Question

A manufacturing firm wants to plan its production and inventory for the months of May, June, July, and
August. The demand to be met at the end of each month is 500, 600, 800, and 1200 units, respectively. The
starting inventory on May 1 is 250 units, and the final inventory on September 1 is to be at least 100 units.
Currently, the monthly production capacity is 600 units at a cost of $30 per unit. However, a new production
line will become operational on August 1 that will increase production capacity to 1200 units per month and unit costs to $35 per unit. Units produced and shipped in the same month do not incur storage costs, but any
units in inventory on the first day of a month must stored at a cost of $5 per unit in a storage facility that can
hold at most 400 units.

Answer 1

need the eqution

Answer 2

supply:
starting inventory = 250 units
base supply = 600units / month ($ 30 each)

so May 31: need 500 units (min)

Answer 3

i think we should work backwards here tho...
31st august: need 100 left over for start of september, and 1200 for sales, so 1300 all up

Answer 4

next part is to use simplex method to solve equation

Answer 5

Use the simplex method for bounded variables to make a production schedule that
minimizes the total production and inventory costs (including carrying costs to be paid on September 1) of
meeting the above requirements

Answer 6

dude i'm out, u know way more than me at this and i've never heard of the simplex method, sorry

Answer 7

it is LP question
we will need to form the equation

Answer 8

you were solving correct only thing is we need the equation

Answer 9

may costs = x*30 ( where x < 601 ) + 100 * 5
june costs = x*30 ( where x < 601 ) + y * 5 (where y < 401 )
july costs = x*30 ( where x < 601 ) + y * 5 (where y < 401 )
august costs = x*35 ( where x < 1201 ) + y * 5 (where y < 401 )

y = number in warehouse on first of month
x = number produced in month
...something like that?

Answer 10

should we take x<601 or x=601?

Answer 11

x must be <601 ie 600 or less as we cant make more than 600 in a month

Answer 12

then choose your values of x to meet demand for each month, working backwards from september 1

Answer 13

demand has to be 500,600 ...

Answer 14

yep
may costs = x*30 ( where x < 601 ) + 100 * 5
june costs = x*30 ( where x < 601 ) + y * 5 (where y < 401 )
july costs = x*30 ( where x < 601 ) + y * 5 (where y < 401 )
august costs = x*35 ( where x < 1201 ) + y * 5 (where y < 401 )
september costs = 100*5

August demand = 1300 so y for August = 100 min
July demand = 800, so y for July = 200 + 100 left over for start of august min
june demand = 600, so y for june = 0 + 300 for start of july min
may demand = 500, so y for may =100 already set + 100 for next month...?
i think...?

Answer 15

yes right

Answer 16

Thanks dude let me now try and form equations

Answer 17

cool, good luck and i hope its right ;D

Answer 18

Thanks")

Answer 19

let me solve and check

Answer 20

Can any one provide me the equation

Answer 21

the production system that starts in August, is essentially =0 before august
so we need 2 equations, or 2 cases for values

Answer 22

can you tell me the equations

Answer 23

I have difficulty to make them

if we have too many units in storage, it is unnecessary cost
however, if we did not save enough, we will not be able to satisfy demand

so, the requirement for "saving"
it doesn't depend on the current month. it also depends on the next ones

Answer 24

we have a function of "spare resources", this is -200 in the last month, and 0 before that, and 100 in the first month

spare(month) = 100 -200*m

Answer 25

savings over all months also has to be at least 200, because we need to deliver 800 at the end of july and can only produce 600

savings >= 200 "at the end of july" <- how to do this???

Answer 26

similarly, some requirements in later months will lead to production in the current month. we should postpone production to as late as possible, because of limited/expensive storage. however, it is certain we WILL have to produce in advance to satisfy the 800 units in last month before new machine is added.

Answer 27

you should make a constraint that tells there must be at least XXX remaining spare, which is the difference in production -200, over the entire time up to that month.
then, we can optimize for cost to postpone as late as possible