OPERATIONS RESEARCH

Case Analysis

SUZ Truck Company

Case Summary SUZ is a truck manufacturing company

The company during the six month period of January to June 2008 has had a

bad financial performance due to which the company’s president expressed his

dissatisfaction at the monthly planning meeting with controller, sales and

production managers July 2008.

The company has been producing 2 models of trucks 101 and 102 in a single

plant at Wheeling, Michigan

Various opinions are placed across the table in the meeting by the managers as

a solution to the problem placed by the president ,some of which are as follows:

o Changing the Product mix

o Purchasing engines from an outside supplier

o Relieving the capacity in the engine assembly department

The manufacturing operations for trucks had been divided into four departments:

engine assembly, metal stamping, Model1 assembly, Model 2 assembly.

Capacity of each department is expressed as the manufacturing machine hours

available .

The output in the 1st 6 months of 2008 had been 1000 Model 101 and 1500

Model 102,at this level the engine assembly and Model 102 assembly capacities

were completely utilized, whereas that of metal stamping and Model 101

assembly were utilized only 83.3% and 40% respectively.

After a keen analysis of the standard costs the sales manager concluded that

Model 101 must be completely stopped from production, as the company was

losing $1205 on each truck of Model 101 sold at $39000, whereas on the other

hand Model 102 sold for $38000 was profitable.

The controller manager suggested that it would be better off to increase

production of Model101 trucks, cutting back if necessary on Model 102

production.

The production manager suggested that Model 101 production could be

increased without compromising on the production of Model 102 by purchasing

engines for Model 101 or Model 102 from an outside supplier.

This could be possible by furnishing the necessary engine and machine

components and reimburse the supplier for labour and overheads.

The main problem facing the company is that it wants a suitable product mix

which can cater to their need of maximizing the profits and improve their financial

position.

Problems

Q1. Find the best product mix for SUZ.

Solution: Using the Linear Programming model we can find the suitable product mix as

follows:

Step 1: Identifying Decision variables

X1 = Number of Model 101 trucks produced

X2 = Number of Model 102 trucks produced

Step 2: Formulate objective Function

Given data

Selling Price of Model 101= $39,000

Selling Price of Model 102= $38,000

Direct material cost (Model101) = $24,000

Direct material cost (Model102) = $20,000

Direct labour cost (Model101) = $4,000

Direct labour cost (Model102) = $4,500

Variable Overhead cost (Per 1 Model101 truck) = $8000

Variable Overhead cost (Per 1 Model102 truck) = $8500

Fixed Overhead Cost (combined)= $8.6 million

Utilizing the above given data we can formulate the objective function as follows

Maximize Z= (39,000X1+ 38,000X2)-(24,000X1+20,000X2+4000X1+4500X2+8000X1+8,500X2+

8.6million

Therefore,on simplification

Maximize Z= 3000x1+5000x2- 8.6 million

Step 3: Formulate constraints

The constraints in the manufacturing of the two trucks is mainly seen in form of the

capacity availability in each manufacturing division in form of net manufacturing

hours available and hours required per truck.

Keeping the pre requisite conditions intact the constraints for the problem are as

follows:

Subject to:

i. X1+ 2X2 ≤ 4000

ii. 2X1+2X2 ≤ 6000

iii. 2X1≤ 5000

iv. 3X2 ≤ 4500

Step 4: Non negativity Variables

X1 , X2 ≥0

Therefore the Linear programming problem for the case is as follows:

Maximize Z= 3000x1+5000x2- 8.6 million

Subject to:

i. X1+ 2X2 ≤ 4000

ii. 2X1+2X2 ≤ 6000

iii. 2X1≤ 5000

iv. 3X2 ≤ 4500

x1 and x2 ≥ 0

*Solving by Simplex method following output is obtained:

*LPP solution output from TORA is also attached here with

Cj -> 3000 5000 0 0 0 0 Bi

X1 X2 S1 S2 S3 S4

3000 X1 1 0 -1 1 0 0 20000 S4 0 0 -3 1.5 0 1 15000 S3 0 0 2 -2 1 0 10005000 X2 0 1 1 -0.5 0 0 1000

Zj -> 3000 5000 2000 500 0 0∆ j -> 0 0 -2000 -500 0 0

Thus the optimal mix given from the above optimal simplex table solution is

X1= 2000 and X2= 1000

Thus substituting the objective function we have

Z= 11, 00,000 –$ 8.6 million = 24, 00,000

Thus with a product mix of 2000 model 101 trucks and 1000 model 102 trucks profit can be maximized to $ 24,00,000.

Q2. What would be the best product mix if engine assembly capacity

were raised by one unit, from 4,000 to 4,001 machine-hours? What is

the extra unit of capacity worth?

