted to establish a preventive maintenance program in this department. In this program, preventive maintenance would be performed on the machines during the evening so that there will be no interruptions to production during the regular shift. The effect of this program is expected to be that the average MTBF will double, and half of the emergency repair time normally accomplished during the day shift will be performed during the evening shift.

The cost of the maintenance crew will be $1500$TL per week. However, a reduction of maintenance personnel on the day shift will result in a savings during the regular shift of $700$TL per week. Assume that machines will never have to wait for the maintenance crew $($ignore the effects of queueing, l$.$e$.$ assume that when a machine breaks down there is always a crew member available to respond and the repair starts immediately$).$

$n=25$ machines

$AT=2000h\frac{r}{y}$ear

$ns=5$ shift$\frac{s}{w}$eek

$CLR=200T\frac{L}{h}\frac{r}{m}$achine

$nh=8h\frac{r}{s}$hift

$CMC=1500T\frac{L}{w}$eek

$CSR=700T\frac{L}{w}$eek

Lost Revenue

$nw=50$ weeks$/$year

Maintenance Crew

Savings in Reg. Shift

The availability of the machines in the department before and after the preventive maintenance program is established:

$\backslash$table$[[,\backslash$table$[[$Before the$],[$Program$]],\backslash$table$[[$After the$],[$Program$]]],[$MTBF$,400,800],[$MTTR$,8,4],[$A$,0,980,0\mathrm{.}995]]$

Total number of times all $25$ machines in the department break down $($need repair$)$ per year both before and after the program is as follows.

$\backslash$table$[[,\backslash$table$[[$Before the$],[$Program$]],\backslash$table$[[$After the$],[$Program$]],],[\backslash$table$[[$Number of Repairs.$],[$per machine$]],5\mathrm{.}0,2\mathrm{.}5,$repair$],[\backslash$table$[[$Number of Repairs$],[$all machines$]],125\mathrm{.}0,62\mathrm{.}5,$repairs$/$year$],[\backslash$table$[[$Time Down$],[$per machine$]],40\mathrm{.}0,10\mathrm{.}0,$hr$/$year$],[\backslash$table$[[$Time Down$],[$all machines$]],1000\mathrm{.}0,250\mathrm{.}0,$hr$/$year$]]$

Note that a break down or a failure have the same meaning in the context of machines and Availability. Each failure requires a repair. Time Down means time spent under repair i$.$e$.$ unavailable to work and produce parts.

Let's calculate if the preventive maintenance program will pay for itself in terms of savings in the cost of lost revenues:

$\backslash$table$[[,\backslash$table$[[$Before the$],[$Program$]],\backslash$table$[[$After the$],[$Program$]],,],[\backslash$table$[[$Cost of Time Down$],[$per machine$]],8,000,2,000,$TL$/$year$,\backslash$table$[[$Note that a break down or a failure$],[$have the same meaning in the$]]],[\backslash$table$[[$Cost of Time Down$],[$all machines$]],200,000,50,000,$TL$/$year$,\backslash$table$[[$Avallability$.$ Each fallure requires a$],[$repair$.$ Time Down means time$]]],[\backslash$table$[[$Cost of$],[$Maintenance Craw$]],0,75,000,$TL$/$year$,$work and produce parts.$],[\backslash$table$[[$Savings in$],[$Regular Shift$]],-0,-35,000,$TL$/$year$,],[$Total Cost,$200,000,90,000,$TL$/$year$,]]$

The preventive maintenance program will pay for itself:

The program will save the company $200,000*90,000**110,000$ TL$/$year$,$

SHOW ME HOW WE FOUND EACH answer MATHmeaticlly $?$