PRODUCTION PROBLEMS

conversion
Yield

In I Out
similar
not to
blending

Goal:maximise revenue costs

Inputs Intermediates Outputs

Raw material valuesareknown focus process in product (j)

20Z
-4016 2/b ↳ I A unlimited
- >
102 demand
16/07
$
30Z
-2 20Z
zW > B > Disposal
$2107
Labour hours 34
-
20 OZ
= 60M $ 14/0Z

Decision variables
Let:

the
pi: number of sessions executed process i,
in where i =
E1,23
xj the
=
amount(inoz) product;produced,
of where j E1,25
=
and j= 1
represents
&US
chemical A and j =
2 represents chemical B

the amount(inoz) product;sold,

of where j 2,25 and j=1 represents
yj=
=

chemical A and j =
2 represents chemical B

objective function
mak z (16y1 14yz) -(xz
=
+
-

yz)z (max revenue less disposal

S.T 241 3p2

+

160 Clabour hours limit)

p1 2pz
+

40 (raw material limit)

y z
= z0 (Chemical B demand limit)
Converting material
to runs
yi =xi ↓j eE1,23 < Cannotsell more than produced)
production)
40
I
=
20 runs
P2 x1 < Chemical A
2p1+ 3P2
=

determine
much we
now
12
(Chemical B production) f'Yonseievation
1p1 2p2
= +

actually produce pi

xj yjz0tj =E1,23
+
0
= fieE1,23
3 (non
negativity)
 -SERA3. 9 ARREN 7

- -

Ba
requires milk to make
crear

cream is notan input

ooyieldcheese
-

sayor
min

or
1000 lb

-High fat milk

60% Fa t >(C1
y
1 $0.811b cream
max so the

·
$1.5011b

I
Unlimited zo,
supply
I ca

~
X
aug fat 135%

h21
min
fat, how fat milk 10001b,2
cottage
30%
$0.4/1b cheese
yz max 2000lb
90%
yield $1.20 11b
$4/lb
(input)
$0.4/1b
Cap I20001b (output)
(input)
cap 13000 lb

(inputs)
 -Decision variable
and Parameters

parameters
pil the
given purchasing cost (in $1b) of
milk
type 1, where i 51,23
=

and i=1 represents high fat milk and i = 2

represents low fat milk

sj the given selling price (in $11b) ofproduct j, where j= 1

represents cream cheese and j 2 represents
=

cheese
cottage
max 21,23(in Ib)
dj given maximum
the of product j, where j
demand =

minethe
dj given minimum demand (in 1b) of product j, wherej:91,23
&the
j
②
given required cream content (in 1.) for product, wherej= 51,23
fj -the given required milk fat content(in) for products, wherej E1,23 =

Variables

= the
xj amount (in (b) of product;produced, wherej E1,23
=

=the (in 1b) ofmilk where i 21,23

yi
amount
type I purchased, =

ei-the amount(inb) ofmilk type; through

sent the evaporator where i E1,25
=

=
hij the amount (in 1b) ofmilk; used in products, where i,j =51,25
↳

cj -the amountof
cream (in 1b) used in product j, where j= E1,25

FACE
-BERNE
purchasing cheese machine
selling evaporator

- yi.pi-o.ei-(u)
2

Ma) Z =

2xj.sj (ma profit)

j 1
=

cost based
based on input
on output
S.T.

ei =
2000 Levaporator capacity)
i 1
=

22

cj. [[ hij13000 (cheese machine

capacity)
j1 =
j 1i 1
= =

j I djmin Fje [1,23 (minimum demand)

xj = djmak Fje [1,23 (maximum demand)

ci -rj Tje [1,23 (min cream requirement)

2

ci +
hij

0.6e1 0.3e2
+

E,j
= I cream
flow conservation)

hijt ei =

Z
yifieE1,25 (flow conservation milk)

3
I flow
x=a
hil + conservation products

xz 0.9(x2 [nij)
=
+

i 1
=

0.6 hij +0.3 hzj -> fij-21,25 (enforce minfat, content)

E hij
 3
yi, ei 20 Fie [1,23
xj, cj20 7j eE1,25 (non-negativity)
hij 20 vi,j eE1,25
Diagram

D
 Decision variables
Let
xi = raw material used
the amountof in inputproduction process i, where i =
E1,23
E 1. the
ysthenumberoftimesmainroductionsis executed,
wher j =
...