Solution: In order to find the effect of the change in the hours in the engine assembly

on the solution we need to undertake the RHS ranging in order to ascertain that along

what range of values for the engine assembly does the basis of the solution remains

unchanged which can be ascertained as follows

Variables Ratio (BI/SJ)

S1 S2 S3 S4 Bi S1 S2 S3 S4

-1 1 0 0 2000 -2000 ∞ ∞ ∞

-3 1.5 0 1 1500 -500 1000 ∞ 1500

2 -2 1 0 1000 500 -500 1000 ∞

1 -0.5 0 0 1000 1000 -2000 ∞ ∞

Therefore the ranges for the constraints are

Constraint

No

Present

Value

Allowable

Increase

Allowable

decrease

Range

1 4000 500 500 3500-4500

2 6000 500 1000 5000-6500

3 5000 ∞ 1000 4000-∞

4 4500 ∞ 1500 3000-∞

Now, as per the RHS ranging table the first constraint i.e engine assembly doesn’t

change over the range 3500-4500,thus if a single unit of engine assembly were added it

would lead to a change in the optimal mix and no change in the basis, the new optimal

mix can be found by solving the following constraint equations simultaneously:

X1+2X2= 4001 and 2X1 +2X2 = 6000

After solving the equations we have

X1= 1999 and X2 = 1001

Z= 24, 02,000

Therefore, the optimal mix changes to 1999 model 101 trucks and 1001 model 102

trucks and profit has increased to $ 24,02,000

The worth of the extra unit of capacity would be equal to the shadow price associated

with the engine assembly hours corresponding to the ∆ j row value of S1 which is

2000,since the resource is scarce an additional unit of the resource would add 2000 to

the profit,which can be see above wherein the new profit for the new mix has increased

by over $2000 than the previous profit obtained from the original optimal mix. This each

unit of capacity is worth $2,000

Q3. Assume that a second, additional unit of engine assembly

capacity worth the same as the first. Verify that if the capacity were

increased to 4,100 machine- hours, then the increase in contribution

would be 100 times that in Q 2.

Solution: As seen from the previous question, a unit increase in the engine assembly

capacity would lead to increase in profits by $2000.Therefore an increase in capacity of

100 units would lead to increase in profits by $2000*100,thus and increase of engine

assembly capacity from 4000 to 4100 would lead to profit increasing 100 times.

Q4.How many units of engine assembly capacity can be added before

there is a change in the value of an additional unit of capacity?

Solution: As seen from Q2, the RHS range calculated for the engine assembly

ranges from 3500-4500 and the present value being 4,000 thus the maximum additional

units up to which the value of the additional units remain unchanged is 4500-4000=500,

thus up to a increase of 500 units from 4000 the value of an additional unit of capacity

remains unchanged since the basis of the solution also remains unchanged.

Q5. SUZ’s production manager suggests purchasing Model 101 or

Model 102 engines from an outside supplier in order to relieve the

capacity problem in the engine assembly department. If Merton

decides to pursue this alternative, it will be effectively “renting”

capacity: furnishing the necessary materials and engine components

and reimbursing the outside supplier for labor and overhead. Should

the company adopt this alternative? If so, what is the maximum rent it

should be willing to pay for a machine-hour of engine assembly

capacity? What is the maximum number of machine-hours it should

rent?

Solution: The company can go ahead for the option of getting the engine assembly

capacity on rent as it is a scarce resource for the company and a additional unit of

capacity would still contribute to the profits, this can be seen as the calculations in Q2

above wherein as per the RHS ranging values the engine assembly capacity can still be

increased by 500 units(Current =4000 ,maximum=4500) without changing the basis of

the solution and also contribute $2000/rented hr to the profits(∆ j row value for S1), since

beyond the permissible range determined the basis would change along with change in

the shadow price ,thus the company must only rent a maximum of 500 hours of

engine assembly in order to contribute predictable additional profits of$2000/hr, this

rented capacity must also not exceed the value(cost) of $2000 per hour rented as

the incentive of renting additional capacity would be lost if the cost of rent is more

than the contribution to the profit by the additional capacity.

Maximum capacity to be rented- 500 hours

Maximum value payable for additional capacity =$2,000

Q6.SUZ is considering the introduction of a new truck, to be called

Model 103. Each Model 103 truck would give a contribution of $2,000.

The total engine assembly capacity would be sufficient to produce

5,000 Model l03s per month, and the total metal stamping capacity

would be sufficient to produce 4,000 Model lO3s. The new truck would

be assembled the Model 101 assembly department, each Model 103

truck requiring only half as much time as a Model 101 truck.

a) Should SUZ produce Model 103 trucks?

b) How high would the contribution on each Model 103 truck

have to be before it became worthwhile to produce the new

model?

Solution:

a. In order to introduce a new model of 103 truck into production the company would

need to assess the cost of its production and utilization of resources as well as the

contribution of the same towards augmentation of profits for the company. In case of

Model 103 in engine assembly if exclusively produced the capacity would be sufficient

for 5000 trucks,

Thus

Time for each Model 103 truck engine assembly= 4000/5000= 0.8 hrs

Similarly

Model 103 in Metal stamping if exclusively produced the capacity would be sufficient for

4000 trucks

Thus

Time for each Model 103 truck metal stamping=6000/4000 = 0.8 hrs

Now as the capacity in metal stamping and engine assembly dept are scarce due to

complete consumption by Model 101 and Model 102 production, thus additional

capacity need to be purchased which must not exceed the value of the marginal

profitability /shadow prices of $2000(engine assembly) and $500 (metal stamping).In

the model 101 since there is additional left over capacity thus it can be utilized for

producing Model 103 truck and need not be purchased and thus shadow price would be

zero.

