A meat market manager for a large grocery store is preparing a processing plan t
ID: 467683 • Letter: A
Question
A meat market manager for a large grocery store is preparing a processing plan to stock the shelves with sausage, ground meat, and jerky, which he can prepare from beef, pork and venison. Sausage and ground meat can be made of any mix of the beef, pork and venison, as long at the fat contents are below 15% for sausage and 10% for ground meat. Sausage sells for $5/pound and ground meat sells for $3/pound. Jerky, which sells or $10/pound, is made in a drying process from beef or venison. In the drying process, there is a 50% loss in weight for jerky made from beef (e.g., one pound of beef yields 0.5 pounds of beef jerky) and a 30% loss in weight for jerky made from venison. The market can sell at most 700 pounds of sausage, 1000 pounds of ground meat, and 100 pounds of jerky before their expiration dates. There are currently 1,000 pounds of beef (10% fat content), 550 pounds of pork (8% fat content), and 225 pounds of venison (2% fat content) available for processing. Determine the optimal processing plan for the meat market.
Explanation / Answer
Decision Variables:
beef
pork
venison
sausage
s
b1
p1
v1
ground meat
g
b2
p2
v2
jerky
j
b3
p3
v3
Constraints:
p3 = 0
(no pork in jerky)
j = 0.5 b3+ 0.7 v3
(jerky constituents by weight )
s = b1+p1+v1
(sausage constituents)
g = b2+ p2+ v2
(ground beef constituents)
0.10 * b1 +0.08 * p1+ 0.02 * v1 >= 0.15 s
(fat content sausage)
0.10 * b2 +0.08 * p2+ 0.02 * v2 >= 0.10 g
(fat content ground meat)
b1+b2+b3<= 1000
(supply constraint)
p1+p2+p3 <=550
v1+v2+v3<=225
s >= 700
(demand constraint)
g >= 1000
j >= 100
s,g,j,b1,b2,b3,p1,p2,p3,v1,v2,v3 > = 0
(non negativity)
Objective function:
Maximize (5 s+3 g + 10 j)
(total sales)
Solving this model in solver we get:
beef
pork
venison
qty stored (lbs)
sausage
0
50
225
275
ground meat
1000
0
0
1000
jerky
0
0
0
0
Constraints:
LHS
Sign
RHS
0
=
0
(no pork in jerky)
0
=
0
(jerky constituents by weight )
275
=
275
(sausage constituents)
1000
=
1000
(ground beef constituents)
8.5
>=
41.25
(fat content sausage)
100
>=
100
(fat content ground meat)
1000
<=
1000
(supply constraint)
50
<=
50
225
<=
225
275
>=
700
(demand constraint)
1000
>=
1000
0
>=
100
Sales
$ 4375
Hence, the meat market manager should store 275 lbs of sausage, 1000 lbs of ground meat, and 0 lbs of jerky, to maximize his sales, given the constraints.
Decision Variables:
beef
pork
venison
sausage
s
b1
p1
v1
ground meat
g
b2
p2
v2
jerky
j
b3
p3
v3
Constraints:
p3 = 0
(no pork in jerky)
j = 0.5 b3+ 0.7 v3
(jerky constituents by weight )
s = b1+p1+v1
(sausage constituents)
g = b2+ p2+ v2
(ground beef constituents)
0.10 * b1 +0.08 * p1+ 0.02 * v1 >= 0.15 s
(fat content sausage)
0.10 * b2 +0.08 * p2+ 0.02 * v2 >= 0.10 g
(fat content ground meat)
b1+b2+b3<= 1000
(supply constraint)
p1+p2+p3 <=550
v1+v2+v3<=225
s >= 700
(demand constraint)
g >= 1000
j >= 100
s,g,j,b1,b2,b3,p1,p2,p3,v1,v2,v3 > = 0
(non negativity)
Objective function:
Maximize (5 s+3 g + 10 j)
(total sales)
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