D the
=

liquid
amountof waste dumped in the river (derived variable)

=the
pi given processing cost ($ /run) ofraw material production process in
where i
= E1,23
pjM = the
given processing cost ($/run) of
main production process j, wherej= E1, ...,
43

ti the
=

given production hour

usage ofraw material production process i, where i 2, ...,23
=

tiM the
=

given production hour

usage of
main production process j, where j =
5,...,43

max z
(+
=

7y4) 11y3
+
-
ici
un
i 1
=
- pii-ys (max profit)
-
in comme
RM cost production cost

= income ($102) of
or
sj the
given selling product(notwastel produced
by processj, where j =El, ...,
45

Mak
=,sjy;
S.T. Exi.ti+ *yj. 3000 (production hour limit)
i =
1
=

j 1

Producing
e

3
W
141 0.842
-
+

using
-
W -

0.843-1.244
-
D ( Calculate and limit wastel
D = 1000

= 5000 jeEl, ...,43 (limited demand)

yj
7
o

ICC 1 =>
zy1
+
1y1
+

2y3 (input flow conservation)

7

3(2 =
(y1 +

2yz zy4
+
(input flow conservation)
a

W,D?
Y'Y'8 es"43 0
3 (non
negativity)

& For(process (i) I

: #LE#2:
 Winston (2003), Chapter 3, Review Problem 50
City 1 produces 500 tons of waste per day, and City 2 produces 400 tons of waste per day.
Waste must be incinerated at incinerator 1 or 2, and each incinerator can process up to 500
tons of waste per day. The cost to incinerate waste is $40/ton at incinerator 1 and $30/ton at
2. Incineration reduces each ton of waste to 0.2 tons of debris, which must be dumped at one
of two landfills. each landfill can receive at most 200 tons of debris per day. It costs $3 per mile
to transport a ton of material (either debris or waste). Distances (in miles) between locations
are shown in Table 1.

Table 1: Distances for the waste incineration problem.

Incinerator
City 1 2
1 30 5
2 36 42
Landfill
Incinerator 1 2
1 5 8
2 9 6

Formulate an LP that can be used to minimise the total cost of disposing of the waste of
both cities.
meportation costII on mile
~
City (il
incenerator (j)
Landfill (K)
1
- -1 - 1

supply

>2
-2 -
2

0.2
of input landfill
Yield
capacity
inceneration
capacity
cost
per ton
of input

References
Winston, W. (2003). Operations Research: Applications and Algorithms. Brooks/Cole —
Thomson Learning, Inc., Pacific Grove, CA, 4th edition.

1
 Decision variables and paramete
Let

the total (intons)

cij= amountof waste moved from
city
i
to incinerator, where
ij[1,25
=

the total (intons)

yik = amountof waste moved from incinerator;
to
landfill K, wherejik 51,25
=

si: the given supply ofcity, where i E1,25 =

cj the
=

given incineration cost($Iton) atincinerators,

where j 51,25
=

dij=the given distance (in miles) between city;and incinerator

j, where ij 21,25
=

mijk (in miles) between incinerator;and

the
=

given distance
landfill K, where j, k E1,25
=

OBJECTIVE
FN
-

>is.wij+ dij+3
2

Min Z = cij. yjk. mik

j 1
=
i 1j
=
1
=

j 1k 1
= =

e - ne
incineration transportation transportation
city -> incinerator incinerator -> landfil

S.T
Exij= si vie E1,25 I ensure

incinerated)
all waste is

Ekij ↳500 7jeE1,2 Cincinerator capacity)

2

EYsK = 200 Ke 31,23 (landfill capacity)