Now, assessing the cost of producing Model 103 we get

Engine Assembly= 0.8*$2000 = $1600

Metal stamping= 1.5*$500 = $750

Model 103 assembly= 1 * 0 = 0

Total cost incurred = $ 2350

Contribution per Model 103 truck= $ 2000

Therefore, since each Model 103 would give a contribution of $ 2000 to the profits but

also incur $2350 as cost in order to produce it, the company should not produce Model

103 trucks as this would be a loss making decision.

b) In order to introduce the Model 103 truck into the production and also be profit

making or no loss the contribution per truck must increase by at least $350 from

current contribution of $2000

Q7. Engines can be assembled on overtime in the engine assembly

department. Suppose production efficiencies do not change and 2,000

machine-hours of engine assembly overtime capacity are available.

Direct labor costs are higher by 50% for overtime production. While

variable overhead would remain the same, monthly fixed overhead in

the engine assembly department would increase by $0.75 million.

Should SUZ assemble engines on overtime?

Solution:

Since the company wants to produce additional trucks in overtime thus the amount of

trucks produced in overtime must be allotted different decision variables as follows:

X3= Number of Model 101 trucks produced in overtime

X4= Number of Model 102 trucks produced in overtime

Now

We need to include the contribution of these overtime produced trucks to the objective

function also, since the labour cost is higher by 50% in overtime thus the intial

contributions by Model 101(3000X1) and Model 102 (5000X2) would reduce in overtime

by the amount of increase in respective labour costs.

Thus

Model 101 overtime object function component= 3000X1 - 600 X1= 2400 X3

Model 102 overtime object function component= 5000X2 -1200 X2 = 3800X4

Therefore the modified objective function is as follows

Z= 3000X1+5000X2+ 2400 X3 +3800X4 –$ 8.6 million-$ 0.75

million

Z= 3000X1+5000X2+ 2400 X3 +3800X4–$ 9.35 million

Subject to:

i. X1+ 2X2 ≤ 4000

ii. 2X1+2X2 ≤ 6000

iii. 2X1≤ 5000

iv. 3X2 ≤ 4500

v. X3+ 2X4 ≤ 2000 (additional capacity for overtime in engine

assembly)

Non negativity

X1 ,X2 ,X3 ,X4 > 0

*Solving the following by simplex method we get

Optimal mix as

Model 101= 2000 trucks

Model 102= 1000 trucks

Model 101(Overtime) = 2000 trucks

Model 102(Overtime) = 0 trucks

The value of the objective function is as follows

Z= 3000(2000) +5000(1000)+2400(2000)+3800(0)- 9.35 million

Z= 1, 58, 00,000-93,50,000= $64, 50,000

Since by conducting overtime production for Model 101 the company gets more profits

than the profits for initial optimal mix calculated in Q1 i.e. 24,00, 000. (64, 50,000-24,

00,000=$40, 50,000) the company should go ahead with the proposed overtime

proposal in order to maximize profits

*LPP solution output from TORA is also attached here with

Q8. SUZ’s president, in arguing that maximizing short-run

contribution was not necessarily good for the company in the long

run, wanted to produce as many Models 101 is as possible. After

some discussion it was agreed to maximize the monthly contribution

as long as the number of model 101 trucks produced was at least

three times the no of model 102. What is the optimal mix in this case?

Solution

The company proposes to add an additional constraint to the production by

stating that it was mandatory to have Model 101 trucks produced at least 3

times that of Model 102 trucks.

Thus from the intial LPP formulation we have

Z= 3000X1+5000X2–$ 8.6 million

Subject to:

i. X1+ 2X2 ≤ 4000

ii. 2X1+2X2 ≤ 6000

iii. 2X1≤ 5000

iv. 3X2 ≤ 4500

To this an additional constraint must be added as follows:

Z= 3000X1+5000X2–$ 8.6 million

Subject to:

i. X1+ 2X2 ≤ 4000

ii. 2X1+2X2 ≤ 6000

iii. 2X1≤ 5000

iv. 3X2 ≤ 4500

v. X1≥ 3X2 (OR) X1- 3X2 ≥ 0

*Solving the problem through simplex method

We get the following optimal mix

X1= 2250

X2= 750

and the value of the objective function as

Z= 3000(2250)+5000(750)-8.6 million

Z= 10500000-86, 00,000=19, 00,000

Thus in this case wherein the organization wants to produce Model 101

trucks at least 3 times the quantity of Model102 ,the optimal mix in the case

is 2250(Model 101 trucks) and 750(Model 102 trucks) with a profit of

$19,00,000.

*LPP solution output from TORA is also attached here with