for all j

0.2Exis=yik
ij K

I
I
2
I I
2
I
sum over

k
vicE1,23 (flow conservation

Sum over

sij?0
yjk =
0
rij E1,25
j, k [1,23 3 I non negativity)
Cold drink production problem
Sunshine Soft Drinks is a company that produces three flavours of cold drinks (Orange, Grape
and Pineapple) and sells these cold drinks directly to the public at price of sj (in R/litre) where j
represents the cold drink flavour such that j 2 J = {1, 2, 3}, and j = 1 represents Orange, j = 2
represents Grape, and j = 3 represents Pineapple. The company’s production process is divided
into two phases. In the first phase water, liquid sweetener, and flavouring are sent through a syrup
production process to produce syrup; and in the second phase the produced syrup and more water
are sent through a main production process to produce the final product.
In the syrup production process, three di↵erent syrup flavours can be produced (Orange, Grape
and Pineapple). It costs the company R 5 to produce a litre of syrup (in any flavour); and 400
ml water, 400 ml liquid sweetener, and 200 ml liquid flavouring must be sent through the syrup
production process to produce 1 litre of a specific syrup flavour (i.e. if 1 litre of Orange syrup is
produced, 400 ml water, 400 ml sweetener and 200 ml Orange flavouring are used; if 1 litre of Grape
syrup is produced, 400 ml water, 400 ml sweetener and 200 ml Grape flavouring are used; and if 1
litre of Pineapple syrup is produced, 400 ml water, 400 ml sweetener and 200 ml Pineapple flavouring
are used).
In the main production process final cold drinks are produced by sending the produced (flavoured)
syrup and more water through the production process at a cost of R 6 per litre of input. Only one
flavour syrup can be used per cold drink flavour produced (i.e. only Orange syrup used for Orange
cold drink ; only Grape syrup used for Grape cold drink ; and only Pineapple syrup used for Pineapple
cold drink ). Each litre of input (syrup and water) into the main production process yields 950 ml of
cold drink, and no more than 15 000 litres of cold drink can be produced per day.
Company policy dictates that inputs sent through the main production process should contain
between 15% and 20% syrup for Orange and Pinapple cold drink, and between 12% and 18% for
Grape cold drink.
The maximum amount of inputs available per day (in litres) are given by ai and the purchasing
cost (in R/litre) of inputs are given by ci , where i represents input type such that i 2 I = {1, . . . , 5},
and i = 1 represents Orange flavouring, i = 2 Grape flavouring, i = 3 Pineapple flavouring, i = 4
Liquid sweetener, and i = 5 Water.
Formulate a Linear Program (LP) that can help Sunshine Soft Drinks to plan daily production
whilst maximising daily profit.

© University of Pretoria 2023 1

 DIAGRAM

Inputs Outputs

E 3
Orange 1 I
Orange sj
selling price
>

Flavouring
2
al Grape >
Grape capacity (15000
flavouring
Ci
3
pineapple
flavouring >3
Pineapple
4
Liquid
sweetner

water
-
S

e
production cost
production cost 95% yield
Rs Il produced Role input

Asian rallies
Let
xi the
= (in litres) ofinput
amount it I
=
yj the amount(in litres) of cooldrink flavour je 5
produced and sold
= the (in litres) cooldrink e5 produced
pj amount of for
syrup flavourj
=
uj the amount (inlitres) of
syrup for
cooldrink flavour;e5
used to produce
final products.
= (in litres) of
water used in main production process of
wj the amount

cooldrink flavour je J

nj = the given minimum syrup percentage requirement for cold drink

flavour je5
=
mj the given maximum syrup percentage requirement for colddrink

flavour jeJ

objective function
max z
=yj.sj
= -

: i.ci -
 assuming each

process 3000 tons of

in put

DIAGRAM

Inputs
Outputs
de-ink
de-ink

E
Bokboard
1
asphalt I
1
sooton

Tissue 2
asphalt
cOSt paper

2 soo ton
pulp content
3
Newsprint
asphalt
Book paper 4
3 600 ton

de-ink

process cost
yield
capacity
Desirable
Let

E
I be the set of inputs, such that it
I= 1 Box board

2 Tissue paper
3 News print
4 Book paper

I be the thatit]
E De-inking
set such 1
of processes =

2
Asphalt dispersion
and K be the setofpaper grades such that
grade keK=[1,2,33

ci= the given cost ($ (ton) of

inputitI

pi= the given pulp content (v) of input it I

ak= the given demand (in t) for papergrade ke

ijk = the amount (in tons) of input it I sent through process;25 to produce

paper grade kek

- BEEEFNE
input purchasing cost

minz Zesijk
-
costin
ocess t (min cost)
it I JEJ KGK

Exiji
= +
Exsjztsujs =
Crestricting inputusage
i 1j 1
for differentgrades)
=
=

paper grade

Eisik -3000 ↓je (production capacity)

0.9
(Epicciik) 0..(Epiccizk)
+
I NKER
ak
Creation
xijk 10 FitE,jE5, kEK (non negativity)
 j k
i
Products

Processing 30Z 107

make demand
>1 >
>
process 10Z $10/07 50000Z
raw material $1/Ib input $20107 50000z
24
$25/ID > 7
=
70Z

0z
1

>Processe

$124 10Z

10Z

30000Z
> Process
107
>
3 >$30/02
3
-

$2 34

K
1 0Z
> Process 4

$6 ih P S

amount sold

102 10$ 10z

mak labour 7
25 000 h
y2 Process 3008
102
107 3 >305102
> 34$2 7
T

x y process 1750

102
zu $1
11b 307, 1 >
,
raw lot
102 107 X
13 Process
process
L
$ 24 S
2 7
in $6

3250

- $2010z